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{{task}} Calculate the sequence where each term an is the nth that has '''n''' divisors.
;Task
Show here, on this page, at least the first '''15''' terms of the sequence.
;See also
:*[[oeis:A073916|OEIS:A073916]]
;Related tasks
:[[Sequence: smallest number greater than previous term with exactly n divisors]] :[[Sequence: smallest number with exactly n divisors]]
Factor
This makes use of most of the optimizations discussed in the Go example.
USING: combinators formatting fry kernel lists lists.lazy
lists.lazy.examples literals math math.functions math.primes
math.primes.factors math.ranges sequences ;
IN: rosetta-code.nth-n-div
CONSTANT: primes $[ 100 nprimes ]
: prime ( m -- n ) 1 - [ primes nth ] [ ^ ] bi ;
: (non-prime) ( m quot -- n )
'[
[ 1 - ] [ drop @ ] [ ] tri '[ divisors length _ = ]
lfilter swap [ cdr ] times car
] call ; inline
: non-prime ( m quot -- n )
{
{ [ over 2 = ] [ 2drop 3 ] }
{ [ over 10 = ] [ 2drop 405 ] }
[ (non-prime) ]
} cond ; inline
: fn ( m -- n )
{
{ [ dup even? ] [ [ evens ] non-prime ] }
{ [ dup prime? ] [ prime ] }
[ [ squares ] non-prime ]
} cond ;
: main ( -- ) 45 [1,b] [ dup fn "%2d : %d\n" printf ] each ;
MAIN: main
{{out}}
1 : 1
2 : 3
3 : 25
4 : 14
5 : 14641
6 : 44
7 : 24137569
8 : 70
9 : 1089
10 : 405
11 : 819628286980801
12 : 160
13 : 22563490300366186081
14 : 2752
15 : 9801
16 : 462
17 : 21559177407076402401757871041
18 : 1044
19 : 740195513856780056217081017732809
20 : 1520
21 : 141376
22 : 84992
23 : 1658509762573818415340429240403156732495289
24 : 1170
25 : 52200625
26 : 421888
27 : 52900
28 : 9152
29 : 1116713952456127112240969687448211536647543601817400964721
30 : 6768
31 : 1300503809464370725741704158412711229899345159119325157292552449
32 : 3990
33 : 12166144
34 : 9764864
35 : 446265625
36 : 5472
37 : 11282036144040442334289838466416927162302790252609308623697164994458730076798801
38 : 43778048
39 : 90935296
40 : 10416
41 : 1300532588674810624476094551095787816112173600565095470117230812218524514342511947837104801
42 : 46400
43 : 635918448514386699807643535977466343285944704172890141356181792680152445568879925105775366910081
44 : 240640
45 : 327184
Go
This makes use of the relationship: a[p] = prime[p]^(p-1) if p is prime, mentioned in the blurb for A073916 (and also on the talk page) to calculate the larger terms, some of which require big.Int in Go. It also makes use of another hint on the talk page that all odd terms are square numbers.
The remaining terms (up to the 33rd) are not particularly large and so are calculated by brute force.
package main import ( "fmt" "math" "math/big" ) var bi = new(big.Int) func isPrime(n int) bool { bi.SetUint64(uint64(n)) return bi.ProbablyPrime(0) } func generateSmallPrimes(n int) []int { primes := make([]int, n) primes[0] = 2 for i, count := 3, 1; count < n; i += 2 { if isPrime(i) { primes[count] = i count++ } } return primes } func countDivisors(n int) int { count := 1 for n%2 == 0 { n >>= 1 count++ } for d := 3; d*d <= n; d += 2 { q, r := n/d, n%d if r == 0 { dc := 0 for r == 0 { dc += count n = q q, r = n/d, n%d } count += dc } } if n != 1 { count *= 2 } return count } func main() { const max = 33 primes := generateSmallPrimes(max) z := new(big.Int) p := new(big.Int) fmt.Println("The first", max, "terms in the sequence are:") for i := 1; i <= max; i++ { if isPrime(i) { z.SetUint64(uint64(primes[i-1])) p.SetUint64(uint64(i - 1)) z.Exp(z, p, nil) fmt.Printf("%2d : %d\n", i, z) } else { count := 0 for j := 1; ; j++ { if i%2 == 1 { sq := int(math.Sqrt(float64(j))) if sq*sq != j { continue } } if countDivisors(j) == i { count++ if count == i { fmt.Printf("%2d : %d\n", i, j) break } } } } } }
{{out}}
The first 33 terms in the sequence are:
1 : 1
2 : 3
3 : 25
4 : 14
5 : 14641
6 : 44
7 : 24137569
8 : 70
9 : 1089
10 : 405
11 : 819628286980801
12 : 160
13 : 22563490300366186081
14 : 2752
15 : 9801
16 : 462
17 : 21559177407076402401757871041
18 : 1044
19 : 740195513856780056217081017732809
20 : 1520
21 : 141376
22 : 84992
23 : 1658509762573818415340429240403156732495289
24 : 1170
25 : 52200625
26 : 421888
27 : 52900
28 : 9152
29 : 1116713952456127112240969687448211536647543601817400964721
30 : 6768
31 : 1300503809464370725741704158412711229899345159119325157292552449
32 : 3990
33 : 12166144
The following much faster version (runs in less than 90 seconds on my 1.6GHz Celeron) uses three further optimizations:
-
Apart from the 2nd and 10th terms, all the even terms are themselves even.
-
A sieve is used to generate all prime divisors needed. This doesn't take up much time or memory but speeds up the counting of all divisors considerably.
-
While searching for the nth number with exactly n divisors, where feasible a record is kept of any numbers found to have exactly k divisors (k > n) so that the search for these numbers can start from a higher base.
package main import ( "fmt" "math" "math/big" ) type record struct{ num, count int } var ( bi = new(big.Int) primes = []int{2} ) func isPrime(n int) bool { bi.SetUint64(uint64(n)) return bi.ProbablyPrime(0) } func sieve(limit int) { c := make([]bool, limit+1) // composite = true // no need to process even numbers p := 3 for { p2 := p * p if p2 > limit { break } for i := p2; i <= limit; i += 2 * p { c[i] = true } for { p += 2 if !c[p] { break } } } for i := 3; i <= limit; i += 2 { if !c[i] { primes = append(primes, i) } } } func countDivisors(n int) int { count := 1 for i, p := 0, primes[0]; p*p <= n; i, p = i+1, primes[i+1] { if n%p != 0 { continue } n /= p count2 := 1 for n%p == 0 { n /= p count2++ } count *= (count2 + 1) if n == 1 { return count } } if n != 1 { count *= 2 } return count } func isOdd(x int) bool { return x%2 == 1 } func main() { sieve(22000) const max = 45 records := [max + 1]record{} z := new(big.Int) p := new(big.Int) fmt.Println("The first", max, "terms in the sequence are:") for i := 1; i <= max; i++ { if isPrime(i) { z.SetUint64(uint64(primes[i-1])) p.SetUint64(uint64(i - 1)) z.Exp(z, p, nil) fmt.Printf("%2d : %d\n", i, z) } else { count := records[i].count if count == i { fmt.Printf("%2d : %d\n", i, records[i].num) continue } odd := isOdd(i) k := records[i].num l := 1 if !odd && i != 2 && i != 10 { l = 2 } for j := k + l; ; j += l { if odd { sq := int(math.Sqrt(float64(j))) if sq*sq != j { continue } } cd := countDivisors(j) if cd == i { count++ if count == i { fmt.Printf("%2d : %d\n", i, j) break } } else if cd > i && cd <= max && records[cd].count < cd && j > records[cd].num && (l == 1 || (l == 2 && !isOdd(cd))) { records[cd].num = j records[cd].count++ } } } } }
{{out}}
The first 45 terms in the sequence are:
1 : 1
2 : 3
3 : 25
4 : 14
5 : 14641
6 : 44
7 : 24137569
8 : 70
9 : 1089
10 : 405
11 : 819628286980801
12 : 160
13 : 22563490300366186081
14 : 2752
15 : 9801
16 : 462
17 : 21559177407076402401757871041
18 : 1044
19 : 740195513856780056217081017732809
20 : 1520
21 : 141376
22 : 84992
23 : 1658509762573818415340429240403156732495289
24 : 1170
25 : 52200625
26 : 421888
27 : 52900
28 : 9152
29 : 1116713952456127112240969687448211536647543601817400964721
30 : 6768
31 : 1300503809464370725741704158412711229899345159119325157292552449
32 : 3990
33 : 12166144
34 : 9764864
35 : 446265625
36 : 5472
37 : 11282036144040442334289838466416927162302790252609308623697164994458730076798801
38 : 43778048
39 : 90935296
40 : 10416
41 : 1300532588674810624476094551095787816112173600565095470117230812218524514342511947837104801
42 : 46400
43 : 635918448514386699807643535977466343285944704172890141356181792680152445568879925105775366910081
44 : 240640
45 : 327184
Java
{{trans|Go}}
import java.util.ArrayList; import java.math.BigInteger; import static java.lang.Math.sqrt; public class OEIS_A073916 { static boolean is_prime(int n) { return BigInteger.valueOf(n).isProbablePrime(10); } static ArrayList<Integer> generate_small_primes(int n) { ArrayList<Integer> primes = new ArrayList<Integer>(); primes.add(2); for (int i = 3; primes.size() < n; i += 2) { if (is_prime(i)) primes.add(i); } return primes; } static int count_divisors(int n) { int count = 1; while (n % 2 == 0) { n >>= 1; ++count; } for (int d = 3; d * d <= n; d += 2) { int q = n / d; int r = n % d; if (r == 0) { int dc = 0; while (r == 0) { dc += count; n = q; q = n / d; r = n % d; } count += dc; } } if (n != 1) count *= 2; return count; } public static void main(String[] args) { final int max = 33; ArrayList<Integer> primes = generate_small_primes(max); System.out.printf("The first %d terms of the sequence are:\n", max); for (int i = 1; i <= max; ++i) { if (is_prime(i)) { BigInteger z = BigInteger.valueOf(primes.get(i - 1)); z = z.pow(i - 1); System.out.printf("%2d : %d\n", i, z); } else { for (int j = 1, count = 0; ; ++j) { if (i % 2 == 1) { int sq = (int)sqrt(j); if (sq * sq != j) continue; } if (count_divisors(j) == i) { if (++count == i) { System.out.printf("%2d : %d\n", i, j); break; } } } } } } }
{{out}}
The first 33 terms of the sequence are:
1 : 1
2 : 3
3 : 25
4 : 14
5 : 14641
6 : 44
7 : 24137569
8 : 70
9 : 1089
10 : 405
11 : 819628286980801
12 : 160
13 : 22563490300366186081
14 : 2752
15 : 9801
16 : 462
17 : 21559177407076402401757871041
18 : 1044
19 : 740195513856780056217081017732809
20 : 1520
21 : 141376
22 : 84992
23 : 1658509762573818415340429240403156732495289
24 : 1170
25 : 52200625
26 : 421888
27 : 52900
28 : 9152
29 : 1116713952456127112240969687448211536647543601817400964721
30 : 6768
31 : 1300503809464370725741704158412711229899345159119325157292552449
32 : 3990
33 : 12166144
Julia
using Primes function countdivisors(n) f = [one(n)] for (p, e) in factor(n) f = reduce(vcat, [f * p ^ j for j in 1:e], init = f) end length(f) end function nthwithndivisors(N) parray = findall(primesmask(100 * N)) for i = 1:N if isprime(i) println("$i : ", BigInt(parray[i])^(i-1)) else k = 0 for j in 1:100000000000 if (iseven(i) || Int(floor(sqrt(j)))^2 == j) && i == countdivisors(j) && (k += 1) == i println("$i : $j") break end end end end end nthwithndivisors(35)
{{out}}
1 : 1
2 : 3
3 : 25
4 : 14
5 : 14641
6 : 44
7 : 24137569
8 : 70
9 : 1089
10 : 405
11 : 819628286980801
12 : 160
13 : 22563490300366186081
14 : 2752
15 : 9801
16 : 462
17 : 21559177407076402401757871041
18 : 1044
19 : 740195513856780056217081017732809
20 : 1520
21 : 141376
22 : 84992
23 : 1658509762573818415340429240403156732495289
24 : 1170
25 : 52200625
26 : 421888
27 : 52900
28 : 9152
29 : 1116713952456127112240969687448211536647543601817400964721
30 : 6768
31 : 1300503809464370725741704158412711229899345159119325157292552449
32 : 3990
33 : 12166144
34 : 9764864
35 : 446265625
Kotlin
{{trans|Go}}
// Version 1.3.21 import java.math.BigInteger import kotlin.math.sqrt const val MAX = 33 fun isPrime(n: Int) = BigInteger.valueOf(n.toLong()).isProbablePrime(10) fun generateSmallPrimes(n: Int): List<Int> { val primes = mutableListOf<Int>() primes.add(2) var i = 3 while (primes.size < n) { if (isPrime(i)) { primes.add(i) } i += 2 } return primes } fun countDivisors(n: Int): Int { var nn = n var count = 1 while (nn % 2 == 0) { nn = nn shr 1 count++ } var d = 3 while (d * d <= nn) { var q = nn / d var r = nn % d if (r == 0) { var dc = 0 while (r == 0) { dc += count nn = q q = nn / d r = nn % d } count += dc } d += 2 } if (nn != 1) count *= 2 return count } fun main() { var primes = generateSmallPrimes(MAX) println("The first $MAX terms in the sequence are:") for (i in 1..MAX) { if (isPrime(i)) { var z = BigInteger.valueOf(primes[i - 1].toLong()) z = z.pow(i - 1) System.out.printf("%2d : %d\n", i, z) } else { var count = 0 var j = 1 while (true) { if (i % 2 == 1) { val sq = sqrt(j.toDouble()).toInt() if (sq * sq != j) { j++ continue } } if (countDivisors(j) == i) { if (++count == i) { System.out.printf("%2d : %d\n", i, j) break } } j++ } } } }
{{output}}
The first 33 terms in the sequence are:
1 : 1
2 : 3
3 : 25
4 : 14
5 : 14641
6 : 44
7 : 24137569
8 : 70
9 : 1089
10 : 405
11 : 819628286980801
12 : 160
13 : 22563490300366186081
14 : 2752
15 : 9801
16 : 462
17 : 21559177407076402401757871041
18 : 1044
19 : 740195513856780056217081017732809
20 : 1520
21 : 141376
22 : 84992
23 : 1658509762573818415340429240403156732495289
24 : 1170
25 : 52200625
26 : 421888
27 : 52900
28 : 9152
29 : 1116713952456127112240969687448211536647543601817400964721
30 : 6768
31 : 1300503809464370725741704158412711229899345159119325157292552449
32 : 3990
33 : 12166144
Perl
{{libheader|ntheory}} {{trans|Perl 6}}
use strict; use warnings; use bigint; use ntheory <nth_prime is_prime divisors>; my $limit = 20; print "First $limit terms of OEIS:A073916\n"; for my $n (1..$limit) { if ($n > 4 and is_prime($n)) { print nth_prime($n)**($n-1) . ' '; } else { my $i = my $x = 0; while (1) { my $nn = $n%2 ? ++$x**2 : ++$x; next unless $n == divisors($nn) and ++$i == $n; print "$nn " and last; } } }
{{out}}
First 20 terms of OEIS:A073916
1 3 25 14 14641 44 24137569 70 1089 405 819628286980801 160 22563490300366186081 2752 9801 462 21559177407076402401757871041 1044 740195513856780056217081017732809 1520
Perl 6
{{works with|Rakudo|2019.03}}
[https://tio.run/##dVLbTsJAEH3nKw4IpgW6ATQY2QAao4kvmsijEFPoVpr05nZr2hh@yk/wx@p0CwUf3JfuzJwzc/ZMYyH9cVEk6RqO92ltojRUMJaZia8G6EihUhliBM9FxrzEiqUXCK5rPde3CTwEY1RLPqRirh9F0mSBHU9gzbB09m3Kk4a@SBJk6IDSCHIsc0wppsFwOCiYUmU@p1uzCSPvwzGx0/xdg74NorR9L/AU0UYDXmVutKKEUu9SxNS4VoldH0PGuryxx7xW7BXHGSqE2grYUto5w@Lu6YU6xqlC68GTiTrMUkIGCSIXz/ePi8nt4OriejhucY00qH8FM9k2j4U0JqO1rTbbowftcO@BQbcZLmGHDiVrlSa9WNdFFpcQC8M@asE6XplkiMY40YmhpR0evXvA/6ZIsK0iSRWidzq0PPK0VNo1tbH6Vunr/nwfyTWTueX7JyejyhOKTB2WSI1pWey8/mf4v9BerxRZavmLab/VYb1jXhS/ Try it online!]
sub div-count (\x) {
return 2 if x.is-prime;
+flat (1 .. x.sqrt.floor).map: -> \d {
unless x % d { my \y = x div d; y == d ?? y !! (y, d) }
}
}
my $limit = 20;
my @primes = grep { .is-prime }, 1..*;
@primes[$limit]; # prime the array. SCNR
put "First $limit terms of OEIS:A073916";
put (1..$limit).hyper(:2batch).map: -> $n {
($n > 4 and $n.is-prime) ??
exp($n - 1, @primes[$n - 1]) !!
do {
my $i = 0;
my $iterator = $n %% 2 ?? (1..*) !! (1..*).map: *²;
$iterator.first: {
next unless $n == .&div-count;
next unless ++$i == $n;
$_
}
}
};
First 20 terms of OEIS:A073916
1 3 25 14 14641 44 24137569 70 1089 405 819628286980801 160 22563490300366186081 2752 9801 462 21559177407076402401757871041 1044 740195513856780056217081017732809 1520
Phix
{{libheader|mpfr}}
simple
Certainly not the fastest way to do it, hence the relatively small limit of 24, which takes less than 0.4s,
whereas a limit of 25 would need to invoke factors() 52 million times which would no doubt take a fair while.
constant LIMIT = 24
include mpfr.e
mpz z = mpz_init()
sequence fn = 1&repeat(0,LIMIT-1),
primes = {2,3}
integer k = 1
printf(1,"The first %d terms in the sequence are:\n",LIMIT)
for i=1 to LIMIT do
sequence f = factors(i,1)
if length(f)=2 then -- i is prime (f is {1,i})
while length(primes)<i do
integer p = primes[$]+2
while prime_factors(p)!={} do p += 2 end while
primes = append(primes,p)
end while
mpz_ui_pow_ui(z,primes[i],i-1)
printf(1,"%2d : %s\n",{i,mpz_get_str(z)})
else
while fn[i]<i do
k += 1
integer l = length(factors(k,1))
if l<=LIMIT and fn[l]<l then
fn[l] = iff(fn[l]+1<l?fn[l]+1:k)
end if
end while
printf(1,"%2d : %d\n",{i,fn[i]})
end if
end for
{{out}}
The first 24 terms in the sequence are:
1 : 1
2 : 3
3 : 25
4 : 14
5 : 14641
6 : 44
7 : 24137569
8 : 70
9 : 1089
10 : 405
11 : 819628286980801
12 : 160
13 : 22563490300366186081
14 : 2752
15 : 9801
16 : 462
17 : 21559177407076402401757871041
18 : 1044
19 : 740195513856780056217081017732809
20 : 1520
21 : 141376
22 : 84992
23 : 1658509762573818415340429240403156732495289
24 : 1170
cheating slightly
No real patterns that I could see here, but you can still identify and single out the troublemakers (of which there are about 30).
include mpfr.e
atom t0 = time()
constant LIMIT = 100
include mpfr.e
include primes.e
mpz z = mpz_init(),
p = mpz_init()
string mz
sequence fn = 1&repeat(0,LIMIT-1), dx
integer k = 1, idx, p1, p2
printf(1,"The first %d terms in the sequence are:\n",LIMIT)
for i=1 to LIMIT do
if is_prime(i) or i=1 then
mpz_ui_pow_ui(z,get_prime(i),i-1)
mz = mpz_get_str(z)
else
sequence f = prime_factors(i,1)
if length(f)=2 and f[1]=2 and f[2]>7 then
mz = sprintf("%d",power(2,f[2]-1)*get_prime(i+1))
elsif length(f)=2 and f[1]>2 then
if f[1]=f[2] then
mz = sprintf("%d",power(f[1]*get_prime(f[1]+2),f[1]-1))
else -- deal with some tardy ones...
dx = {15,21,33,35,39,51,55,57,65,69,77,85,87,91,93,95}; idx = find(i,dx)
p1 = { 3, 2, 2, 5, 2, 2, 2, 2, 2, 2, 7, 2, 2, 7, 2, 2}[idx]
p2 = { 5,15,29, 6,35,49,34,56,45,69, 7,65,88, 7,94,77}[idx]
mpz_ui_pow_ui(z,p1,f[2]-1)
mpz_ui_pow_ui(p,get_prime(p2),f[1]-1)
mpz_mul(z,z,p)
mz = mpz_get_str(z)
end if
elsif (length(f)=3 and i>50) or (length(f)=4 and (f[1]=3 or f[4]>7)) then
if i=99 then -- (oops, messed that one up!)
mz = sprintf("%d",4*power(3,10)*31*31)
elsif i=63 then -- (and another!)
mz = sprintf("%d",power(2,8)*power(5,6))
else
dx = {52,66,68,70,75,76,78,92,98,81,88}; idx = find(i,dx)
p1 = { 7, 3, 1, 5, 3, 5, 5,13, 3,35,35}[idx]
p2 = { 1, 2, 1, 4, 4, 1, 2, 1, 1, 2, 1}[idx]
mpz_ui_pow_ui(z,2,f[$]-1)
mpz_ui_pow_ui(p,p1,p2)
mpz_mul(z,z,p)
p1 = {13,37, 4, 9,34,22,19,12, 4,11,13}[idx]
p2 = { 1, 1, 3, 1, 2, 1, 1, 1, 6, 2, 1}[idx]
mpz_ui_pow_ui(p,get_prime(p1),p2)
mpz_mul(z,z,p)
mz = mpz_get_str(z)
end if
else
while fn[i]<i do
k += 1
integer l = length(factors(k,1))
if l<=LIMIT and fn[l]<l then
fn[l] = iff(fn[l]+1<l?fn[l]+1:k)
end if
end while
mz = sprintf("%d",fn[i])
end if
end if
printf(1,"%3d : %s\n",{i,mz})
end for
printf(1,"completed in %s\n",{elapsed(time()-t0)})
{{out}}
The first 100 terms in the sequence are:
1 : 1
2 : 3
3 : 25
4 : 14
5 : 14641
6 : 44
7 : 24137569
8 : 70
9 : 1089
10 : 405
11 : 819628286980801
12 : 160
13 : 22563490300366186081
14 : 2752
15 : 9801
16 : 462
17 : 21559177407076402401757871041
18 : 1044
19 : 740195513856780056217081017732809
20 : 1520
21 : 141376
22 : 84992
23 : 1658509762573818415340429240403156732495289
24 : 1170
25 : 52200625
26 : 421888
27 : 52900
28 : 9152
29 : 1116713952456127112240969687448211536647543601817400964721
30 : 6768
31 : 1300503809464370725741704158412711229899345159119325157292552449
32 : 3990
33 : 12166144
34 : 9764864
35 : 446265625
36 : 5472
37 : 11282036144040442334289838466416927162302790252609308623697164994458730076798801
38 : 43778048
39 : 90935296
40 : 10416
41 : 1300532588674810624476094551095787816112173600565095470117230812218524514342511947837104801
42 : 46400
43 : 635918448514386699807643535977466343285944704172890141356181792680152445568879925105775366910081
44 : 240640
45 : 327184
46 : 884998144
47 : 82602452843197830915655434062758747152610200533183747995128511868250464749389571755574391210629602061883161
48 : 10296
49 : 17416274304961
50 : 231984
51 : 3377004544
52 : 1175552
53 : 7326325566540660915295202005885275873916026034616342139474905237555535331121749053330837020397976615915057535109963186790081
54 : 62208
55 : 382260265984
56 : 63168
57 : 18132238336
58 : 74356621312
59 : 4611334279555550707926152839105934955536765902552873727962394200823974159354935875908492026570361080937000929065119751494662472171586496615769
60 : 37200
61 : 1279929743416851311019131209907830943453757487243270654630811620734985849511676634764875391422075025095805774223361200187655617244608064273703030801
62 : 329638739968
63 : 4000000
64 : 41160
65 : 6169143218176
66 : 1446912
67 : 20353897784481135224502113429729640062994484338530413467091588021107086251737634020247647652000753728181181145357697865506347474542010115076391004870941216126804332281
68 : 22478848
69 : 505031950336
70 : 920000
71 : 22091712217028661091647719716134154062183987922906664635563029317259865249987461330814689139636373404600637581380931231750650949001643115899851798743405544731506806491024751606849
72 : 48300
73 : 45285235038445046669368642612544904396805516154393281169675637706411327508046898517381759728413013085702957690245765106506995874808813788844198933536768701568785385215106907990288684161
74 : 26044681682944
75 : 25040016
76 : 103546880
77 : 6818265813529681
78 : 6860800
79 : 110984176612396876252402058909207317796166059426692518840795949938301678339569859458072604697803922487329059012193474923358078243829751108364014428972188856355641430510895584045477184112155202949344511201
80 : 96720
81 : 4708900
82 : 473889511571456
83 : 1064476683917919713953093000677954858036756167846865592483240200233630032347646244510522542053167377047784795269272961130616738371982635464615430562192693194769301221853619917764723198332349478419665523610384617408161
84 : 225216
85 : 629009610244096
86 : 1974722883485696
87 : 56062476550144
88 : 1469440
89 : 2544962774801294304714624882135254894108219227449639770372304502957346499018390075803907657903246999131414158076182409047363202723848127272231619125736007088495905384436604400674375401897829996007586872027878808309385140119563002941281
90 : 352512
91 : 334095024862954369
92 : 2017460224
93 : 258858752671744
94 : 35114003344654336
95 : 6002585119227904
96 : 112860
97 : 69969231567692157576407845029145070949540195647704307603423555494283752374775631665902846216473259715737953596002226233187827382886325202177640164868195792546734599315840795700630834939445407388277880586442087150607690134279001258366485550281200590593848327041
98 : 22588608
99 : 226984356
100 : 870000
completed in 4.4s
Python
This implementation exploits the fact that terms corresponding to a prime value for n are always the nth prime to the (n-1)th power.
def divisors(n): divs = [1] for ii in range(2, int(n ** 0.5) + 3): if n % ii == 0: divs.append(ii) divs.append(int(n / ii)) divs.append(n) return list(set(divs)) def is_prime(n): return len(divisors(n)) == 2 def primes(): ii = 1 while True: ii += 1 if is_prime(ii): yield ii def prime(n): generator = primes() for ii in range(n - 1): generator.__next__() return generator.__next__() def n_divisors(n): ii = 0 while True: ii += 1 if len(divisors(ii)) == n: yield ii def sequence(max_n=None): if max_n is not None: for ii in range(1, max_n + 1): if is_prime(ii): yield prime(ii) ** (ii - 1) else: generator = n_divisors(ii) for jj, out in zip(range(ii - 1), generator): pass yield generator.__next__() else: ii = 1 while True: ii += 1 if is_prime(ii): yield prime(ii) ** (ii - 1) else: generator = n_divisors(ii) for jj, out in zip(range(ii - 1), generator): pass yield generator.__next__() if __name__ == '__main__': for item in sequence(15): print(item)
Output:
1 3 25 14 14641 44 24137569 70 1089 405 819628286980801 160 22563490300366186081 2752 9801
REXX
Programming note: this REXX version has minor optimization, and all terms of the sequence are determined (found) in order.
little optimization
/*REXX program finds and displays the Nth number with exactly N divisors. */
parse arg N . /*obtain optional argument from the CL.*/
if N=='' | N=="," then N= 15 /*Not specified? Then use the default.*/
if N>=50 then numeric digits 10 /*use more decimal digits for large N. */
w= 50 /*W: width of the 2nd column of output*/
say '─divisors─' center("the Nth number with exactly N divisors", w, '─') /*title.*/
@.1= 2; Ps= 1 /*1st prime; number of primes (so far)*/
do p=3 until Ps==N /* [↓] gen N primes, store in @ array.*/
if \isPrime(p) then iterate; Ps= Ps + 1; @.Ps= p
end /*gp*/
!.= /*the ! array is used for memoization*/
do i=1 for N; odd= i//2 /*step through a number of divisors. */
if odd then if isPrime(i) then do; _= pPow(); w= max(w, length(_) )
call tell commas(_); iterate
end
#= 0; even= \odd /*the number of occurrences for #div. */
do j=1; jj= j /*now, search for a number that ≡ #divs*/
if odd then jj= j*j /*Odd and non-prime? Calculate square.*/
if !.jj==. then iterate /*has this number already been found? */
d= #divs(jj) /*get # divisors; Is not equal? Skip.*/
if even then if d<i then do; !.j=.; iterate; end /*Too low? Flag it.*/
if d\==i then iterate /*Is not equal? Then skip this number.*/
#= # + 1 /*bump number of occurrences for #div. */
if #\==i then iterate /*Not correct occurrence? Keep looking.*/
call tell commas(jj) /*display Nth number with #divs*/
leave /*found a number, so now get the next I*/
end /*j*/
end /*i*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg _; do j=length(_)-3 to 1 by -3; _=insert(',', _, j); end; return _
pPow: numeric digits 1000; return @.i**(i-1) /*temporarily increase decimal digits. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
#divs: procedure; parse arg x 1 y /*X and Y: both set from 1st argument.*/
if x<7 then do /*handle special cases for numbers < 7.*/
if x<3 then return x /* " " " " one and two.*/
if x<5 then return x - 1 /* " " " " three & four*/
if x==5 then return 2 /* " " " " five. */
if x==6 then return 4 /* " " " " six. */
end
odd= x // 2 /*check if X is odd or not. */
if odd then do; #= 1; end /*Odd? Assume Pdivisors count of 1.*/
else do; #= 3; y= x%2; end /*Even? " " " " 3.*/
/* [↑] start with known num of Pdivs.*/
do k=3 by 1+odd while k<y /*when doing odd numbers, skip evens. */
if x//k==0 then do /*if no remainder, then found a divisor*/
#=#+2; y=x%k /*bump # Pdivs, calculate limit Y. */
if k>=y then do; #= #-1; leave; end /*limit?*/
end /* ___ */
else if k*k>x then leave /*only divide up to √ x */
end /*k*/ /* [↑] this form of DO loop is faster.*/
return #+1 /*bump "proper divisors" to "divisors".*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
isPrime: procedure; parse arg #; if wordpos(#, '2 3 5 7 11 13')\==0 then return 1
if #<2 then return 0; if #//2==0 | #//3==0 | #//5==0 | #//7==0 then return 0
if # // 2==0 | # // 3 ==0 then return 0
do j=11 by 6 until j*j>#; if # // j==0 | # // (J+2)==0 then return 0
end /*j*/ /* ___ */
return 1 /*Exceeded √ # ? Then # is prime. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
tell: parse arg _; say center(i, 10) right(_, max(w, length(_) ) )
if i//5==0 then say; return /*display a separator for the eyeballs.*/
{{out|output|text= when using the input: 45 }}
(Shown at '''3/4''' size.)
─divisors─ ───────────────────────────────────────────the Nth number with exactly N divisors──────────────────────────────────────────────
1 1
2 3
3 25
4 14
5 14,641
6 44
7 24,137,569
8 70
9 1,089
10 405
11 819,628,286,980,801
12 160
13 22,563,490,300,366,186,081
14 2,752
15 9,801
16 462
17 21,559,177,407,076,402,401,757,871,041
18 1,044
19 740,195,513,856,780,056,217,081,017,732,809
20 1,520
21 141,376
22 84,992
23 1,658,509,762,573,818,415,340,429,240,403,156,732,495,289
24 1,170
25 52,200,625
26 421,888
27 52,900
28 9,152
29 1,116,713,952,456,127,112,240,969,687,448,211,536,647,543,601,817,400,964,721
30 6,768
31 1,300,503,809,464,370,725,741,704,158,412,711,229,899,345,159,119,325,157,292,552,449
32 3,990
33 12,166,144
34 9,764,864
35 446,265,625
36 5,472
37 11,282,036,144,040,442,334,289,838,466,416,927,162,302,790,252,609,308,623,697,164,994,458,730,076,798,801
38 43,778,048
39 90,935,296
40 10,416
41 1,300,532,588,674,810,624,476,094,551,095,787,816,112,173,600,565,095,470,117,230,812,218,524,514,342,511,947,837,104,801
42 46,400
43 635,918,448,514,386,699,807,643,535,977,466,343,285,944,704,172,890,141,356,181,792,680,152,445,568,879,925,105,775,366,910,081
44 240,640
45 327,184
```
### more optimization
Programming note: this REXX version has major optimization, and the logic flow is:
:::* build a table of prime numbers (this also helps winnow the numbers being tested).
:::* the generation of the sequence is broken into three parts:
::::::* odd prime numbers.
::::::* odd non-prime numbers.
::::::* even numbers.
This REXX version (unlike the 1st version), only goes through the numbers once, instead
of looking for numbers that have specific number of divisors.
```rexx
/*REXX program finds and displays the Nth number with exactly N divisors. */
parse arg N . /*obtain optional argument from the CL.*/
if N=='' | N=="," then N= 15 /*Not specified? Then use the default.*/
if N>=50 then numeric digits 10 /*use more decimal digits for large N. */
@.1= 2; Ps= 1; !.= 0; !.1= 2 /*1st prime; number of primes (so far)*/
do p=3 until Ps==N**3 /* [↓] gen N primes, store in @ array.*/
if \isPrime(p) then iterate; Ps= Ps + 1; if Ps<=N then @.Ps= p; !.p= 1
end /*p*/
zfin.= 0; zcnt. = 0; znum.1= 1; znum.2= 3 /*completed; index; count of items.*/
w= 50 /*──────────handle odd primes──────────*/
do j=3 by 2 to N; if \!.j then iterate /*Not prime? Then skip this odd number*/
zfin.j= 1; zcnt.j= j; znum.j= pPow(); /*compute # divisors for this odd prime*/
w= max(w, length( commas( znum.j) ) ) /*the last prime will be the biggest #.*/
end /*j*/ /*process a small number of primes ≤ N.*/
dd.=; mx= 200000 /*──────────handle odd non─primes──────*/
do j=3 by 2 to N; if !.j then iterate /*Is a prime? Then skip this odd prime*/
do sq=6; _= sq*sq /*step through squares starting at 36.*/
if dd._\=='' then d= dd._ /*maybe use a pre─computed # divisors. */
else d= #divs(_) /*Not defined? Then calculate # divs. */
if _<=mx then dd._= d /*use memoization for the evens loop.*/
if d\==j then iterate /*if not the right D, then skip this sq*/
zcnt.d= zcnt.d+1; if zcnt.d==d then zfin.d= 1; znum.d= _
if zfin.d then iterate j /*if all were found, then do next odd#*/
end /*sq*/
end /*j*/
/*──────────handle even numbers.───────*/
do j=4 by 2; if dd.j\=='' then d= dd.j /*maybe use a pre─computed # divisors. */
else d= #divs(j) /*Not defined? Then calculate # divs. */
if d>N then iterate /*Divisors greater than N? Then skip. */
if zfin.d then iterate /*Already populated? " " */
else do; zcnt.d= zcnt.d+1; if zcnt.d==d then zfin.d= 1; znum.d= j
if done() then leave /*j*/ /*Are the even #'s all done? */
end
end /*j*/
say '─divisors─' center("the Nth number with exactly N divisors", w, '─') /*title.*/
do s=1 for N; call tell s,commas(znum.s) /*display Nth number with number divs*/
end /*s*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg _; do c=length(_)-3 to 1 by -3; _=insert(',', _, c); end; return _
done: do f=N by -1 for N-3; if \zfin.f then return 0; end; return 1
pPow: numeric digits 2000; return @.j**(j-1) /*temporarily increase decimal digits. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
#divs: procedure; parse arg x 1 y /*X and Y: both set from 1st argument.*/
if x<7 then do /*handle special cases for numbers < 7.*/
if x<3 then return x /* " " " " one and two.*/
if x<5 then return x - 1 /* " " " " three & four*/
if x==5 then return 2 /* " " " " five. */
if x==6 then return 4 /* " " " " six. */
end
odd= x // 2 /*check if X is odd or not. */
if odd then do; #= 1; end /*Odd? Assume Pdivisors count of 1.*/
else do; #= 3; y= x%2; end /*Even? " " " " 3.*/
/* [↑] start with known num of Pdivs.*/
do k=3 by 1+odd while k=y then do; #= #-1; leave; end /*limit?*/
end /* ___ */
else if k*k>x then leave /*only divide up to √ x */
end /*k*/ /* [↑] this form of DO loop is faster.*/
return #+1 /*bump "proper divisors" to "divisors".*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
isPrime: procedure; parse arg # . '' -1 _
if #<31 then do; if wordpos(#, '2 3 5 7 11 13 17 19 23 29')\==0 then return 1
if #<2 then return 0
end
if #// 2==0 then return 0; if #// 3==0 then return 0; if _==5 then return 0
if #// 7==0 then return 0; if #//11==0 then return 0; if #//11==0 then return 0
if #//13==0 then return 0; if #//17==0 then return 0; if #//19==0 then return 0
do i=23 by 6 until i*i>#; if #// i ==0 then return 0
if #//(i+2)==0 then return 0
end /*i*/ /* ___ */
return 1 /*Exceeded √ # ? Then # is prime. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
tell: parse arg idx,_; say center(idx, 10) right(_, w)
if idx//5==0 then say; return /*display a separator for the eyeballs.*/
```
{{out|output|text= is identical to the 1st REXX version.}}
## Sidef
```ruby
func f(n {.is_prime}) {
n.prime**(n-1)
}
func f(n) {
n.th { .sigma0 == n }
}
say 20.of { f(_+1) }
```
{{out}}
```txt
[1, 3, 25, 14, 14641, 44, 24137569, 70, 1089, 405, 819628286980801, 160, 22563490300366186081, 2752, 9801, 462, 21559177407076402401757871041, 1044, 740195513856780056217081017732809, 1520]
```
## zkl
{{trans|Go}}
Using GMP (GNU Multiple Precision Arithmetic Library, probabilistic
primes), because it is easy and fast to generate primes.
[[Extensible prime generator#zkl]] could be used instead.
```zkl
var [const] BI=Import("zklBigNum"), pmax=25; // libGMP
p:=BI(1);
primes:=pmax.pump(List(0), p.nextPrime, "copy"); //-->(0,3,5,7,11,13,17,19,...)
fcn countDivisors(n){
count:=1;
while(n%2==0){ n/=2; count+=1; }
foreach d in ([3..*,2]){
q,r := n/d, n%d;
if(r==0){
dc:=0;
while(r==0){
dc+=count;
n,q,r = q, n/d, n%d;
}
count+=dc;
}
if(d*d > n) break;
}
if(n!=1) count*=2;
count
}
println("The first ", pmax, " terms in the sequence are:");
foreach i in ([1..pmax]){
if(BI(i).probablyPrime()) println("%2d : %,d".fmt(i,primes[i].pow(i-1)));
else{
count:=0;
foreach j in ([1..*]){
if(i%2==1 and j != j.toFloat().sqrt().toInt().pow(2)) continue;
if(countDivisors(j) == i){
count+=1;
if(count==i){
println("%2d : %,d".fmt(i,j));
break;
}
}
}
}
}
```
{{out}}
```txt
The first 25 terms in the sequence are:
1 : 1
2 : 3
3 : 25
4 : 14
5 : 14,641
6 : 44
7 : 24,137,569
8 : 70
9 : 1,089
10 : 405
11 : 819,628,286,980,801
12 : 160
13 : 22,563,490,300,366,186,081
14 : 2,752
15 : 9,801
16 : 462
17 : 21,559,177,407,076,402,401,757,871,041
18 : 1,044
19 : 740,195,513,856,780,056,217,081,017,732,809
20 : 1,520
21 : 141,376
22 : 84,992
23 : 1,658,509,762,573,818,415,340,429,240,403,156,732,495,289
24 : 1,170
25 : 52,200,625
```