Iterated digits squaring

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If you add the square of the digits of a Natural number (an integer bigger than zero), you always end with either 1 or 89:

15 -> 26 -> 40 -> 16 -> 37 -> 58 -> 89
7 -> 49 -> 97 -> 130 -> 10 -> 1

An example in Python:

 step = lambda x: sum(int(d) ** 2 for d in str(x))
>>> iterate = lambda x: x if x in [1, 89] else iterate(step(x))
>>> [iterate(x) for x in xrange(1, 20)]
[1, 89, 89, 89, 89, 89, 1, 89, 89, 1, 89, 89, 1, 89, 89, 89, 89, 89, 1]

;Task: : Count how many number chains for integers 1 <= n < 100_000_000 end with a value 89. Or, for much less credit - (showing that your algorithm and/or language is slow): : Count how many number chains for integers 1 <= n < 1_000_000 end with a value 89.

This problem derives from the [https://projecteuler.net/problem=92 Project Euler problem 92].

For a quick algorithm for this task see [[Talk:Iterated_digits_squaring|the talk page]]

;Related tasks:

[[Combinations with repetitions]]
[[Digital root]]
[[Digital root/Multiplicative digital root]]

ALGOL 68

Brute-force with some caching.

# count the how many numbers up to 100 000 000 have squared digit sums of 89            #

# compute a table of the sum of the squared digits of the numbers 00 to 99              #
[ 0 : 99 ]INT    digit pair square sum;
FOR d1 FROM 0 TO 9 DO
    FOR d2 FROM 0 TO 9 DO
        digit pair square sum[ ( d1 * 10 ) + d2 ] := ( d1 * d1 ) + ( d2 * d2 )
    OD
OD;

# returns the sum of the squared digits of n                                            #
PROC squared digit sum = ( INT n )INT:
     BEGIN
         INT   result := 0;
         INT   rest   := n;
         WHILE rest /= 0 DO
             INT    digit pair = rest MOD 100;
             result PLUSAB digit pair square sum[ digit pair ];
             rest   OVERAB 100
         OD;
         result
     END # squared digit sum # ;

# for values up to 100 000 000, the largest squred digit sum will be that of 99 999 999 #
# i.e. 81 * 8 = 648, we will cache the values of the squared digit sums                 #
INT   cache max = 81 * 8;
[ 1 : cache max ]INT cache;
FOR i TO cache max DO cache[ i ] := 0 OD;

INT count 89 := 0;

# fill in the cache                                                                     #
FOR value FROM 2 TO cache max DO cache[ value ] := squared digit sum( value ) OD;
# we "know" that 89 and 1 are the terminal values                                       #
cache[  1 ] :=  1;
cache[ 89 ] := 89;
FOR value FROM 2 TO cache max DO
    INT sum := cache[ value ];
    WHILE sum /= 1 AND sum /= 89 DO
        sum := cache[ sum ]
    OD;
    cache[ value ] := sum
OD;

FOR value FROM 1 TO 100 000 000 DO
    IF cache[ squared digit sum( value ) ] = 89 THEN count 89 +:= 1 FI
OD;

print( ( "Number of values whose squared digit sum is 89: ", whole( count 89, -10 ), newline ) )


Number of values whose squared digit sum is 89:   85744333

AWK

We use a brute-force approach with buffering for better performance. Numbers are assumed to be double precision floats, which is true for most implementations. It runs in about 320 s on an Intel i5.

# Usage: GAWK -f ITERATED_DIGITS_SQUARING.AWK
BEGIN {
    # Setup buffer for results up to 9*9*8
    for (i = 1; i <= 648; i++) {
        k = i
        do {
            k = squaredigitsum(k)
        } while ((k != 1) && (k != 89))
        if (k == 1) # This will give us 90 entries
            buffer[i] = ""
    }
    # Check sequence for every number
    pow10 = 1
    for (i = 1; i <= 100000000; i++) {
        count += (squaredigitsum(i) in buffer) ? 0 : 1
        if (i == pow10) {
            printf("1->10^%d: %d\n", length(i) - 1, count)
            pow10 *= 10
        }
    }
}
function squaredigitsum(n,    r) {
    while (n) {
        r += (n % 10) ^ 2
        n = int(n / 10)
    }
    return r
}


1->10^0: 0
1->10^1: 7
1->10^2: 80
1->10^3: 857
1->10^4: 8558
1->10^5: 85623
1->10^6: 856929
1->10^7: 8581146
1->10^8: 85744333

BBC BASIC

{{works with|BBC BASIC for Windows}} Three versions timed on a 2.50GHz Intel Desktop.

      REM Version 1: Brute force
      REM ---------------------------------------------------------
      T%=TIME
      N%=0
      FOR I%=1 TO 100000000
        J%=I%
        REPEAT
          K%=0:REPEAT K%+=(J%MOD10)^2:J%=J%DIV10:UNTIL J%=0
          J%=K%
        UNTIL J%=89 OR J%=1
        IF J%>1 N%+=1
      NEXT
      PRINT "Version 1: ";N% " in ";(TIME-T%)/100 " seconds."

      REM Version 2: Brute force + building lookup table
      REM ---------------------------------------------------------
      T%=TIME
      DIM B% 9*9*8,H%(9)
      N%=0
      FOR I%=1 TO 100000000
        J%=I%
        H%=0
        REPEAT
          K%=0:REPEAT K%+=(J%MOD10)^2:J%=J%DIV10:UNTIL J%=0
          H%(H%)=K%:H%+=1
          J%=K%
          IF B%?J%=1 EXIT REPEAT
        UNTIL J%=89 OR J%=1
        IF J%>1 N%+=1:WHILE H%>0:H%-=1:B%?H%(H%)=1:ENDWHILE
      NEXT
      PRINT "Version 2: ";N% " in ";(TIME-T%)/100 " seconds."

      REM Version 3: Calc possible combinations (translation of C)
      REM ---------------------------------------------------------
      T%=TIME
      DIM B%(9*9*8):B%(0)=1
      FOR N%=1 TO 8
        FOR I%=9*9*N% TO 1 STEP -1
          FOR J%=1 TO 9
            S%=J%*J%
            IF S%>I% EXIT FOR
            B%(I%)+=B%(I%-S%)
          NEXT
        NEXT
      NEXT

      N%=0
      FOR I%=1 TO 9*9*8
        J%=I%
        REPEAT
          K%=0:REPEAT K%+=(J%MOD10)^2:J%=J%DIV10:UNTIL J%=0
          J%=K%
        UNTIL J%=89 OR J%=1
        IF J%>1 N%+=B%(I%)
      NEXT
      PRINT "Version 3: ";N% " in ";(TIME-T%)/100 " seconds."

      END


Version 1: 85744333 in 1447.08 seconds.
Version 2: 85744333 in 718.04 seconds.
Version 3: 85744333 in 0.02 seconds.

Befunge

This is just a brute force solution, so it's not very fast. A decent interpreter will probably take a minute or two for a 1,000,000 iterations. If you want to test with 100,000,000 iterations, change the ::** (100³) near the end of the first line to :: (100²^²). With that many iterations, though, you'll almost certainly want to be using a compiler, otherwise you'll be waiting a long time for the result.

1-1\10v!:/+55\<>::**>>-!|
v0:\+<_:55+%:*^^"d":+1$<:
>\`!#^ _$:"Y"-#v_$\1+\:^0
>01-\0^ @,+55.<>:1>-!>#^_
>,,,$." >=",,,^ >>".1">#<

1..1000000 => 856929

1..100000000 => 85744333

C

C99, tested with "gcc -std=c99". Record how many digit square sum combinations there are. This reduces $10^n$ numbers to $81n$ , and the complexity is about $O(n^2)$ . The 64 bit integer counter is good for up to $10^{19}$ , which takes practically no time to run.

#include <stdio.h>

typedef unsigned long long ull;

int is89(int x)
{
	while (1) {
		int s = 0;
		do s += (x%10)*(x%10); while ((x /= 10));

		if (s == 89) return 1;
		if (s == 1) return 0;
		x = s;
	}
}


int main(void)
{
	// array bounds is sort of random here, it's big enough for 64bit unsigned.
	ull sums[32*81 + 1] = {1, 0};

	for (int n = 1; ; n++) {
		for (int i = n*81; i; i--) {
			for (int j = 1; j < 10; j++) {
				int s = j*j;
				if (s > i) break;
				sums[i] += sums[i-s];
			}
		}

		ull count89 = 0;
		for (int i = 1; i < n*81 + 1; i++) {
			if (!is89(i)) continue;

			if (sums[i] > ~0ULL - count89) {
				printf("counter overflow for 10^%d\n", n);
				return 0;
			}
			count89 += sums[i];
		}

		printf("1->10^%d: %llu\n", n, count89);
	}

	return 0;
}


1->10^1: 7
1->10^2: 80
1->10^3: 857
1->10^4: 8558
1->10^5: 85623
1->10^6: 856929
1->10^7: 8581146
1->10^8: 85744333
1->10^9: 854325192
1->10^10: 8507390852
1->10^11: 84908800643
1->10^12: 850878696414
1->10^13: 8556721999130
1->10^14: 86229146720315
1->10^15: 869339034137667
1->10^16: 8754780882739336
1->10^17: 87975303595231975
1->10^18: 881773944919974509
1->10^19: 8816770037940618762
counter overflow for 10^20

Fast C implementation (<1 second my machine), which performs iterated digits squaring only once for each unique 8 digit combination. The cases 0 and 100,000,000 are ignored since they don't sum to 89:


#include <stdio.h>

const int digits[] = { 0,1,2,3,4,5,6,7,8,9 };

// calculates factorial of a number
int factorial(int n) {
    return n == 0 ? 1 : n * factorial(n - 1);
}

// returns sum of squares of digits of n
unsigned int sum_square_digits(unsigned int n) {
        int i,num=n,sum=0;
        // process digits one at a time until there are none left
        while (num > 0) {
                // peal off the last digit from the number
                int digit=num % 10;
                num=(num - digit)/10;
                // add it's square to the sum
                sum=sum+digit*digit;
        }
        return sum;
}

// builds all combinations digits 0-9 of length len
// for each of these it will perform iterated digit squaring
// and for those which result in 89 add to a counter which is
// passed by pointer.
long choose_sum_and_count_89(int * got, int n_chosen, int len, int at, int max_types, int *count89)
{
        int i;
        long count = 0;
        int digitcounts[10];
        for (i=0; i < 10; i++) {
                digitcounts[i]=0;
        }
        if (n_chosen == len) {
                if (!got) return 1;

                int sum=0;
                for (i = 0; i < len; i++) {
                        int digit=digits[got[i]];
                        digitcounts[digit]++;
                        sum=sum + digit * digit;
                }
                if (sum == 0) {
                        return 1;
                }
                if ((sum != 1) && (sum != 89)) {
                        while ((sum != 1) && (sum != 89)) {
                                sum=sum_square_digits(sum);
                        }
                }
                if (sum == 89) {
                        int count_this_comb=factorial(len);
                        for (i=0; i<10; i++) {
                                count_this_comb/=factorial(digitcounts[i]);
                        }
                        (*count89)+=count_this_comb;
                }

                return 1;
        }

        for (i = at; i < max_types; i++) {
                if (got) got[n_chosen] = i;
                count += choose_sum_and_count_89(got, n_chosen + 1, len, i, max_types, count89);
        }
        return count;
}

int main(void)
{
        int chosen[10];
        int count=0;
        // build all unique 8 digit combinations which represent
        // numbers 0-99,999,999 and count those
        // whose iterated digit squaring sum to 89
        // case 0, 100,000,000 are ignored since they don't sum to 89
        choose_sum_and_count_89(chosen, 0, 8, 0, 10, &count);
        printf("%d\n",count);
        return 0;
}

85744333

C++

Slow (~10 seconds on my machine) brute force C++ implementation:


#include <iostream>

// returns sum of squares of digits of n
unsigned int sum_square_digits(unsigned int n) {
        int i,num=n,sum=0;
        // process digits one at a time until there are none left
        while (num > 0) {
                // peal off the last digit from the number
                int digit=num % 10;
                num=(num - digit)/10;
                // add it's square to the sum
                sum+=digit*digit;
        }
        return sum;
}
int main(void) {
        unsigned int i=0,result=0, count=0;
        for (i=1; i<=100000000; i++) {
                // if not 1 or 89, start the iteration
                if ((i != 1) || (i != 89)) {
                        result = sum_square_digits(i);
                }
                // otherwise we're done already
                else {
                        result = i;
                }
                // while we haven't reached 1 or 89, keep iterating
                while ((result != 1) && (result != 89)) {
                        result = sum_square_digits(result);
                }
                if (result == 89) {
                        count++;
                }
        }
        std::cout << count << std::endl;
        return 0;
}

85744333

C#

The largest sum possible for any number is 999, so the first 730 numbers are calculated and stored in an array.
The rest is then looked up. A limit of 100 million takes about 6 seconds. int.MaxValue takes about 2 and a half minutes.

using System;
public static class IteratedDigitsSquaring
{
    public static void Main() {
        Console.WriteLine(Count89s(1_000_000));
        Console.WriteLine(Count89s(100_000_000));
    }

    public static int Count89s(int limit) {
        if (limit < 1) return 0;
        int[] end = new int[Math.Min(limit, 9 * 9 * 9 + 2)];
        int result = 0;

        for (int i = 1; i < end.Length; i++) {
            for (end[i] = i; end[i] != 1 && end[i] != 89; end[i] = SquareDigitSum(end[i])) { }
            if (end[i] == 89) result++;
        }
        for (int i = end.Length; i < limit; i++) {
            if (end[SquareDigitSum(i)] == 89) result++;
        }
        return result;

        int SquareDigitSum(int n) {
            int sum = 0;
            while (n > 0) {
                int digit = n % 10;
                sum += digit * digit;
                n /= 10;
            }
            return sum;
        }
    }

}


856929
85744333

BigInteger version

{{trans|C}} {{Libheader|System.Numerics}} Translation of the first C version, with BigIntegers. This can get pretty far in six seconds, even on Tio.run.

using System;
using System.Numerics;

class Program {

  const int MaxPow = 301;
  static int [] sq = {1, 4, 9, 16, 25, 36, 49, 64, 81};
  static BigInteger [] sums;

  static bool is89(int x) {
    while (true) {
      int s = 0, t;
      do if ((t = (x % 10) - 1) >= 0) s += sq[t]; while ((x /= 10) > 0);
      if (s == 89) return true;
      if (s == 1) return false;
      x = s;
    }
  }

  static BigInteger count89(int n) {
      BigInteger result = 0;
      for (int i = n * 81; i > 0; i--) {
        foreach (int s in sq) { if(s > i) break; sums[i] += sums[i - s]; }
        if (is89(i)) result += sums[i];
      }
      return result;
  }

  static void Main(string[] args) {
    BigInteger [] t = new BigInteger[2] {1, 0}; sums = new BigInteger[MaxPow * 81]; Array.Copy(t, sums, t.Length);
    DateTime st = DateTime.Now;
    for (int n = 1; n < MaxPow; n++) {
      Console.Write("1->10^{0,-3}: {1}\n", n, count89(n));
      if ((DateTime.Now - st).TotalSeconds > 6) break;
    }
    Console.WriteLine("{0} seconds elapsed.", (DateTime.Now - st).TotalSeconds);
  }
}

1->10^1  : 7
1->10^2  : 80
1->10^3  : 857
1->10^4  : 8558
1->10^5  : 85623
1->10^6  : 856929
1->10^7  : 8581146
1->10^8  : 85744333
1->10^9  : 854325192
1->10^10 : 8507390852
1->10^11 : 84908800643
1->10^12 : 850878696414
1->10^13 : 8556721999130
1->10^14 : 86229146720315
1->10^15 : 869339034137667
1->10^16 : 8754780882739336
1->10^17 : 87975303595231975
1->10^18 : 881773944919974509
1->10^19 : 8816770037940618762
1->10^20 : 87994965555707002706
1->10^21 : 877214809753814412449
1->10^22 : 8740475212714948184620
1->10^23 : 87086767569032964273481
1->10^24 : 867912763131207135645491
1->10^25 : 8652685884347431487002838
1->10^26 : 86292591735549905389544085
1->10^27 : 860834491746260610360036431
1->10^28 : 8589383648492973833587962133
1->10^29 : 85719021282987319689186339605
1->10^30 : 855551075003449256539175506135
1->10^31 : 8539846767881104092122936276127
1->10^32 : 85245373514507207808857201531419
1->10^33 : 850921798797738318678358430121498
1->10^34 : 8493602724656082624921256124945709
1->10^35 : 84775765928320499747460839463166887
1->10^36 : 846127234701773214379999133850790428
1->10^37 : 8445101119798901092741398494615146552
1->10^38 : 84297231641833173945386163054551847907
1->10^39 : 841596309978956515337376882969248454407
1->10^40 : 8404688192812158407616126296428757287918
1->10^41 : 83966751636707267524727665346136900559808
1->10^42 : 839249062380369832617111284115323596416189
1->10^43 : 8392404334111393647768734710144578436411820
1->10^44 : 83963458265257975880706035079312646291089162
1->10^45 : 840390620671402119260216748725664301844515595
1->10^46 : 8414380030090502032224993998030998898525187113
1->10^47 : 84268378296544752164579356732419005387066100619
1->10^48 : 844021806190251380758758476585216084473498054164
1->10^49 : 8453427257465803796850958549692384862623307213954
1->10^50 : 84654382110763756920355712358557288888652143589824
1->10^51 : 847537750217936548550698726085731005366031699187697
1->10^52 : 8482595213704622541116090344851904585191448008008698
1->10^53 : 84867114171087369978017651353669784240040553506347863
1->10^54 : 848763596449838290475849513610494144653829069301555744
1->10^55 : 8485560484449848898784875907345401899210439410548661905
1->10^56 : 84809241613331707710051455489300240267084096119421192555
1->10^57 : 847435762855526547824875506635678396375585724580676401281
1->10^58 : 8466611716350744168054316461343227422117005648835357501042
1->10^59 : 84584794275749872157313978459784905712596125963065261887087
1->10^60 : 845072003706634444132974487900963836188835216550117810064042
1->10^61 : 8443993493344896883975002240891650144660444520675597442455846
1->10^62 : 84388114632696697235622301117847815843833846024601175739193818
1->10^63 : 843550873686677877815689986525235580589881417969543147823955468
1->10^64 : 8434235773893302085490040550865199018569738474571414542759091420
1->10^65 : 84349704267294170985441634483996250787548886013951222960902661326
1->10^66 : 843754473866041852258692025296354310048924258957916675280270848383
1->10^67 : 8441681459520956437757926334685945498961397097395903879157454648255
1->10^68 : 84470346186515545447015226246395087364669380975589204241143094714939
1->10^69 : 845322904478358163301387625325558514367250680088175647529431667836984
1->10^70 : 8459894886866423548013102102792807954026974401198186041921374671243280
1->10^71 : 84666946651672233790199404593444235036570835062615369577576893503846807
1->10^72 : 847334186689171226140326334304533885325304121282085041331744831976064350
1->10^73 : 8479611353077227882670645671814119173057689480878540635598355377828682030
1->10^74 : 84853427154299528465645782338465354008595064980299709100384597454288492601
1->10^75 : 849043816798387454378741585441510256257380507579571260789337785619852662347
1->10^76 : 8494856623561856764693867992374031564067482173292744174563186543401165633828
1->10^77 : 84986053947449161776274594245379439290411038180766518290148488651233497246177
1->10^78 : 850173060522433967868762132659331655304647077578564191004961947049948444228365
1->10^79 : 8504307025056582630550030771632855466797975344124296918801658700334809780702180
1->10^80 : 85064277463177819389470781000622284519150163101058649012431344960771964694651571
1->10^81 : 850819394401164074041611461274974545538571046366755739155832724445684685332968364
1->10^82 : 8509704926350516030551921703403722812551632988675827388709515859909259694851647834
1->10^83 : 85110487385572993176128742106677248626395367895834794577983294930763809503319428014
1->10^84 : 851229369385484114012959335791260147114430529630506351180704055826749898923602659534
1->10^85 : 8513481307977862280230328507187688326498927519035854456240791004629338052316252944962
1->10^86 : 85146211618619610403764635439165512633290370187486987569410348341895803492714570487527
1->10^87 : 851568818184158929602427279369759555161328403821517914312791180281061345268793327734747
1->10^88 : 8516622700452095396072535230209391313806399904195398975658231816363535402709890663189362
1->10^89 : 85173334571538953250258781660855404353386662001559167780912035060599928280076465921472154
1->10^90 : 851770344573699439297915741793926110106776082569441364749653121400250416057006702446954384
1->10^91 : 8517598250615252557316601155045090636911412094296867452464613221521534622516854479200111167
1->10^92 : 85168762749001784598676830952569236245621633296172168977732463493080042840432604954811508256
1->10^93 : 851540008029830880300994057100452233115864754809360142651983875738294861965389543527957072529
1->10^94 : 8513047953516561208480829418619110455666666761484728667018453838522192578086491705360981209049
1->10^95 : 85097247949654801901491432145587952810159950536920233184462924578646829963281410434249631392723
1->10^96 : 850537149051806438025447330906036177346231271896967500380026706058200824325895506679143940414076
1->10^97 : 8499972063859782028890346555815409171233558167321697969098574428852568869015102044124776256646518
1->10^98 : 84935580971302195864938049552661540297299353108396246935346418128128262766226035981031392864950607
1->10^99 : 848621012961290009929016807175952691990221212867104144408323613154869710828195585404753698625209870
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1->10^207: 853659202819987541247160120887834521284760886339623274004394773066680299538480440121545335974469636403243323858304782951228852284398676153012813966360440529328380943838264684632008844950545647229432408419763
1->10^208: 8544952483138354272479452096326871415134170280533803383323382424225457971898051136208471099688648406254320867700931409630655569469573762186473633746768288944756293766713800079402839068636993722539464464014343
1->10^209: 85533733751997391022351921893082947709535591919068528290332852883967864327556645461585949789541151338595934256435388217939077881109820952288423214884499900454917289123598440879211618935608105485422232130901905
1->10^210: 856178628495280232427926168171704244059740501955488094383750755874097867668689843542569526012394028099851590049094767507865596216897633456922557606621505076739639542878814850925266187250164585329019683542629923
1->10^211: 8570131419693618760966628804173079377257638567195555240493311797353797176489667120305647330718511690292693777674170531508951053735086796114378779311384689107558641543989958869638931709314824117850451000380143021
1->10^212: 85783575440047109414461631493979767235315177615543985312544659159056610182875530781537114391430809057845669077275617329819355730217516493005290650519511006772034807212815853844586588988998227020982705473135728765
1->10^213: 858642157725030348223249509804854637186812832528125424148240160404301905107061460749243511335395491708297289251008381529398185566019895635547713725083517506594429645460660280089932247894935780442351898303313140945
1->10^214: 8594287902664891352913658993716294858626490086433812797669386361846753520746735854191540089415236757792436360765472598893490276479705293275725004645785633334970346652528315678317371791476569985195862855380417401542
1->10^215: 86019274419595998866603848298462306888943095199436257368964632161654631977139765349618577697513221456293489609436575083556837195728003126429131084775756767513323846022083954060095590650961801996674310643811350499369
1->10^216: 860931655618474553063989578082933761682457487457489431016612890143003640365600635223968806347174143526789184001346670039335478957198921437817108109746919660278785057686592225736631570790200460497279046021138265882633
1->10^217: 8616435837729622438277275972924774728615421337934287445283316837787898004557496433610326251817310796184207726389126647218963569950986844300319174321919640256132751161493356727133356946791128243709978104039611221929848
1->10^218: 86232688597037059510650496264090044650582001854485728865997396326431091394758611347974329812789632648546842598755241789989941130068748668444311151669575147062603211288550024198424668424830168061712646711077510494361276
1->10^219: 862980094753456140909290085044350220214633094993933779019549067941866017572800599010552481547901256759922208947964802924199156525319895138745088257779570795905170294076748630599291864563865074304312620545350757654949167
1->10^220: 8636018033559847409410591107268210323615373240747163973441778682144682091271457867694794590752142037466191114879096481073228275415949941743493243536215269195973060634460249426462226091308923903368436341962027769078827045
1->10^221: 86419056489171874993789903489060892796455772565320723803637734001174467305715397474011039116397479926656101081127323729931224029537185192310507718579127400181923630610893970171709931933408787373551201437680240028108283958
1->10^222: 864744809847891662279169612523698598292488130571716275496525629546147011808378355197408294680827690818106146785428930909260093817332410616456120307490986571108036081906704392578762581319448395342866886631949686317950079685
1->10^223: 8652627854354814142021372274236655285412150520596931780504767483938734335443732380812376622197005225828947530209843225214814025134173901180081812700123059474516130142427254496259962223594226063501914926447312107963923255816
1->10^224: 86574251920745435851516366426371437236500297164193572646840936487152446338309838340335914152667764515000963423743096085244510673418048367379825329583910629778255879869131857325574114859632491805222223301052145753142373402038
1->10^225: 866181808148306233153035741070162251220868553540999914228623747312278276373367606857723265686649243714013350573228296681022846168327651205292941587217634862968933831862883106033509203213506996294508078581405379889390718900884
1->10^226: 8665782332173490493624385682733920863071548417814598549460323221230417435701507196058986103580683262649034387808689514519733914026030490404939317647480340690773539215255948949250489348489232493792231877166169779541100947510992
1->10^227: 86692919814710675783589991859619703285754546114420206472621039807364718363865261702270566986525194821577798634423416768340074070191330534224131123412508988375899005313698308076230449814177068560441644180880325653197748136923185
1->10^228: 867231990206473247052871536000445102504192081962389666619381949130498077603334870783904602961079237586504849633880397962186289424840869477113707180313165669320870472468535571056811158182157681436281093030894644992830298477981244
1->10^229: 8674838596801362677923547970017972903037995901012236697795083419765821418590394410295854079897879829098141309577554554785416139269074776459368918266410283727698808695686584095897235693981445795558726984321580750493111552025864339
1->10^230: 86768211402812128806590576564537513494737520987736487082881857738963221877281843731844788716420658593474347727365894819526796319707828593251356370569187398794672340428112756386987781701631240923503544557476729747177320351749598558
6.0396929 seconds elapsed.
```
It doesn't always get to 10^230 in six seconds at Tio.run, sometimes it only gets to 10^201 or so.


## Ceylon


```ceylon
shared void run() {

	function digitsSquaredSum(variable Integer n) {
		variable value total = 0;
		while(n > 0) {
			total += (n % 10) ^ 2;
			n /= 10;
		}
		return total;
	}

	function lastSum(variable Integer n) {
		while(true) {
			n = digitsSquaredSum(n);
			if(n == 89 || n == 1) {
				return n;
			}
		}
	}

	variable value eightyNines = 0;
	for(i in 1..100M - 1) {
		if(lastSum(i) == 89) {
			eightyNines++;
		}
	}
	print(eightyNines);
}
```



## Clojure


### Direct Method


```lisp
(ns async-example.core
  (:require [clojure.math.numeric-tower :as math])
  (:use [criterium.core])
  (:gen-class))
(defn sum-sqr [digits]
  " Square sum of list of digits "
  (let [digits-sqr (fn [n]
                     (apply + (map #(* % %) digits)))]
    (digits-sqr digits)))

(defn get-digits [n]
  " Converts a digit to a list of digits (e.g. 545 -> ((5) (4) (5)) (used for squaring digits) "
  (map #(Integer/valueOf (str %)) (String/valueOf n)))

(defn -isNot89 [x]
  " Returns nil on 89 "
  (cond
    (= x 0) 0
    (= x 89) nil
    (= x 1) 0
    (< x 10) (recur (* x x))
    :else (recur (sum-sqr (get-digits x)))))

;; Cached version of isNot89 (i.e. remembers prevents inputs, and returns result by looking it up when input repeated)
(def isNot89 (memoize -isNot89))

(defn direct-method [ndigits]
  " Simple approach of looping through all the numbers from 0 to 10^ndigits - 1 "
  (->>
    (math/expt 10 ndigits)
    (range 0)									; 0 to 10^ndigits
    (filter #(isNot89 (sum-sqr (get-digits %))))	; filters out 89
    (count)										; count non-89
    (- (math/expt 10 ndigits))))				; count 89 (10^ndigits - (count 89))


(time (println (direct-method 8)))

```

{{Output}}

```txt

85744333
Time: 335 seconds

```


### Using Combinations


(def DIGITS (range 0 10))

(defn -factorial [n]
  (apply * (take n (iterate inc 1))))
;   Cached version of factorial
(def factorial (memoize -factorial))

(defn -combinations [coll k]
  " From http://rosettacode.org/wiki/Combinations_with_repetitions#Clojure "
  (when-let [[x & xs] coll]
    (if (= k 1)
      (map list coll)
      (concat (map (partial cons x) (-combinations coll (dec k)))
              (-combinations xs k)))))
;   Cached version of combinations
(def combinations (memoize -combinations))

(defn comb [n r]
  " count of n items select r "
  (/ (/ (factorial n) (factorial r)) (factorial (- n r))))

(defn count-digits [digit-list]
  " count nunmber of occurences of digit in list "
  (reduce (fn [m v] (update-in m [v] (fnil inc 0))) {} digit-list))

(defn count-patterns [c]
  " Count of number of patterns with these digits "
  (->>
    c
    (count-digits)
    (reduce (fn [accum [k v]]
              (* accum (factorial v)))
            1)
    (/ (factorial (count c)))))

(defn itertools-comb [ndigits]
  (->>
    ndigits
    (combinations DIGITS)
    (filter #(is89 (sum-sqr %)))                 ; items which are not 89 (i.e. 1 since lower count)
    (reduce (fn [acc c]
              (+ acc (count-patterns c)))
            0)
    (- (math/expt 10 ndigits))))

(println (itertools-comb 8))
;; Time obtained using benchmark library (i.e. (bench (itertools-comb 8))  )

```

{{{Output}}

```txt

85744333
Time: 78 ms  (i.e. using combinations was over 4,000 times faster
                   both tested on i7 CPU [email protected])

```


## Common Lisp


```lisp

(defun square (number)
  (expt number 2))

(defun list-digits (number)
  "Return the `number' as a list of its digits."
  (loop
    :for (rest digit) := (multiple-value-list (truncate number 10))
                      :then (multiple-value-list (truncate rest 10))
    :collect digit
    :until (zerop rest)))

(defun next (number)
  (loop
    :for digit :in (list-digits number)
    :sum (square digit)))

(defun chain-end (number)
  "Return the ending number after summing the squaring of the digits of
`number'.  Either 1 or 89."
  (loop
    :for next := (next number) :then (next next)
    :until (or (eql next 1)
               (eql next 89))
    :finally (return next)))

(time
 (loop
   :with count := 0
   :for candidate :from 1 :upto 100000000
   :do (when (eql 89 (chain-end candidate))
         (incf count))
   :finally (return count)))

```


{{out}}

```txt

Evaluation took:
  1128.773 seconds of real time
  1126.231095 seconds of total run time (1117.296987 user, 8.934108 system)
  [ Run times consist of 56.419 seconds GC time, and 1069.813 seconds non-GC time. ]
  99.77% CPU
  2,815,545,509,836 processor cycles
  580,663,356,272 bytes consed

*

```



## D

A simple memoizing partially-imperative brute-force solution:

```d
import std.stdio, std.algorithm, std.range, std.functional;

uint step(uint x) pure nothrow @safe @nogc {
    uint total = 0;
    while (x) {
        total += (x % 10) ^^ 2;
        x /= 10;
    }
    return total;
}

uint iterate(in uint x) nothrow @safe {
    return (x == 89 || x == 1) ? x : x.step.memoize!iterate;
}

void main() {
    iota(1, 100_000_000).filter!(x => x.iterate == 89).count.writeln;
}
```

{{out}}

```txt
85744333
```

The run-time is about 10 seconds compiled with ldc2.

A fast imperative brute-force solution:

```d
void main() nothrow @nogc {
    import core.stdc.stdio: printf;

    enum uint magic = 89;
    enum uint limit = 100_000_000;
    uint[(9 ^^ 2) * 8 + 1] lookup = void;

    uint[10] squares;
    foreach (immutable i, ref x; squares)
        x = i ^^ 2;

    foreach (immutable uint i; 1 .. lookup.length) {
        uint x = i;

        while (x != magic && x != 1) {
            uint total = 0;
            while (x) {
                total += squares[(x % 10)];
                x /= 10;
            }
            x = total;
        }

        lookup[i] = x == magic;
    }

    uint magicCount = 0;
    foreach (immutable uint i; 1 .. limit) {
        uint x = i;
        uint total = 0;

        while (x) {
            total += squares[(x % 10)];
            x /= 10;
        }

        magicCount += lookup[total];
    }

    printf("%u\n", magicCount);
}
```

The output is the same.
The run-time is less than 3 seconds compiled with ldc2.

A more efficient solution:

```d
import core.stdc.stdio, std.algorithm, std.range;

enum factorial = (in uint n) pure nothrow @safe @nogc
    => reduce!q{a * b}(1u, iota(1u, n + 1));

uint iLog10(in uint x) pure nothrow @safe @nogc
in {
    assert(x > 0);
} body {
    return (x >= 1_000_000_000) ? 9 :
           (x >=   100_000_000) ? 8 :
           (x >=    10_000_000) ? 7 :
           (x >=     1_000_000) ? 6 :
           (x >=       100_000) ? 5 :
           (x >=        10_000) ? 4 :
           (x >=         1_000) ? 3 :
           (x >=           100) ? 2 :
           (x >=            10) ? 1 : 0;
}

uint nextStep(uint x) pure nothrow @safe @nogc {
    typeof(return) result = 0;

    while (x > 0) {
        result += (x % 10) ^^ 2;
        x /= 10;
    }
    return result;
}

uint check(in uint[] number) pure nothrow @safe @nogc {
    uint candidate = reduce!((tot, n) => tot * 10 + n)(0, number);

    while (candidate != 89 && candidate != 1)
        candidate = candidate.nextStep;

    if (candidate == 89) {
        uint[10] digitsCount;
        foreach (immutable d; number)
            digitsCount[d]++;

        return reduce!((r, c) => r / c.factorial)
                      (number.length.factorial, digitsCount);
    }

    return 0;
}

void main() nothrow @nogc {
    enum uint limit = 100_000_000;
    immutable uint cacheSize = limit.iLog10;

    uint[cacheSize] number;
    uint result = 0;
    uint i = cacheSize - 1;

    while (true) {
        if (i == 0 && number[i] == 9)
            break;
        if (i == cacheSize - 1 && number[i] < 9) {
            number[i]++;
            result += number.check;
        } else if (number[i] == 9) {
            i--;
        } else {
            number[i]++;
            number[i + 1 .. $] = number[i];
            i = cacheSize - 1;
            result += number.check;
        }
    }

    printf("%u\n", result);
}
```

The output is the same.
The run-time is about 0.04 seconds or less.
This third version was ported to D and improved from:
mathblog.dk/project-euler-92-square-digits-number-chain/

A purely functional version, from the Haskell code. It includes two functions currently missing in Phobos used in the Haskell code.
{{trans|Haskell}}

```d
import std.stdio, std.typecons, std.traits, std.typetuple, std.range, std.algorithm;

auto divMod(T)(T x, T y) pure nothrow @safe @nogc {
    return tuple(x / y, x % y);
}

auto expand(alias F, B)(B x) pure nothrow @safe @nogc
if (isCallable!F &&
    is(ParameterTypeTuple!F == TypeTuple!B)
    && __traits(isSame, TemplateOf!(ReturnType!F), Nullable)
    && isTuple!(TemplateArgsOf!(ReturnType!F)[0])
    && is(TemplateArgsOf!(TemplateArgsOf!(ReturnType!F)[0])[1] == B)) {

    alias NAB = ReturnType!F;
    alias AB = TemplateArgsOf!NAB[0];
    alias A = AB.Types[0];

    struct Expand {
        bool first;
        NAB last;

        @property bool empty() pure nothrow @safe @nogc {
            if (first) {
                first = false;
                popFront;
            }
            return last.isNull;
        }

        @property A front() pure nothrow @safe @nogc {
            if (first) {
                first = false;
                popFront;
            }
            return last.get[0];
        }

        void popFront() pure nothrow @safe @nogc { last = F(last.get[1]); }
    }

    return Expand(true, NAB(AB(A.init, x)));
}

//------------------------------------------------

uint step(uint x) pure nothrow @safe @nogc {
    Nullable!(Tuple!(uint, uint)) f(uint n) pure nothrow @safe @nogc {
        return (n == 0) ? typeof(return)() : typeof(return)(divMod(n, 10u).reverse);
    }

    return expand!f(x).map!(x => x ^^ 2).sum;
}

uint iter(uint x) pure nothrow @nogc {
    return x.recurrence!((a, n) => step(a[n - 1])).filter!(y => y.among!(1, 89)).front;
}

void main() {
    iota(1u, 100_000u).filter!(n => n.iter == 89).count.writeln;
}
```

With a small back-porting (to run it with the Phobos of LDC2 2.065) it runs in about 15.5 seconds.


## ERRE


```ERRE

PROGRAM ITERATION

BEGIN
   PRINT(CHR$(12);) ! CLS
   INPUT(N)
   LOOP
      N$=MID$(STR$(N),2)
      S=0
      FOR I=1 TO LEN(N$) DO
         A=VAL(MID$(N$,I,1))
         S=S+A*A
      END FOR
      IF S=89 OR S=1 THEN PRINT(S;)  EXIT END IF
      PRINT(S;)
      N=S
   END LOOP
   PRINT
END PROGRAM

```

This program verifies a number only. With a FOR..END FOR loop it's possible to verify a number range.


## Factor

A brute-force approach with some optimizations. It uses the fact that the first digit-square-sum of any number < 100,000,000 is, at most, 648. These few chains are rapidly memoized as the results for all hundred-million numbers are calculated for the first time or looked up.
USING: kernel math math.ranges math.text.utils memoize
prettyprint sequences tools.time ;
IN: rosetta-code.iterated-digits-squaring

: sum-digit-sq ( n -- m ) 1 digit-groups [ sq ] map-sum ;

MEMO: 1or89 ( n -- m )
    [ dup [ 1 = ] [ 89 = ] bi or ] [ sum-digit-sq ] until ;

[
    0 1
    [
        dup sum-digit-sq 1or89 89 = [ [ 1 + ] dip ] when
        1 + dup 100,000,000 <
    ] loop
    drop .
] time
```

{{out}}

```txt

85744333
Running time: 55.76544594 seconds

```



## FreeBASIC


```freebasic
' FB 1.05.0 Win64

' similar to C Language (first approach)
' timing for i3 @ 2.13 GHz

Function endsWith89(n As Integer) As Boolean
  Dim As Integer digit, sum = 0
  Do
    While n > 0
      digit = n Mod 10
      sum += digit * digit
      n \= 10
    Wend
    If sum = 89 Then Return True
    If sum = 1 Then Return False
    n = sum
    sum  = 0
  Loop
End Function

Dim As Double start = timer
Dim sums(0 To 8 * 81) As UInteger
sums(0) = 1
sums(1) = 0
Dim s As Integer
For n As Integer = 1 To 8
  For i As Integer = n * 81 To 1 Step -1
    For j As Integer = 1 To 9
      s = j * j
      If s > i Then Exit For
      sums(i) += sums(i - s)
    Next j
  Next i

  If n = 8 Then
    Dim As UInteger count89 = 0
    For i As Integer = 1 To n * 81
      If Not endsWith89(i) Then Continue For
      count89 += sums(i)
    Next i
    Print "There are";count89; " numbers from 1 to 100 million ending with 89"
  End If
Next
Print "Elapsed milliseconds ="; Int((timer - start) * 1000 + 0.5)
Print
Print "Press any key to quit"
Sleep
```


{{out}}

```txt

There are 85744333 numbers from 1 to 100 million ending with 89
Elapsed milliseconds = 2

```



## Frink


```frink

total = 0
d = new dict
var sum

for n = 1 to 100 million - 1
{
   sum = n
   do
   {
      if sum < 1000 and d@sum != undef
      {
         sum = d@sum
         break
      }

      c = sum

      sum = 0
      for digit = integerDigits[c]
         sum = sum + digit^2
   } while (sum != 89) and (sum != 1)

   if (n < 1000)
      d@n = sum

   if (sum == 89)
      total = total + 1
}

println[total]

```


{{out}}

```txt

85744333

```



## Go

It's basic. Runs in about 30 seconds on an old laptop.


```go
package main

import (
	"fmt"
)

func main() {
	var d, n, o, u, u89 int64

	for n = 1; n < 100000000; n++ {
		o = n
		for {
			u = 0
			for {
				d = o%10
				o = (o - d) / 10
				u += d*d
				if o == 0 {
					break
				}
			}
			if u == 89 || u == 1 {
				if u == 89 { u89++ }
				break
			}
			o = u
		}
	}
	fmt.Println(u89)
}
```

{{out}}

```txt

85744333

```



## Haskell

Basic solution that contains just a little more than the essence of this computation. This runs in less than eight minutes:

```haskell
import Data.List (unfoldr)
import Data.Tuple (swap)

step :: Int -> Int
step = sum . map (^ 2) . unfoldr f where
    f 0 = Nothing
    f n = Just . swap $ n `divMod` 10

iter :: Int -> Int
iter = head . filter (`elem` [1, 89]) . iterate step

main = do
    print $ length $ filter ((== 89) . iter) [1 .. 99999999]
```

{{out}}

```txt
85744333
```



## J


Here's an expression to turn a number into digits:


```J
digits=: 10&#.inv
```


And here's an expression to square them and find their sum:

```J
sumdigsq=: +/"1@:*:@digits
```


But note that while the task description claims "you always end with either 1 or 89", that claim is somewhat arbitrary.
:But only somewhat the loop is 89 → 145 → 42 → 20 → 4 → 16 → 37 → 58 → 89, so it only ends with 1 or one of the numbers in this loop. 42 is of course far more significant and the one I would choose!!--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 10:12, 16 September 2014 (UTC)


```J
   sumdigsq^:(i.16) 15
15 26 40 16 37 58 89 145 42 20 4 16 37 58 89 145
```


You could just as easily claim that you always end with either 1 or 4. So here's a routine which repeats the sum-square process until the sequence converges, or until it reaches the value 4:


```J
itdigsq4=:4 = sumdigsq^:(0=e.&4)^:_"0
```


But we do not actually need to iterate. The largest value after the first iteration would be:


```J
   sumdigsq 99999999
648
```


So we could write a routine which works for the intended range, and stops after the first iteration:

```J
itdigsq1=:1 = sumdigsq^:(0=e.&4)^:_"0
digsq1e8=:(I.itdigsq1 i.649) e.~ sumdigsq
```


In other words, if the result after the first iteration is any of the numbers in the range 0..648 which converges to 1, it's not a result which would converge to the other loop. This is considerably faster than trying to converge 1e8 sequences, and also evades having to pick an arbitrary stopping point for the sequence which loops for the bulk computation.

And this is sufficient to find our result. We don't want to compute the entire batch of values in one pass, however, so let's break this up into 100 batches of one million each:


```J
   +/+/@:-.@digsq1e8"1(1+i.100 1e6)
85744333
```


Of course, there are faster ways of obtaining that result. The fastest is probably this:

```J
   85744333
85744333
```


This might be thought of as representing the behavior of a highly optimized compiled program. We could abstract this further by using the previous expression at compile time, so we would not have to hard code it.


## Julia

{{works with|Julia|0.6}}
'''Brute force solution''':

```julia
function iterate(m::Integer)
    while m != 1 && m != 89
        s = 0
        while m > 0 # compute sum of squares of digits
            m, d = divrem(m, 10)
            s += d ^ 2
        end
        m = s
    end
    return m
end
itercount(k::Integer) = count(x -> iterate(x) == 89, 1:k)
```


'''More clever solution''':

```julia
using Combinatorics
function itercountcombos(ndigits::Integer)
    cnt = 0
    f = factorial(ndigits)
    # loop over all combinations of ndigits decimal digits:
    for comb in combinations(1:(10+ndigits-1), ndigits)
        s = 0
        perms = 1
        prevd = -1
        rep = 1
        for k = eachindex(comb) # sum digits ^ 2 and count permutations
            d = comb[k] - k
            s += d ^ 2
            # accumulate number of permutations of repeated digits
            if d == prevd
                rep += 1
                perms *= rep
            else
                prevd = d
                rep = 1
            end
        end
        if s > 0 && iterate(s) == 89
            cnt += f ÷ perms # numbers we can get from digits
        end
    end
    return cnt
end
```


'''Benchmarks'''

```julia
@time itercount(100_000_000)
@time itercountcombos(8)
@time itercountcombos(17)
```

{{out}}

```txt
  8.866063 seconds (4.32 k allocations: 232.908 KiB)
  0.053470 seconds (101.05 k allocations: 8.729 MiB)
  1.588977 seconds (12.50 M allocations: 1.536 GiB, 16.94% gc time)
```



## Java

{{works with|Java|8}}

```java
import java.util.stream.IntStream;

public class IteratedDigitsSquaring {

    public static void main(String[] args) {
        long r = IntStream.range(1, 100_000_000)
                .parallel()
                .filter(n -> calc(n) == 89)
                .count();
        System.out.println(r);
    }

    private static int calc(int n) {
        while (n != 89 && n != 1) {
            int total = 0;
            while (n > 0) {
                total += Math.pow(n % 10, 2);
                n /= 10;
            }
            n = total;
        }
        return n;
    }
}
```



```txt
85744333
```



## jq

{{works with|jq|1.4}}
The algorithm presented here caches the results for 1 ... D*81
(where D is the relevant number of digits) and uses the
combinatorial approach, but to keep things relatively brief, the
factorials themselves are not cached.

'''Part 1: Foundations'''

```jq
def factorial: reduce range(2;.+1) as $i (1; . * $i);

# Pick n items (with replacement) from the input array,
# but only consider distinct combinations:
def pick(n):
  def pick(n; m):  # pick n, from m onwards
    if n == 0 then []
    elif m == length then empty
    elif n == 1 then (.[m:][] | [.])
    else ([.[m]] + pick(n-1; m)), pick(n; m+1)
    end;
  pick(n;0) ;

# Given any array, produce an array of [item, count] pairs for each run.
def runs:
  reduce .[] as $item
    ( [];
      if . == [] then [ [ $item, 1] ]
      else  .[length-1] as $last
            | if $last[0] == $item then (.[0:length-1] + [ [$item, $last[1] + 1] ] )
              else . + [[$item, 1]]
              end
      end ) ;
```

'''Part 2: The Generic Task'''

Count how many number chains beginning with n (where 0 < n < 10^D) end with a value 89.

```jq
def terminus:
  # sum of the squared digits
  def ssdigits: tostring | explode | map(. - 48 | .*.) | add;

  if . == 1 or . == 89 then .
  else ssdigits | terminus
  end;

# Count the number of integers i in [1... 10^D] with terminus equal to 89.
def task(D):
  # The max sum of squares is D*81 so return an array that will instantly
  # reveal whether n|terminus is 89:
  def cache:
    reduce range(1; D*81+1) as $d ([false]; . + [$d|terminus == 89]);

  # Compute n / (i1! * i2! * ... ) for the given combination,
  # which is assumed to be in order:
  def combinations(n):
    runs | map( .[1] | factorial) | reduce .[] as $i (n; ./$i);

  cache as $cache
  | (D|factorial) as $Dfactorial
  | reduce ([range(0;10)] | pick(D)) as $digits
      (0;
       ($digits | map(.*.) | add) as $ss
       | if $cache[$ss] then . + ($digits|combinations($Dfactorial))
         else .
         end) ;
```

'''Part 3: D=8'''

```jq
task(8)
```

{{out}}

```sh
$ jq -M -n -f Iterated_digits_squaring_using_pick.jq
85744333

# Using jq>1.4:
# user	0m2.595s
# sys	0m0.010s

# Using jq 1.4:
# user	0m3.942s
# sys	0m0.009s
```



## Kotlin

{{trans|FreeBASIC}}

```scala
// version 1.0.6

fun endsWith89(n: Int): Boolean {
    var digit: Int
    var sum = 0
    var nn = n
    while (true) {
        while (nn > 0) {
            digit = nn % 10
            sum += digit * digit
            nn /= 10
        }
        if (sum == 89) return true
        if (sum == 1) return false
        nn = sum
        sum  = 0
    }
}

fun main(args: Array) {
    val sums = IntArray(8 * 81 + 1)
    sums[0] = 1
    sums[1] = 0
    var s: Int
    for (n in 1 .. 8)
        for (i in n * 81 downTo 1)
            for (j in 1 .. 9) {
                s = j * j
                if (s > i) break
                sums[i] += sums[i - s]
            }
    var count89 = 0
    for (i in 1 .. 8 * 81)
        if (endsWith89(i)) count89 += sums[i]
    println("There are $count89 numbers from 1 to 100 million ending with 89")
}
```


{{out}}

```txt

There are 85744333 numbers from 1 to 100 million ending with 89

```



## Lua


```lua
squares = {}

for i = 0, 9 do
    for j = 0, 9 do
        squares[i * 10 + j] = i * i + j * j
    end
end

for i = 1, 99 do
    for j = 0, 99 do
        squares[i * 100 + j] = squares[i] + squares[j]
    end
end

function sum_squares(n)
    if n < 9999.5 then
        return squares[n]
    else
        local m = math.floor(n / 10000)
        return squares[n - 10000 * m] + sum_squares(m)
    end
end

memory = {}

function calc_1_or_89(n)
    local m = {}
    n = memory[n] or n
    while n ~= 1 and n ~= 89 do
        n = memory[n] or sum_squares(n)
        table.insert(m, n)
    end
    for _, i in pairs(m) do
        memory[i] = n
    end
    return n
end

counter = 0

for i = 1, 100000000 do
    if calc_1_or_89(i) == 89 then
        counter = counter + 1
    end
end

print(counter)
```

{{out}}

```txt
85744333
```


=={{header|Mathematica}} / {{header|Wolfram Language}}==

```Mathematica
sumDigitsSquared[n_Integer] := Total[IntegerDigits[n]^2]
stopValues = Join[{1}, NestList[sumDigitsSquared, 89, 7]];
iterate[n_Integer] :=
 NestWhile[sumDigitsSquared, n, Intersection[stopValues, {#}] == {} &]

numberOfDigits = 8;
maxSum = numberOfDigits 9^2;
loopVariables =
  ToExpression@Table["i" <> ToString[n], {n, numberOfDigits}];
iteratesToOne = Cases[Range@maxSum, _?(iterate[#] == 1 &)];
allIterators =
  Flatten[{Reverse@#, 9}] & /@ Partition[loopVariables, 2, 1];
maxCombinations = numberOfDigits!;

ssd =
  SparseArray[Table[n^2 -> numberOfDigits, {n, 9}], {maxSum}];

Do[
  variables = loopVariables[[;; digitCount]];
  iterators = allIterators[[;; digitCount - 1]];

  Do[ssd +=
    SparseArray[
     Total[variables^2] ->
      maxCombinations/
       Times @@ (Tally[PadRight[variables, numberOfDigits]][[All,
            2]]!), {maxSum}], {i, 9}, Evaluate[Sequence @@ iterators]],

  {digitCount, 2, numberOfDigits}];

onesCount =
 Total[Cases[
    ArrayRules[ssd] /.
     HoldPattern[{a_} -> b_] :> {a,
       b}, {_?(MemberQ[iteratesToOne, #] &), _}][[All, 2]]];

(10^numberOfDigits - 1) - onesCount
```


{{out}}
```txt
85744333
```



## Objeck


```objeck
class Abbreviations {
  function : Main(args : String[]) ~ Nil {
    Count89s(1000000)->PrintLine();
    Count89s(100000000)->PrintLine();
  }

  function : Count89s(limit : Int) ~ Int {
    if(limit < 1) {
      return 0;
    };

    result := 0;
    ends := Int->New[Int->Min(limit, 9 * 9 * 9 + 2)];
    for(i := 1; i < ends->Size(); i++;) {
      ends[i] := i;
      while(ends[i] <> 1 & ends[i] <> 89) {
        ends[i] := SquareDigitSum(ends[i]);
      };

      if(ends[i] = 89) {
        result++;
      };
    };

    for(i := ends->Size(); i < limit; i++;) {
      if(ends[SquareDigitSum(i)] = 89) {
        result++;
      };
    };

    return result;
  }

  function : SquareDigitSum(n : Int) ~ Int {
    sum := 0;
    while(n > 0) {
      digit := n % 10;
      sum += digit * digit;
      n /= 10;
    };

    return sum;
  }
}
```


Output:

```txt

856,929
85,744,333

```


=={{header|Oberon-2}}==
{{works with|oo2c Version 2}

```oberon2

MODULE DigitsSquaring;
IMPORT
  Out;

VAR
  i,hits89: LONGINT;

  PROCEDURE Squaring(n: LONGINT): LONGINT;
  VAR
    d, sum: LONGINT;
  BEGIN
    LOOP
      sum := 0;
      WHILE n > 0 DO
        d := n MOD 10;
        INC(sum,d * d);
        n := n DIV 10
      END;
      IF (sum = 1) OR (sum = 89) THEN EXIT END;
      n := sum;
    END;

    RETURN sum
  END Squaring;

BEGIN
  hits89 := 0;
  FOR i := 1 TO 100000000 DO
    IF Squaring(i) = 89 THEN INC(hits89) END
  END;
  Out.LongInt(hits89,0);Out.Ln
END DigitsSquaring.


```

{{out}}

```txt

85744333

real    0m12.201s
user    0m12.179s
sys     0m0.001s


```



## Oforth


Brute force implementation


```Oforth
: sq_digits(n)
   while (n 1 <> n 89 <> and ) [
      0 while(n) [ n 10 /mod ->n dup * + ]
      ->n
      ] n ;

: iterDigits  | i | 0 100000000 loop: i [ i sq_digits 89 &= + ] . ;
```


{{out}}

```txt

85744333

```



## PARI/GP


```parigp
ssd(n)=n=digits(n); sum(i=1, #n, n[i]^2);
happy(n)=while(n>6, n=ssd(n)); n==1;
ct(n)=my(f=n!,s=10^n-1,d); forvec(v=vector(9,i,[0,n]), d=vector(9,i, if(i>8,n,v[i+1])-v[i]); if(happy(sum(i=1,9,d[i]*i^2)), s-=f/prod(i=1,9,d[i]!)/v[1]!), 1); s;
ct(8)
```

{{out}}

```txt
%1 = 85744333
```



## Pascal

A limited, but fast implementation (up to 10e14).
It calculates first all the possible sums up to cM= sqrt(MAX), with the drawback that cM must be 10^n and can only count from 10^n to 10^(2*n).
Runtime:  n-> 100*n => t(100*n)-> ~10*t(n) O(n) sqrt(n)


```txt

1E8 -> runtime 0..4 ms // not really measureable
1E12-> runtime 0.22 secs
1E14 -> runtime 2,7 secs
1E16 -> runtime 31,0 secs
1E18 -> runtime  354 secs // 2GByte
```


Tested with freepascal.

```pascal
program Euler92;
const
  maxdigCnt = 14;
  //2* to use the sum of two square-sums without access violation
  maxPoss = 2* 9*9*maxdigCnt;// every digit is 9
  cM  = 10*1000*1000;// 10^(maxdigCnt div 2)
  IdxSqrSum = cM;//MaxPoss;//max(cM,MaxPoss);
type
  tSqrSum   = array[0..IdxSqrSum] of Word;
  tEndsIn   = array[0..maxPoss]of Byte;
  tresCache = array[0..maxPoss]of Uint64;

var
  aSqrDigSum : tSqrSum;
  aEndsIn: tEndsIn;
  aresCache : tresCache;

procedure CreateSpuareDigitSum;
var
  i,j,k,l : integer;
begin
  For i := 0 to 9 do
    aSqrDigSum[i] := sqr(i);
  k := 10;
  l := k;
  while k < cM do
  begin
    For i := 1 to 9 do
      For j := 0 to k-1 do
      begin
        aSqrDigSum[l]:=aSqrDigSum[i]+aSqrDigSum[j];
        inc(l);
      end;
    k := l;
  end;
  aSqrDigSum[l] := 1;
end;

function InitEndsIn(n:LongWord):longWord;
{fill aEndsIN recursive}
var
  d,s:LongWord;
begin
  IF n in [0..1] then
  begin
    InitEndsIn := n;
    EXIT;
  end;
  s := aSqrDigSum[n];
  {if unknown}
  IF aEndsIN[s] = byte(-1) then
  begin
    d := InitEndsIn(s);
    aEndsIN[s]:= d;
    InitEndsIn := d;
  end
  else
    InitEndsIn := aEndsIN[s];
end;

function CntSmallOnes(s:longWord;
                      n:longWord=cM-1):NativeUint;
var
  i: longword;
begin
  result := 0;
  For i := cM-1 downto 0 do
    result := result+aEndsIN[aSqrDigSum[i]+s];
end;

procedure Init;
var
  i,j,cnt : integer;
begin
  CreateSpuareDigitSum;
  fillchar(aEndsIN,Sizeof(aEndsIN) ,#255);
  aEndsIN[0] := 0;
  aEndsIN[1]:= 1;
  aEndsIN[89]:= 0;// no need to use 89
  For i := 1 to maxPoss do
    aEndsIN[i]:= InitEndsIN(i);

  cnt := 0;
  fillchar(aresCache,SizeOf(aresCache),#0);
  For i := Low(tSqrSum) to high(tSqrSum) do
  begin
    j := aSqrDigSum[i];
    If aresCache[j] = 0 then
    begin
//      write(i,',');
      aresCache[j] := CntSmallOnes(j);
      inc(cnt);
    end;
  end;
//  writeln;  writeln(cnt,' small counts out of ',cM);
end;
{
function EndsIn(n:LongWord):Word;
var
  d,s:LongWord;
begin
  d := n;
  s := 0;
  while d > High(tSqrSum) do
  begin
    s := s+aSqrDigSum[d Mod cM];
    d := d Div cM
  end;
  s :=s+aSqrDigSum[d];
  EndsIn := aEndsIN[s];
end;
}

function CntOnes(s: longWord;n:Int64):Int64;
var
  i : Int64;
begin
writeln;
  result := 0;
  i := n div cM;
  repeat
    result := result+aresCache[s+aSqrDigSum[i]];
    dec(i)
  until i < 0
end;

const
  upperlimit = cM*cM ;
var
  Res : Int64;
begin
  Init;
  Res := CntOnes(0,upperlimit-1)+1;
  writeln('there are ',res,'  1s ');
  writeln('there are ',upperlimit-res,' 89s ');
end.

```
output  i3 3.5 Ghz
```txt

10e18
//658 small counts out of 1000000000
there are 118226055080025491  1s
there are 881773944919974509 89s

real	5m54.431s
user	5m53.977s

10e14
there are 13770853279685  1s
there are 86229146720315 89s

real  0m2.699s
user  0m2.693s

```



## Perl

{{trans|perl6}}


```perl
use warnings;
use strict;

my @sq = map { $_ ** 2 } 0 .. 9;
my %cache;
my $cnt = 0;

sub Euler92 {
    my $n = 0 + join( '', sort split( '', shift ) );
    $cache{$n} //= ($n == 1 || $n == 89) ? $n :
    Euler92( sum( @sq[ split '', $n ] ) )
}

sub sum {
   my $sum;
   $sum += shift while @_;
   $sum;
}

for (1 .. 100_000_000) {
   ++$cnt if Euler92( $_ ) == 89;
}

print $cnt;
```


    85744333


## Perl 6

This fairly abstract version does caching and filtering to reduce the number of values it needs to check and moves calculations out of the hot loop, but is still interminably slow... even for just up to 1,000,000.


```perl6
constant @sq = ^10 X** 2;
my $cnt = 0;

sub Euler92($n) {
    $n == any(1,89) ?? $n !!
    (state %){$n} //= Euler92( [+] @sq[$n.comb] )
}

for 1 .. 1_000_000 -> $n {
   my $i = +$n.comb.sort.join;
   ++$cnt if Euler92($i) == 89;
}

say $cnt;
```

{{out}}

```txt
856929
```

All is not lost, however. Through the use of gradual typing, Perl 6 scales down as well as up, so this jit-friendly version is performant enough to brute force the larger calculation:

```perl6
my @cache;
@cache[1] = 1;
@cache[89] = 89;

sub Euler92(int $n) {
    $n < 649  # 99,999,999 sums to 648, so no point remembering more
        ?? (@cache.AT-POS($n) //= ids($n))
        !! ids($n)
}

sub ids(int $num --> int) {
    my int $n = $num;
    my int $ten = 10;
    my int $sum = 0;
    my int $t;
    my int $c;
    repeat until $n == 89 or $n == 1 {
        $sum = 0;
        repeat {
            $t = $n div $ten;
            $c = $n - $t * $ten;
            $sum = $sum + $c * $c;
        } while $n = $t;
        $n = @cache.AT-POS($sum) // $sum;
    }
    $n;
}

my int $cnt = 0;
for 1 .. 100_000_000 -> int $n {
   $cnt = $cnt + 1 if Euler92($n) == 89;
}
say $cnt;
```

{{out}}

```txt
85744333
```

Which is better, but is still pretty slow.

The biggest gains are often from choosing the right algorithm.


```perl6
sub endsWithOne($n --> Bool) {
    my $digit;
    my $sum = 0;
    my $nn = $n;
    loop {
        while $nn > 0 {
            $digit = $nn % 10;
            $sum += $digit²;
            $nn = $nn div 10;
        }
        return True  if $sum == 1;
        return False if $sum == 89;
        $nn = $sum;
        $sum = 0;
    }
}

my $k = 8; # 10**8
my @sums is default(0) = 1,0;
my $s;
for 1 .. $k -> $n {
    for $n*81 … 1 -> $i {
        for 1 .. 9 -> $j {
            $s = $j²;
            last if $s > $i;
            @sums[$i] += @sums[$i - $s];
        }
    }
}

my $ends-with-one = sum flat @sums[(1 .. $k*81).grep: { endsWithOne($_) }], +endsWithOne(10**$k);

say 10**$k - $ends-with-one;
```


{{out}}

```txt
85744333
```



## Phix

{{trans|C}}

```Phix
constant MAXINT = iff(machine_bits()=32?9007199254740992
                                       :9223372036854775807)

procedure main(integer limit)
sequence sums = repeat(0,limit*81+1)
    sums[1] = 1
    for n=1 to limit do
        for i=n*81 to 1 by -1 do
            for j=1 to 9 do
                integer s = j*j
                if s>i then exit end if
                sums[i+1] += sums[i+1-s]
            end for
        end for
        atom count89 = 0
        for i=1 to n*81 do
            integer r, digit, w = i
            while w!=1 do
                r = 0
                while w!=0 do
                    digit = mod(w,10)
                    r += digit*digit
                    w = floor(w/10)
                end while
                if r=89 then
                    count89 += sums[i+1]
                    if count89>MAXINT then
                        printf(1,"counter overflow for 10^%d\n",n)
                        return
                    end if
                    exit
                end if
                w = r
            end while
        end for
        printf(1,"There are %d numbers from 1 to 10^%d ending with 89\n",{count89,n})
    end for
end procedure

atom t0 = time()
main(20)
?time()-t0
```

{{out}}
on 32 bit:

```txt

There are 7 numbers from 1 to 10^1 ending with 89
There are 80 numbers from 1 to 10^2 ending with 89
There are 857 numbers from 1 to 10^3 ending with 89
There are 8558 numbers from 1 to 10^4 ending with 89
There are 85623 numbers from 1 to 10^5 ending with 89
There are 856929 numbers from 1 to 10^6 ending with 89
There are 8581146 numbers from 1 to 10^7 ending with 89
There are 85744333 numbers from 1 to 10^8 ending with 89
There are 854325192 numbers from 1 to 10^9 ending with 89
There are 8507390852 numbers from 1 to 10^10 ending with 89
There are 84908800643 numbers from 1 to 10^11 ending with 89
There are 850878696414 numbers from 1 to 10^12 ending with 89
There are 8556721999130 numbers from 1 to 10^13 ending with 89
There are 86229146720315 numbers from 1 to 10^14 ending with 89
There are 869339034137667 numbers from 1 to 10^15 ending with 89
There are 8754780882739336 numbers from 1 to 10^16 ending with 89
counter overflow for 10^17
0.031

```

same on 64-bit, but ending

```txt

There are 87975303595231975 numbers from 1 to 10^17 ending with 89
There are 881773944919974509 numbers from 1 to 10^18 ending with 89
There are 8816770037940618762 numbers from 1 to 10^19 ending with 89
counter overflow for 10^20
0.109

```


### Combinatorics version

Following the steps outlined on the talk page.

I realised I needed to do this in two stages.

Phase 1. Make sure we can count.

```Phix
function comb(sequence res, from, integer n, at=1, sequence chosen={})
    if length(chosen)=n then
        sequence digits = repeat(0,10)
        for i=1 to length(chosen) do
            digits[chosen[i]+1]+=1
        end for
        atom p = factorial(length(chosen))
        for i=1 to 10 do
            if digits[i] then
                p /= factorial(digits[i])
            end if
        end for
        res = sq_add(res,{p,1})
    else
        for i=at to length(from) do
            res = comb(res,from,n,i,append(chosen,from[i]))
        end for
    end if
    return res
end function

constant nums = {0,1,2,3,4,5,6,7,8,9}
for i=1 to 8 do
    ?comb({0,0},nums,i)
end for
```

Starting with the combinations method from http://rosettacode.org/wiki/Combinations_with_repetitions#Phix converted to a function, make sure we
are covering all the numbers correctly by checking that we have indeed found power(10,n) of them, and show we are looking at significantly fewer combinations.
{{out}}

```txt

{10,10}
{100,55}
{1000,220}
{10000,715}
{100000,2002}
{1000000,5005}
{10000000,11440}
{100000000,24310}

```

Phase 2. Add in the rest of the logic, as suggested count 1's in preference to 89's and subtract from 10^n to get the answer.

[PS There is an eerie similarity between this and the 2nd C version, but I swear it is not a copy, and I noticed that later.]

```Phix
sequence is1

function comb(atom res, sequence from, integer n, at=1, sequence chosen={})
    if length(chosen)=n then
        sequence digits = repeat(0,10)
        atom sumsq = 0
        for i=1 to length(chosen) do
            integer ci = chosen[i]
            sumsq += ci*ci
            digits[ci+1]+=1
        end for
        if sumsq=0 or is1[sumsq] then
            atom perms = factorial(length(chosen))
            for i=1 to 10 do
                if digits[i] then
                    perms /= factorial(digits[i])
                end if
            end for
            res += perms
        end if
    else
        for i=at to length(from) do
            res = comb(res,from,n,i,append(chosen,from[i]))
        end for
    end if
    return res
end function

procedure setis1(integer n)
    is1 = repeat(0,n*81)
    for i=1 to length(is1) do
        integer r, digit, w = i
        while 1 do
            r = 0
            while w!=0 do
                digit = mod(w,10)
                r += digit*digit
                w = floor(w/10)
            end while
            if r=89 then exit end if
            if r=1 then
                is1[i] = 1
                exit
            end if
            w = r
        end while
    end for
end procedure

constant nums = {0,1,2,3,4,5,6,7,8,9}
for i=1 to 16 do
    atom t0 = time()
    setis1(i)
    printf(1,"There are %d numbers from 1 to 10^%d ending with 89 (%3.2fs)\n",{power(10,i)-comb(0,nums,i),i,time()-t0})
end for
```

{{out}}
Sadly, while still very much faster than brute force, several times slower than the translated from C version.

```txt

There are 7 numbers from 1 to 10^1 ending with 89 (0.00s)
There are 80 numbers from 1 to 10^2 ending with 89 (0.00s)
There are 857 numbers from 1 to 10^3 ending with 89 (0.00s)
There are 8558 numbers from 1 to 10^4 ending with 89 (0.00s)
There are 85623 numbers from 1 to 10^5 ending with 89 (0.02s)
There are 856929 numbers from 1 to 10^6 ending with 89 (0.00s)
There are 8581146 numbers from 1 to 10^7 ending with 89 (0.02s)
There are 85744333 numbers from 1 to 10^8 ending with 89 (0.02s)
There are 854325192 numbers from 1 to 10^9 ending with 89 (0.05s)
There are 8507390852 numbers from 1 to 10^10 ending with 89 (0.09s)
There are 84908800643 numbers from 1 to 10^11 ending with 89 (0.17s)
There are 850878696414 numbers from 1 to 10^12 ending with 89 (0.34s)
There are 8556721999130 numbers from 1 to 10^13 ending with 89 (0.58s)
There are 86229146720315 numbers from 1 to 10^14 ending with 89 (0.92s)
There are 869339034137667 numbers from 1 to 10^15 ending with 89 (1.66s)
There are 8754780882739337 numbers from 1 to 10^16 ending with 89 (2.75s)

```



## PicoLisp

Brute force with caching

```PicoLisp
(de *Idx1or89 (89 . 89) ((1 . 1)))

(de 1or89 (N)
   (let L (mapcar format (chop N))
      (if (lup *Idx1or89 (setq N (sum * L L)))
         (cdr @)
         (prog1
            (1or89 N)
            (idx '*Idx1or89 (cons N @) T) ) ) ) )
```

Test:

```PicoLisp
(let Ones 0
   (for I 100000000
      (and (=1 (1or89 I)) (inc 'Ones)) )
   (println (- 100000000 Ones)) )
```

Output:

```txt
85744333
```



## PL/I


```PL/I

test: procedure options (main, reorder); /* 6 August 2015 */

   declare (m, n) fixed decimal (10);
   declare (i, j, p, s, tally initial (0) ) fixed binary (31);
   declare d fixed binary (7);
   declare (start_time, finish_time, elapsed_time) float (15);

   start_time = secs();

   do m = 1 to 1000000;
      n = m;
      do until ((n = 1) | (n = 89));
         p = n; s = 0;
         do while (p > 0);
            d = mod(p, 10);
            p = p/10;
            s = s + d*d;
         end;
         n = s;
      end;
      if n = 89 then tally = tally + 1;
   end;

   finish_time = secs();
   put skip edit (Tally, ' numbers iterated to 89') (f(10), A);
   elapsed_time = finish_time - start_time;
   put skip edit ('Elapsed time=', elapsed_time, ' secs') (A, F(10,3));

end test;

```

Output:

```txt

    856929 numbers iterated to 89
Elapsed time=    39.280 secs

```



## PureBasic

{{Trans|C}}

```purebasic
OpenConsole()
Procedure is89(x)
  Repeat
    s=0
    While x : s+ x%10*x%10 : x/10 : Wend
    If s=89 : ProcedureReturn 1 : EndIf
    If s=1  : ProcedureReturn 0 : EndIf
    x=s
  ForEver
EndProcedure

Procedure main()
  Dim sums(32*81+1) : sums(0)=1 : sums(1)=0

  For n=1 To n+1
    For i=n*81 To 1 Step -1
      For j=1 To 9
        s=j*j : If s>i : Break : EndIf
        sums(i)+sums(i-s)
      Next
    Next
    count89=0
    For i=1 To n*81+1
      If Not is89(i) : Continue : EndIf
      If sums(i)>9223372036854775807-count89
        PrintN("counter overflow for 10^"+Str(n))
        ProcedureReturn 0
      EndIf
      count89+sums(i)
    Next
    PrintN("1->10^"+LSet(Str(n),2,Chr(32))+": "+Str(count89))
  Next
EndProcedure

start=ElapsedMilliseconds()
main()
Print("elapsed milliseconds= "+Str(ElapsedMilliseconds()-start))
Input()
```

{{out}}

```txt
1->10^1 : 7
1->10^2 : 80
1->10^3 : 857
1->10^4 : 8558
1->10^5 : 85623
1->10^6 : 856929
1->10^7 : 8581146
1->10^8 : 85744333
1->10^9 : 854325192
1->10^10: 8507390852
1->10^11: 84908800643
1->10^12: 850878696414
1->10^13: 8556721999130
1->10^14: 86229146720315
1->10^15: 869339034137667
1->10^16: 8754780882739336
1->10^17: 87975303595231975
1->10^18: 881773944919974509
1->10^19: 8816770037940618762
counter overflow for 10^20
elapsed milliseconds= 9
```

{{Trans|C++}}

```purebasic
OpenConsole()
Procedure sum_square_digits(n)
  num=n : sum=0
  While num>0
    digit=num%10
    num=(num-digit)/10
    sum+ digit*digit
  Wend
  ProcedureReturn sum
EndProcedure

Procedure main()
  i=0 : result=0 : count=0
  For i=1 To 1e8
    If Not i=1 Or Not i=89
      result=sum_square_digits(i)
    Else
      result=i
    EndIf
    While Not result=1 And Not result=89
      result=sum_square_digits(result)
    Wend
    If result=89 : count+1 : EndIf
  Next
  PrintN(Str(count))
EndProcedure

start=ElapsedMilliseconds()
main()
Print("elapsed milliseconds: "+Str(ElapsedMilliseconds()-start))
Input()
```

{{out}}

```txt
85744333
elapsed milliseconds: 65553
```



## Python


### Combinatorics


### =Translation of D=

{{trans|D}}

```python
from math import ceil, log10, factorial

def next_step(x):
    result = 0
    while x > 0:
        result += (x % 10) ** 2
        x /= 10
    return result

def check(number):
    candidate = 0
    for n in number:
        candidate = candidate * 10 + n

    while candidate != 89 and candidate != 1:
        candidate = next_step(candidate)

    if candidate == 89:
        digits_count = [0] * 10
        for d in number:
            digits_count[d] += 1

        result = factorial(len(number))
        for c in digits_count:
            result /= factorial(c)
        return result

    return 0

def main():
    limit = 100000000
    cache_size = int(ceil(log10(limit)))
    assert 10 ** cache_size == limit

    number = [0] * cache_size
    result = 0
    i = cache_size - 1

    while True:
        if i == 0 and number[i] == 9:
            break
        if i == cache_size - 1 and number[i] < 9:
            number[i] += 1
            result += check(number)
        elif number[i] == 9:
            i -= 1
        else:
            number[i] += 1
            for j in xrange(i + 1, cache_size):
                number[j] = number[i]
            i = cache_size - 1
            result += check(number)

    print result

main()
```

{{out}}
 85744333
The run-time is less than half a second.

### =Translation of Ruby=

{{trans|Ruby}}

```ruby

from itertools import combinations_with_replacement
from array import array
from time import clock
D = 8
F = [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800, 87178291200, 1307674368000, 20922789888000, 355687428096000]
def b(n):
    yield 1
    for g in range(1,n+1):
        gn = g
        res = 0
        while gn > 0:
            gn,rem = divmod(gn,10)
            res += rem**2
        if res==89:
            yield 0
        else:
            yield res
N = array('I',b(81*D))
for n in range(2,len(N)):
    q = N[n]
    while q>1:
        q = N[q]
    N[n] = q

es = clock()
z = 0
for n in combinations_with_replacement(range(10),D):
    t = 0
    for g in n:
        t += g*g
    if N[t] == 0:
        continue
    t = [0,0,0,0,0,0,0,0,0,0]
    for g in n:
        t[g] += 1
    t1 = F[D]
    for g in t:
        t1 /= F[g]
    z += t1
ee = clock() - es
print "\nD==" + str(D) + "\n  " + str(z) + " numbers produce 1 and " + str(10**D-z) + " numbers produce 89"
print "Time ~= " + str(ee) + " secs"

```

{{out}}

```txt

D==8
 14255667 numbers produce 1 and 85744333 numbers produce 89
Time ~= 0.14 secs

D==11
15091199357 numbers produce 1 and 84908800643 numbers produce 89
Time ~= 1.12 secs

D==14
13770853279685 numbers produce 1 and 86229146720315 numbers produce 89
Time ~= 7.46 secs

D==17
12024696404768025 numbers produce 1 and 87975303595231975 numbers produce 89
Time ~= 34.16 secs

```



### Python: Simple caching


```python>>>
 from functools import lru_cache
>>> @lru_cache(maxsize=1024)
def ids(n):
	if n in {1, 89}: return n
	else: return ids(sum(int(d) ** 2 for d in str(n)))


>>> ids(15)
89
>>> [ids(x) for x in range(1, 21)]
[1, 89, 89, 89, 89, 89, 1, 89, 89, 1, 89, 89, 1, 89, 89, 89, 89, 89, 1, 89]
>>> sum(ids(x) == 89 for x in range(1, 100000000))
85744333
>>>
```


This took a much longer time, in the order of hours.


### Python: Enhanced caching

Notes that the order of digits in a number does not affect the final result so caches the digits of the number in sorted order for more hits.

```python>>>
 from functools import lru_cache
>>> @lru_cache(maxsize=1024)
def _ids(nt):
	if nt in {('1',), ('8', '9')}: return nt
	else: return _ids(tuple(sorted(str(sum(int(d) ** 2 for d in nt)))))


>>> def ids(n):
	return int(''.join(_ids(tuple(sorted(str(n))))))

>>> ids(1), ids(15)
(1, 89)
>>> [ids(x) for x in range(1, 21)]
[1, 89, 89, 89, 89, 89, 1, 89, 89, 1, 89, 89, 1, 89, 89, 89, 89, 89, 1, 89]
>>> sum(ids(x) == 89 for x in range(1, 100000000))
85744333
>>> _ids.cache_info()
CacheInfo(hits=99991418, misses=5867462, maxsize=1024, currsize=1024)
>>>
```


This took tens of minutes to run.


### Count digit sums

If we always count up to powers of 10, it's faster to just record how many different numbers have the same digit square sum. The check89() function is pretty simple-minded, because it doesn't need to be fancy here.

```python
from __future__ import print_function
from itertools import count

def check89(n):
    while True:
        n, t = 0, n
        while t: n, t = n + (t%10)**2, t//10
        if n <= 1: return False
        if n ==89: return True

a, sq, is89 = [1], [x**2 for x in range(1, 10)], [False]
for n in range(1, 500):
    b, a = a, a + [0]*81
    is89 += map(check89, range(len(b), len(a)))

    for i,v in enumerate(b):
        for s in sq: a[i + s] += v

    x = sum(a[i] for i in range(len(a)) if is89[i])
    print("10^%d" % n, x)
```



## Racket


This contains two versions (in one go). The naive version which can (and should, probably) be used for investigating a single number. The second version can count the IDSes leading to 89 for powers of 10.


```racket
#lang racket
;; Tim-brown 2014-09-11

;; The basic definition.
;; It is possible to memoise this or use fixnum (native) arithmetic, but frankly iterating over a
;; hundred million, billion, trillion numbers will be slow. No matter how you do it.
(define (digit^2-sum n)
  (let loop ((n n) (s 0))
    (if (= 0 n) s (let-values ([(q r) (quotient/remainder n 10)]) (loop q (+ s (sqr r)))))))

(define (iterated-digit^2-sum n)
  (match (digit^2-sum n) [0 0] [1 1] [89 89] [(app iterated-digit^2-sum rv) rv]))

;; Note that: ids(345) = ids(354) = ids(435) = ids(453) = ids(534) = ids(543) = 50 --> 89
;; One calculation does for 6 candidates.
;; The plan:
;;  - get all the ordered combinations of digits including 0's which can be used both as digits and
;;    "padding" digits in the most significant digits. (n.b. all-zeros is not in the range to be
;;    tested and should be dropped)
;;  - find the digit sets that have an IDS of 89
;;  - find out how many combinations there are of these digits

;; output: a list of n-digits long lists containing all of the digit combinations.
;;         a smart bunny would figure out the sums of the digits as they're generated but I'll plod
;;         along step-by-step. a truly smart bunny would also count the combinations. that said, I
;;         don't think I do much unnecessary computation here.
(define (all-digit-lists n-digits)
  (define (inner remain acc least-digit)
    (cond
      [(zero? remain) (list (list))]
      [(= least-digit 10) null]
      [else
       (for*/list
           ((ld+ (in-range least-digit 10))
            (rgt (in-list (inner (sub1 remain) empty ld+))))
         (append acc (cons ld+ rgt)))]))
  (inner n-digits '() 0))

;; We calculate IDS differently since we're presented with a list of digits rather than a number
(define (digit-list-IDS c)
  (define (digit-combo-IDS c)
    (apply + (map sqr c)))
  (iterated-digit^2-sum (digit-combo-IDS c)))

;; ! (factiorial) -- everyone's favourite combinatorial function! (that's just an exclamation mark)
;; there's one in (require math/number-theory) for any heavy lifting, but we're not or I could import
;; it from math/number-theory -- but this is about all I need. A lookup table is going to be faster
;; than a more general function.
(define (! n)
  (case n [(0 1) 1] [(2) 2] [(3) 6] [(4) 24] [(5) 120] [(6) 720] [(7) 5040] [(8) 40320] [(9) 362880]
    [else (* n (! (sub1 n)))] ; I expect this clause'll never be called
    ))

;; We need to count the permutations -- digits are in order so we can use the tail (cdr) function for
;; determining my various k's. See: https://en.wikipedia.org/wiki/Combination
(define (count-digit-list-permutations c #:length (l (length c)) #:length! (l! (! l)))
  (let loop ((c c) (i 0) (prev -1 #;"never a digit") (p l!))
    (match c
      [(list) (/ p (! i))]
      [(cons (== prev) d) (loop d (+ i 1) prev p)]
      [(cons a d) (loop d 1 a (/ p (! i)))])))

;; Wrap it all up in a neat function
(define (count-89s-in-100... n)
  (define n-digits (order-of-magnitude n))
  (define combos (drop (all-digit-lists n-digits) 1)) ; don't want first one which is "all-zeros"
  (for/sum ((c (in-list combos)) #:when (= 89 (digit-list-IDS c)))
    (count-digit-list-permutations c #:length n-digits)))

(displayln "Testing permutations:")
(time (printf "1000000:\t~a~%"        (count-89s-in-100...       1000000)))
(time (printf "100000000:\t~a~%"      (count-89s-in-100...     100000000)))
(time (printf "1000000000:\t~a~%"     (count-89s-in-100...    1000000000)))
(time (printf "1000000000000:\t~a~%"  (count-89s-in-100... 1000000000000)))
(newline)
;; Do these last, since the 10^8 takes longer than my ADHD can cope with
(displayln "Testing one number at a time (somewhat slower):")
(time (printf "1000000:\t~a~%"   (for/sum ((n (in-range 1   1000000))
                                           #:when (= 89 (iterated-digit^2-sum n))) 1)))
(time (printf "100000000:\t~a~%" (for/sum ((n (in-range 1 100000000))
                                           #:when (= 89 (iterated-digit^2-sum n))) 1)))

{module+ test
  (require tests/eli-tester)
  [test
   (iterated-digit^2-sum 15) => 89
   (iterated-digit^2-sum 7) => 1
   (digit-combo-perms '()) => 1
   (digit-combo-perms '(1 2 3)) => 6
   (digit-combo-perms '(1 1 3)) => 3
   (for/sum ((n (in-range 1 1000000)) #:when (= 89 (iterated-digit^2-sum n))) 1) => 856929
   (all-digit-lists 1) => '((0) (1) (2) (3) (4) (5) (6) (7) (8) (9))
   (length (all-digit-lists 2)) => 55
   (length (all-digit-lists 3)) => 220
   (count-89s-in-100... 1000000) => 856929]
  }
```


{{out}}


```txt
Testing permutations:
1000000:	856929
cpu time: 8 real time: 8 gc time: 0
100000000:	85744333
cpu time: 44 real time: 43 gc time: 0
1000000000:	854325192
cpu time: 112 real time: 110 gc time: 20
1000000000000:	850878696414
cpu time: 1108 real time: 1110 gc time: 472

Testing one number at a time (somewhat slower):
1000000:	856929
cpu time: 1168 real time: 1171 gc time: 0
100000000:	85744333
cpu time: 130720 real time: 130951 gc time: 0
```


Ok, so maybe 131 seconds is not so flattering -- but I have not memoised or anything fancy like that, because even doing that isn't going to come anywhere near competing with 44ms.


## REXX

{Both REXX versions don't depend on a specific end─number.}

### with memoization


```rexx
/*REXX program performs the squaring of iterated digits  (until the sum equals 1 or 89).*/
parse arg n .                                    /*obtain optional arguments from the CL*/
if n=='' | n==","  then n=10 * 1000000           /*Not specified?  Then use the default.*/
!.=0;   do m=1  for 9;  !.m=m**2;  end  /*m*/    /*build a short─cut for the squares.   */
a.=.;                                #.=!.       /*intermediate counts of some numbers. */
     do j=1  for n;                  x=j         /* [↓] process the numbers in the range*/
       do q=1  until s==89 | s==1;   s=0         /*add sum of the squared decimal digits*/
             do  until x==''                     /*process each of the dec. digits in X.*/
             parse var x _ +1 x;     s=s + !._   /*get a digit;  sum the fast square,   */
             end   /*until x ··· */              /* [↑]  S≡is sum of the squared digits.*/
       z.q=s                                     /*assign sum to a temporary auxiliary. */
       if a.s\==.  then do;  s=a.s;  leave;  end /*Found a previous sum?  Then use that.*/
       x=s                                       /*substitute the sum for the "new"  X. */
       end   /*q*/                               /* [↑]  keep looping 'til   S= 1 or 89.*/
                   do f=1  for q;  _=a.f;  a._=s /*use the auxiliary arrays (for lookup)*/
                   end   /*f*/
     #.s=#.s + 1                                 /*bump the counter for the 1's or 89's.*/
     end   /*j*/

  do k=1  by 88  for 2;   @k=right('"'k'"', 5)   /*display two results; define a literal*/
  say 'count of'   @k   " chains for all natural numbers up to "     n     ' is:'      #.k
  end   /*k*/                                    /*stick a fork in it,  we're all done. */
```

{{out|output|text=  when using the default input:}}

(ten million)

```txt

count of   "1"  chains for all natural numbers up to  10000000  is: 1418854
count of  "89"  chains for all natural numbers up to  10000000  is: 8581146

```



### process in chunks

This version is about    2½    times faster than the previous REXX version.

```rexx
/*REXX program performs the squaring of iterated digits  (until the sum equals 1 or 89).*/
parse arg n .                                    /*obtain optional arguments from the CL*/
if n=='' | n==","  then n=10 * 1000000           /*Not specified?  Then use the default.*/
!.=0;      do m=1  for 9;  !.m=m**2;  end /*m*/  /*build a short─cut for the squares.   */
$.=.; $.0=0;  $.00=0;  $.000=0;  $.0000=0; @.=$. /*short-cuts for sub-group summations. */
#.=0                                             /*count of  1  and  89  results so far.*/
     do j=1  for n;       s=sumDs(j)             /* [↓]  process each number in a range.*/
     #.s=#.s + 1                                 /*bump the counter for  1's  or  89's. */
     end   /*j*/

  do k=1  by 88  for 2;   @=right('"'k'"', 5)    /*display two results; define a literal*/
  say 'count of'   @    " chains for all natural numbers up to "     n     ' is:'      #.k
  end   /*k*/                                    /*stick a fork in it,  we're all done. */
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
sumDs: parse arg z;  chunk=3                     /*obtain the number (for adding digits)*/
       p=0                                       /*set partial sum of the decimal digits*/
          do m=1  by chunk  to length(z)         /*process the number, in chunks of four*/
          y=substr(z, m, chunk)                  /*extract a 4─byte chunk of the number.*/
          if @.y==.  then do;   oy=y;      a=0   /*Not done before?  Then sum the number*/
                            do  until y==''      /*process each of the dec. digits in Y.*/
                            parse var y  _ +1 y  /*obtain a decimal digit; add it to  A.*/
                            a=a + !._            /*obtain a decimal digit; add it to  A.*/
                            end  /*until y ···*/ /* [↑]   A ≡ is the sum of squared digs*/
                          @.oy=a                 /*mark original  Y  as being summed.   */
                          end
                     else [email protected]                  /*use the  pre─summed  digits of  Y.   */
          p=p + a                                /*add all the parts of number together.*/
          end   /*m*/

       if $.p\==.  then return $.p               /*Computed before?  Then use the value.*/
       y=p                                       /*use a new copy of  P.                */
              do  until s==1 | s==89;  s=0       /*add the squared decimal digits of  P.*/
                 do  until y==''                 /*process each  decimal digits in    X.*/
                 parse var y  _ +1 y;  s=s + !._ /*get a dec. digit; sum the fast square*/
                 end   /*until y ···*/           /* [↑]  S ≡ is sum of the squared digs.*/
              y=s                                /*substitute the sum for a  "new"  X.  */
              end      /*until s ···*/           /* [↑]  keep looping 'til  S=1  or  89.*/
       $.p=s                                     /*use this for memoization for the sum.*/
       return s
```

{{out|output|text=  when using the input of:    100000000 }}

(one hundred million)

```txt

count of   "1"  chains for all natural numbers up to  100000000  is  14255667
count of  "89"  chains for all natural numbers up to  100000000  is  85744333

```



## Ring


```ring

nr = 1000
num = 0
for n = 1 to nr
   sum = 0
   for m = 1 to len(string(n))
       sum += pow(number(substr(string(n),m,1)),2)
       if sum = 89 num += 1 see "" + n + " " + sum + nl ok
   next
next
see "Total under 1000 is : " + num + nl

```

Output:

```txt

58 89
85 89
229 89
267 89
276 89
292 89
348 89
384 89
438 89
483 89
508 89
580 89
581 89
582 89
583 89
584 89
585 89
586 89
587 89
588 89
589 89
627 89
672 89
726 89
762 89
805 89
834 89
843 89
850 89
851 89
852 89
853 89
854 89
855 89
856 89
857 89
858 89
859 89
922 89
Total under 1000 is : 41

```



## Ruby


```ruby
# Count how many number chains for Natural Numbers < 10**d end with a value of 1.
def iterated_square_digit(d)
  f = Array.new(d+1){|n| (1..n).inject(1, :*)}      #Some small factorials
  g = -> (n) { res = 0
               while n>0
                 n, mod = n.divmod(10)
                 res += mod**2
               end
               res==89 ? 0 : res
             }

  #An array: table[n]==0 means that n translates to 89 and 1 means that n translates to 1
  table = Array.new(d*81+1){|n| n.zero? ? 1 : (i=g.call(n))==89 ? 0 : i}
  table.collect!{|n| n = table[n] while n>1; n}
  z = 0                                             #Running count of numbers translating to 1
  [*0..9].repeated_combination(d) do |rc|           #Iterate over unique digit combinations
    next if table[rc.inject(0){|g,n| g+n*n}].zero?  #Count only ones
    nn = [0] * 10                                   #Determine how many numbers this digit combination corresponds to
    rc.each{|n| nn[n] += 1}
    z += nn.inject(f[d]){|gn,n| gn / f[n]}          #Add to the count of numbers terminating in 1
  end
  puts "\nd=(#{d}) in the range 1 to #{10**d-1}",
       "#{z} numbers produce 1 and #{10**d-1-z} numbers produce 89"
end

[8, 11, 14, 17].each do |d|
  t0 = Time.now
  iterated_square_digit(d)
  puts "  #{Time.now - t0} sec"
end
```


{{out}}

```txt

d=(8) in the range 1 to 99999999
14255667 numbers produce 1 and 85744332 numbers produce 89
  0.116007 sec

d=(11) in the range 1 to 99999999999
15091199357 numbers produce 1 and 84908800642 numbers produce 89
  0.921052 sec

d=(14) in the range 1 to 99999999999999
13770853279685 numbers produce 1 and 86229146720314 numbers produce 89
  5.503315 sec

d=(17) in the range 1 to 99999999999999999
12024696404768025 numbers produce 1 and 87975303595231974 numbers produce 89
  24.337392 sec

```




## Rust

These are two naive solutions, one with lots of redundant calculations (memoizationless recursion) and one with a few precomputed values. All digit square sums are no greater than 648 for numbers < 100_000_000.

Both are slow algorithms, however, Rust is among faster languages, so this doesn't take minutes or hours.

'''Naive Recursion'''

```rust
fn digit_square_sum(mut num: usize) -> usize {
    let mut sum = 0;
    while num != 0 {
        sum += (num % 10).pow(2);
        num /= 10;
    }
    sum
}

fn last_in_chain(num: usize) -> usize {
    match num {
        1 | 89 => num,
        _ => last_in_chain(digit_square_sum(num)),
    }
}

fn main() {
    let count = (1..100_000_000).filter(|&n| last_in_chain(n) == 89).count();
    println!("{}", count);
}
```

{{out}}

```txt
85744333
```

Runtime: 6s on a 2500k @ 4Ghz

'''With precomputation'''

```rust
fn dig_sq_sum(mut num : usize ) -> usize {
    let mut sum = 0;
    while num != 0 {
        sum += (num % 10).pow(2);
        num /= 10;
    }
    sum
}

fn last_in_chain(num: usize) -> usize {
    match num {
        0 => 0,
        1 | 89 => num,
        _ => last_in_chain(dig_sq_sum(num)),
    }
}

fn main() {
    let prec: Vec<_> = (0..649).map(|n| last_in_chain(n)).collect();
    let count = (1..100_000_000).filter(|&n| prec[dig_sq_sum(n)] == 89).count();
    println!("{}", count);
}
```

Runtime: 1.7s on a 2500k @ 4Ghz
{{out}}

```txt
85744333
```



## Scala

===Naïve Version, conventional iteration and (tail) recursive in one===
{{Out}}Best seen running in your browser either by [https://scalafiddle.io/sf/3XRtgEE/0 ScalaFiddle (ES aka JavaScript, non JVM)] or [https://scastie.scala-lang.org/6HeszTpxTXOqvrytzShUzg Scastie (remote JVM)] to compare the run times.

```Scala
import scala.annotation.tailrec

object Euler92 extends App {
  override val executionStart = compat.Platform.currentTime

  @tailrec
  private def calcRec(i: Int): Int = {

    @tailrec
    def iter0(n: Int, total: Int): Int =
      if (n > 0) {
        val rest = n % 10
        iter0(n / 10, total + rest * rest)
      }
      else total


    if (i == 89 || i == 1) i else calcRec(iter0(i, 0))
  }


  private def calcConv(i: Int) = {
    var n: Int = i
    while (n != 89 && n != 1) {
      var total = 0
      while (n > 0) {
        val x = n % 10
        total += (x * x)
        n /= 10
      }
      n = total
    }
    n
  }

  println((1 until 100000000).par.count(calcConv(_) == 89))
  println(s"Runtime conventional loop.[total ${compat.Platform.currentTime - executionStart} ms]")

  val executionStart0 = compat.Platform.currentTime
  println((1 until 100000000).par.count(calcRec(_) == 89))
  println(s"Runtime recursive loop.   [total ${compat.Platform.currentTime - executionStart0} ms]")

}
```



## Sidef


```ruby
func digit_square_sum_iter(n) is cached {

    if ((n == 1) || (n == 89)) {
        return n
    }

    __FUNC__(n.digits.sum { .sqr })
}

say (1..1e6 -> count_by { digit_square_sum_iter(_) == 89 })
```

{{out}}

```txt

856929

```



## Swift

{{trans|C}}
{{libheader|Attaswift BigInt}}

With nth power support.


```swift
import BigInt

func is89(_ n: Int) -> Bool {
  var n = n

  while true {
    var s = 0

    repeat {
      s += (n%10) * (n%10)
      n /= 10
    } while n > 0

    if s == 89 {
      return true
    } else if s == 1 {
      return false
    }

    n = s
  }
}

func iterSquare(upToPower pow: Int) {
  var sums = [BigInt](repeating: 0, count: pow * 81 + 1)
  sums[0] = 1

  for n in 1...pow {
    var i = n * 81

    while i > 0 {
      for j in 1..<10 {
        let s = j * j
        guard s <= i else { break }
        sums[i] += sums[i-s]
      }

      i -= 1
    }

    var count89 = BigInt(0)

    for x in 1..10^\(n): \(count89)")
  }
}

iterSquare(upToPower: 8)
```


{{out}}

```txt
1->10^1: 7
1->10^2: 80
1->10^3: 857
1->10^4: 8558
1->10^5: 85623
1->10^6: 856929
1->10^7: 8581146
1->10^8: 85744333
```



## Tcl

All three versions below produce identical output (85744333), but the third is fastest and the first is the slowest, both by substantial margins.
===''Very'' Naïve Version===

```tcl
proc ids n {
    while {$n != 1 && $n != 89} {
	set n [tcl::mathop::+ {*}[lmap x [split $n ""] {expr {$x**2}}]]
    }
    return $n
}
for {set i 1} {$i <= 100000000} {incr i} {
    incr count [expr {[ids $i] == 89}]
}
puts $count
```


### Intelligent Version

Conversion back and forth between numbers and strings is slow. Using math operations directly is much faster (around 4 times in informal testing).

```tcl
proc ids n {
    while {$n != 1 && $n != 89} {
	for {set m 0} {$n} {set n [expr {$n / 10}]} {
	    incr m [expr {($n%10)**2}]
	}
	set n $m
    }
    return $n
}
for {set i 1} {$i <= 100000000} {incr i} {
    incr count [expr {[ids $i] == 89}]
}
puts $count
```


### Substantially More Intelligent Version

Using the observation that the maximum value after 1 step is obtained for 999999999, which is  $9^3=729$ . Thus, running one step of the reduction and then using a lookup table (which we can construct quickly at the start of the run, and which has excellent performance) is much faster overall, approximately 3–4 times than the second version above (and over 12 times faster than the first version).

:Donald, you have 1 too many 9's the value after step 1 is 81*8 = 648. Not that that is the problem here, you can not afford to go around this loop 100 million times.  Notice that IDS[21] == IDS[12], IDS[123] == IDS[132] == IDS[213} ... etc, etc. The Ruby version takes about a tenth of a second.--[[User:Nigel Galloway|Nigel Galloway]] ([[User talk:Nigel Galloway|talk]]) 12:47, 31 August 2014 (UTC)

```tcl
# Basic implementation
proc ids n {
    while {$n != 1 && $n != 89} {
	for {set m 0} {$n} {set n [expr {$n / 10}]} {
	    incr m [expr {($n%10)**2}]
	}
	set n $m
    }
    return $n
}

# Build the optimised version
set body {
    # Microoptimisation to avoid an unnecessary alloc in the loop
    for {set m 0} {$n} {set n [expr {"$n[unset n]" / 10}]} {
	incr m [expr {($n%10)**2}]
    }
}
set map 0
for {set i 1} {$i <= 729} {incr i} {
    lappend map [ids $i]
}
proc ids2 n [append body "return \[lindex [list $map] \$m\]"]

# Put this in a lambda context for a little extra speed.
apply {{} {
    set count 0
    for {set i 1} {$i <= 100000000} {incr i} {
	incr count [expr {[ids2 $i] == 89}]
    }
    puts $count
}}
```



## VBScript


```vb

start_time = Now
cnt = 0
For i = 1 To 100000000
	n = i
	sum = 0
	Do Until n = 1 Or n = 89
		For j = 1 To Len(n)
			sum = sum + (CLng(Mid(n,j,1))^2)
		Next
		n = sum
		sum = 0
	Loop
	If n = 89 Then
		cnt = cnt + 1
	End If
Next
end_time = Now

WScript.Echo "Elapse Time: " & DateDiff("s",start_time,end_time) &_
			vbCrLf & "Count: " & cnt

```


{{Out}}
Elapse time is in seconds.  Friends don't let friends do this in VBScript. :-)

```txt

Elapse Time: 2559
Count: 85744333

```



## zkl

Using brute force is a never ending process so need to be clever, which takes under a second.
{{trans|Python}}
{{trans|D}}
{{trans|http://www.mathblog.dk/project-euler-92-square-digits-number-chain/}}

```zkl
fcn check(number){  // a list of digits: 13 is L(0,0,0,0,0,0,1,3)
   candidate:=number.reduce(fcn(sum,n){ sum*10 + n },0);  // digits to int

   while(candidate != 89 and candidate != 1)  // repeatedly sum squares of digits
      { candidate = candidate.split().reduce(fcn(sum,c){ sum + c*c },0); }

   if(candidate == 89){ // count permutations of these digits, they all sum to 89
      digitsCount:=List(0,0,0,0,0,0,0,0,0,0);
      foreach d in (number){ digitsCount[d] += 1; }
      return(digitsCount.reduce(fcn(r,c){ r/factorial(c) },cacheBang)); // cacheBang==number.len()!
   }
   0 // this number doesn't sum to 89 (ie sums to 1)
}
fcn factorial(n) { (1).reduce(n,fcn(N,n){ N*n },1) }

limit:=0d100_000_000;  cacheSize:=limit.toFloat().log10().ceil().toInt();
number:=(0).pump(cacheSize,List().write,0); // list of zeros
result:=0; i:=cacheSize - 1;
var cacheBang=factorial(cacheSize);  //== number.len()!

while(True){  // create numbers s.t. no set of digits is repeated
   if(i == 0 and number[i] == 9) break;
   if(i == cacheSize - 1 and number[i] < 9){ number[i] += 1; result += check(number); }
   else if(number[i] == 9) i -= 1;
   else{
      number[i] += 1;
      foreach j in ([i + 1 .. cacheSize - 1]){ number[j] = number[i]; }
      i = cacheSize - 1;
      result += check(number);
   }
}
println(result);
```

{{out}}

```txt
85744333
```



## ZX Spectrum Basic

{{trans|BBC_BASIC}}
Very, very slow. Use a ZX Spectrum emulator and run with maximum speed option enabled.

```zxbasic
10 LET n=0
20 FOR i=1 TO 1000
30 LET j=i
40 LET k=0
50 LET k=INT (k+FN m(j,10)^2)
60 LET j=INT (j/10)
70 IF j<>0 THEN GO TO 50
80 LET j=k
90 IF j=89 OR j=1 THEN GO TO 100
95 GO TO 40
100 IF j>1 THEN LET n=n+1
110 NEXT i
120 PRINT "Version 1: ";n
200 DEF FN m(a,b)=a-INT (a/b)*b: REM modulo
```

{{out}}

```txt
Version 1: 857
```