Sequence: nth number with exactly n divisors
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
- Related tasks
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.
The remaining terms (up to the 24th) are not particularly large and so are calculated by brute force. <lang go>package main
import (
"fmt" "math/big"
)
const max = 24
var primes = [max]int{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,
41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89}
func isSmallPrime(n int) bool {
for _, prime := range primes {
if prime == n {
return true
}
}
return false
}
func countDivisors(n int) int {
count := 0
for i := 1; i*i <= n; i++ {
if n%i == 0 {
if i == n/i {
count++
} else {
count += 2
}
}
}
return count
}
func main() {
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 isSmallPrime(i) {
z.SetInt64(int64(primes[i-1]))
p.SetInt64(int64(i - 1))
z.Exp(z, p, nil)
fmt.Printf("%2d : %d\n", i, z)
} else {
count := 0
for j := 1; ; j++ {
if countDivisors(j) == i {
count++
if count == i {
fmt.Printf("%2d : %d\n", i, j)
break
}
}
}
}
}
}</lang>
- Output:
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
Java
<lang java>import java.math.BigInteger;
public class OEIS_A073916 {
static int[] primes = new int[]{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89};
static boolean is_small_prime(int n) {
for (int prime : primes) {
if (prime == n) return true;
}
return false;
}
static int count_divisors(int n) {
int count = 0;
for (int i = 1; i * i <= n; ++i) {
if (n % i == 0) {
if (i == n / i)
count++;
else
count += 2;
}
}
return count;
}
public static void main(String[] args) {
final int max = 24;
System.out.printf("The first %d terms of the sequence are:\n", max);
for (int i = 1; i <= max; ++i) {
if (is_small_prime(i)) {
BigInteger z = BigInteger.valueOf(primes[i - 1]);
z = z.pow(i - 1);
System.out.printf("%2d : %d\n", i, z);
} else {
for (int j = 1, count = 0; ; ++j) {
if (count_divisors(j) == i) {
if (++count == i) {
System.out.printf("%2d : %d\n", i, j);
break;
}
}
}
}
}
}
}</lang>
- Output:
The first 24 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
Kotlin
<lang scala>// Version 1.3.21
import java.math.BigInteger
const val MAX = 24
var primes = intArrayOf(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37,
41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89)
fun isSmallPrime(n: Int) = primes.contains(n)
fun countDivisors(n: Int): Int {
var count = 0
var i = 1
while (i * i <= n) {
if (n % i == 0) {
count += if (i == n / i) 1 else 2
}
i++
}
return count
}
fun main() {
println("The first $MAX terms in the sequence are:")
for (i in 1..MAX) {
if (isSmallPrime(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 (countDivisors(j) == i) {
if (++count == i) {
System.out.printf("%2d : %d\n", i, j)
break
}
}
j++
}
}
}
}</lang>
- Output:
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
Perl 6
<lang perl6>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;
$_
}
}
};</lang>
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
REXX
Programming note: this REXX program automatically right justifies the output.
If the output is wider than 100 decimal digits (with commas), it is left justified.
<lang 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.*/
say '─divisors─' center("the Nth number with exactly N divisors", 100, '─') /*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; say center(i, 10) pPow(); iterate
end
#= 0 /*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); if d\==i then iterate /*get # divisors; Is not equal? Skip.*/
!.jj=. /*mark as having found #divs for this J*/
#= # + 1 /*bump number of occurrences for #div. */
if #\==i then iterate /*Not correct occurrence? Keep looking.*/
say center(i, 10) right( commas(jj), 100) /*display Nth number with #divs*/
if i//4==0 then say /*display a separator for eyes.*/
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 500; _= commas(@.i ** (i-1) ); return right(_, max(100, length(_))) /*──────────────────────────────────────────────────────────────────────────────────────*/
- 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 for x%2-3 by 1+odd while k<y /*for 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. */</lang>
- output when using the input: 33
─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
Sidef
<lang ruby>func f(n {.is_prime}) {
n.prime**(n-1)
}
func f(n) {
n.th { .sigma0 == n }
}
say 20.of { f(_+1) }</lang>
- Output:
[1, 3, 25, 14, 14641, 44, 24137569, 70, 1089, 405, 819628286980801, 160, 22563490300366186081, 2752, 9801, 462, 21559177407076402401757871041, 1044, 740195513856780056217081017732809, 1520]