Sequence: nth number with exactly n divisors: Difference between revisions
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=={{header|Phix}}== |
=={{header|Phix}}== |
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{{libheader|mpfr}} |
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Certainly not the fastest way to do it, hence the relatively small limit of 24, which takes less than 0.4s,<br> |
Certainly not the fastest way to do it, hence the relatively small limit of 24, which takes less than 0.4s,<br> |
||
whereas a limit of 25 would need to invoke factors() 52 million times which would no doubt take a fair while. |
whereas a limit of 25 would need to invoke factors() 52 million times which would no doubt take a fair while. |
||
Revision as of 21:37, 2 June 2019
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. 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. <lang go>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
}
}
}
}
}
}</lang>
- 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
The following much faster version (runs in less than 90 seconds on my 1.6GHz Celeron) uses three further optimizations:
1. Apart from the 2nd and 10th terms, all the even terms are themselves even.
2. 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.
3. 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.
<lang go>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++
}
}
}
}
}</lang>
- Output:
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
<lang java>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;
}
}
}
}
}
}
}</lang>
- Output:
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
Kotlin
<lang scala>// 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++
}
}
}
}</lang>
- 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
<lang perl>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;
}
}
}</lang>
- Output:
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
<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;
$_
}
}
};</lang>
First 20 terms of OEIS:A073916 1 3 25 14 14641 44 24137569 70 1089 405 819628286980801 160 22563490300366186081 2752 9801 462 21559177407076402401757871041 1044 740195513856780056217081017732809 1520
Phix
Certainly not the fastest way to do it, hence the relatively small limit of 24, which takes less than 0.4s,
whereas a limit of 25 would need to invoke factors() 52 million times which would no doubt take a fair while.
<lang Phix>constant LIMIT = 24
include mpfr.e
mpz z = mpz_init()
sequence fn = 1&repeat(0,LIMIT-1),
primes = {2,3}
integer k = 1 printf(1,"The first %d terms in the sequence are:\n",LIMIT) for i=1 to LIMIT do
sequence f = factors(i,1)
if length(f)=2 then -- i is prime (f is {1,i})
while length(primes)<i do
integer p = primes[$]+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</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
REXX
Programming note: this REXX version has minor optimization, and all terms of the sequence are determined (found) in order.
little optimization
<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.*/ 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 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. */
/*──────────────────────────────────────────────────────────────────────────────────────*/ tell: parse arg _; say center(i, 10) right(_, max(w, length(_) ) )
if i//5==0 then say; return /*display a separator for the eyeballs.*/</lang>
- output when using the input: 45
─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. <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.*/ @.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 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 # . -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.*/</lang>
- output is identical to the 1st REXX version.
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]
zkl
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. <lang 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; } }
} }
}</lang>
- Output:
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