Sequence: nth number with exactly n divisors
You are encouraged to solve this task according to the task description, using any language you may know.
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
Factor
This makes use of most of the optimizations discussed in the Go example. <lang factor>USING: combinators formatting fry kernel lists lists.lazy lists.lazy.examples math math.functions math.primes math.primes.factors math.ranges sequences ; IN: rosetta-code.nth-n-div
- prime ( m -- n ) 1 - [ nprimes last ] [ ^ ] 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</lang>
- Output:
1 : 1 2 : 3 3 : 9 4 : 14 5 : 2401 6 : 44 7 : 4826809 8 : 70 9 : 1089 10 : 405 11 : 420707233300201 12 : 160 13 : 6582952005840035281 14 : 2752 15 : 9801 16 : 462 17 : 3876269050118516845397872321 18 : 1044 19 : 136753052840548005895349735207881 20 : 1520 21 : 141376 22 : 84992 23 : 559494740587480879172162808385362976196641 24 : 1170 25 : 52200625 26 : 421888 27 : 52900 28 : 9152 29 : 664883836375406516433586250657527435394385613490936917201 30 : 6768 31 : 39115897957341730208009052194412390955771656152457329084502049 32 : 3990 33 : 12166144 34 : 9764864 35 : 446265625 36 : 5472 37 : 2774406559237649514373597245387098526607420602299133512933658553632174767930401 38 : 43778048 39 : 90935296 40 : 10416 41 : 332540178180616289194509686159107063111500537850642030367642480439906210197179331394602401 42 : 46400 43 : 66450789401396266861985003360578832766171215212266995967429417229964093893359523047678782463961 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. <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
Julia
<lang 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)
</lang>
- Output:
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
<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;
$_
}
}
};</lang>
First 20 terms of OEIS:A073916 1 3 25 14 14641 44 24137569 70 1089 405 819628286980801 160 22563490300366186081 2752 9801 462 21559177407076402401757871041 1044 740195513856780056217081017732809 1520
Phix
simple
Certainly not the fastest way to do it, hence the relatively small limit of 24, which takes less than 0.4s,
whereas a limit of 25 would need to invoke factors() 52 million times which would no doubt take a fair while.
<lang Phix>constant LIMIT = 24
include mpfr.e
mpz z = mpz_init()
sequence fn = 1&repeat(0,LIMIT-1),
primes = {2,3}
integer k = 1 printf(1,"The first %d terms in the sequence are:\n",LIMIT) for i=1 to LIMIT do
sequence f = factors(i,1)
if length(f)=2 then -- i is prime (f is {1,i})
while length(primes)<i do
integer p = primes[$]+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
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). <lang Phix>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)})</lang>
- Output:
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. <lang Python> 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)
</lang> Output: <lang Python> 1 3 25 14 14641 44 24137569 70 1089 405 819628286980801 160 22563490300366186081 2752 9801 </lang>
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.*/ 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.*/</lang>
- output 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. <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.*/ 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 /*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 # . -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