I'm working on modernizing Rosetta Code's infrastructure. Starting with communications. Please accept this time-limited open invite to RC's Slack.. --Michael Mol (talk) 20:59, 30 May 2020 (UTC)

# Powerful numbers

Powerful numbers is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

A k-powerful (or k-full) number is a positive integer n such that for every prime number p dividing n, p^k also divides n.

These are numbers of the form:

2-powerful = a^2 * b^3,             for a,b >= 1
3-powerful = a^3 * b^4 * c^5,       for a,b,c >= 1
4-powerful = a^4 * b^5 * c^6 * d^7, for a,b,c,d >= 1
...
k-powerful = a^k * b^(k+1) * c^(k+2) *...* ω^(2*k-1), for a,b,c,...,ω >= 1

Write a function that generates all the k-powerful numbers less than or equal to n.

• For k = 2..10, generate the set of k-powerful numbers <= 10^k and show the first 5 and the last 5 terms, along with the length of the set.

Write a function that counts the number of k-powerful numbers less than or equal to n. (optional)

• For k = 2..10, show the number of k-powerful numbers less than or equal to 10^j, for 0 <= j < k+10.

## C++

Translation of: Go
Translation of: Perl
#include <algorithm>
#include <cmath>
#include <cstdint>
#include <iostream>
#include <numeric>
#include <vector>

bool is_square_free(uint64_t n) {
static constexpr uint64_t primes[] {
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,
43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97
}; // seems to be enough
for (auto p : primes) {
auto p2 = p * p;
if (p2 > n)
break;
if (n % p2 == 0)
return false;
}
return true;
}

uint64_t iroot(uint64_t n, uint64_t r) {
// adjustment to ensure f/p square root exact for perfect integer squares
static constexpr double adj = 1e-6;
}

uint64_t ipow(uint64_t n, uint64_t p) {
uint64_t prod = 1;
for (; p > 0; p >>= 1) {
if (p & 1)
prod *= n;
n *= n;
}
return prod;
}

std::vector<uint64_t> powerful(uint64_t n, uint64_t k) {
std::vector<uint64_t> result;
std::function<void(uint64_t, uint64_t)> f = [&](uint64_t m, uint64_t r) {
if (r < k) {
result.push_back(m);
return;
}
uint64_t root = iroot(n/m, r);
for (uint64_t v = 1; v <= root; ++v) {
if (r > k && (!is_square_free(v) || std::gcd(m, v) != 1))
continue;
f(m * ipow(v, r), r - 1);
}
};
f(1, 2*k - 1);
std::sort(result.begin(), result.end());
return result;
}

uint64_t powerful_count(uint64_t n, uint64_t k) {
uint64_t count = 0;
std::function<void(uint64_t, uint64_t)> f = [&](uint64_t m, uint64_t r) {
if (r <= k) {
count += iroot(n/m, r);
return;
}
uint64_t root = iroot(n/m, r);
for (uint64_t v = 1; v <= root; ++v) {
if (is_square_free(v) && std::gcd(m, v) == 1)
f(m * ipow(v, r), r - 1);
}
};
f(1, 2*k - 1);
return count;
}

int main() {
const size_t max = 5;
for (uint64_t k = 2, p = 100; k <= 10; ++k, p *= 10) {
auto result = powerful(p, k);
std::cout << result.size() << " " << k
<< "-powerful numbers <= 10^" << k << ":";
for (size_t i = 0; i < result.size(); ++i) {
if (i == max)
std::cout << " ...";
else if (i < max || i + max >= result.size())
std::cout << ' ' << result[i];
}
std::cout << '\n';
}
std::cout << '\n';
for (uint64_t k = 2; k <= 10; ++k) {
std::cout << "Count of " << k << "-powerful numbers <= 10^j for 0 <= j < "
<< k + 10 << ":";
for (uint64_t j = 0, p = 1; j < k + 10; ++j, p *= 10)
std::cout << ' ' << powerful_count(p, k);
std::cout << '\n';
}
}
Output:
14 2-powerful numbers <= 10^2: 1 4 8 9 16 ... 49 64 72 81 100
20 3-powerful numbers <= 10^3: 1 8 16 27 32 ... 625 648 729 864 1000
25 4-powerful numbers <= 10^4: 1 16 32 64 81 ... 5184 6561 7776 8192 10000
32 5-powerful numbers <= 10^5: 1 32 64 128 243 ... 65536 69984 78125 93312 100000
38 6-powerful numbers <= 10^6: 1 64 128 256 512 ... 559872 746496 823543 839808 1000000
46 7-powerful numbers <= 10^7: 1 128 256 512 1024 ... 7558272 8388608 8957952 9765625 10000000
52 8-powerful numbers <= 10^8: 1 256 512 1024 2048 ... 60466176 67108864 80621568 90699264 100000000
59 9-powerful numbers <= 10^9: 1 512 1024 2048 4096 ... 644972544 725594112 816293376 967458816 1000000000
68 10-powerful numbers <= 10^10: 1 1024 2048 4096 8192 ... 7739670528 8589934592 8707129344 9795520512 10000000000

Count of 2-powerful numbers <= 10^j for 0 <= j < 12: 1 4 14 54 185 619 2027 6553 21044 67231 214122 680330
Count of 3-powerful numbers <= 10^j for 0 <= j < 13: 1 2 7 20 51 129 307 713 1645 3721 8348 18589 41136
Count of 4-powerful numbers <= 10^j for 0 <= j < 14: 1 1 5 11 25 57 117 235 464 906 1741 3312 6236 11654
Count of 5-powerful numbers <= 10^j for 0 <= j < 15: 1 1 3 8 16 32 63 117 211 375 659 1153 2000 3402 5770
Count of 6-powerful numbers <= 10^j for 0 <= j < 16: 1 1 2 6 12 21 38 70 121 206 335 551 900 1451 2326 3706
Count of 7-powerful numbers <= 10^j for 0 <= j < 17: 1 1 1 4 10 16 26 46 77 129 204 318 495 761 1172 1799 2740
Count of 8-powerful numbers <= 10^j for 0 <= j < 18: 1 1 1 3 8 13 19 32 52 85 135 211 315 467 689 1016 1496 2191
Count of 9-powerful numbers <= 10^j for 0 <= j < 19: 1 1 1 2 6 11 16 24 38 59 94 145 217 317 453 644 919 1308 1868
Count of 10-powerful numbers <= 10^j for 0 <= j < 20: 1 1 1 1 5 9 14 21 28 43 68 104 155 227 322 447 621 858 1192 1651

## Go

Translation of: Perl

Curiously, the 'powerful' function is producing duplicates which the other existing solutions don't seem to suffer from. It's easily dealt with by using a set (rather than a slice) but means that we're unable to take advantage of the counting shortcut - not that it matters as the whole thing still runs in less than 0.4 seconds!

package main

import (
"fmt"
"math"
"sort"
)

const adj = 0.0001 // adjustment to ensure f/p square root exact for perfect integer squares

var primes = []uint64{
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97,
} // seems to be enough

func gcd(x, y uint64) uint64 {
for y != 0 {
x, y = y, x%y
}
return x
}

func isSquareFree(x uint64) bool {
for _, p := range primes {
p2 := p * p
if p2 > x {
break
}
if x%p2 == 0 {
return false
}
}
return true
}

func iroot(x, p uint64) uint64 {
}

func ipow(x, p uint64) uint64 {
prod := uint64(1)
for p > 0 {
if p&1 != 0 {
prod *= x
}
p >>= 1
x *= x
}
return prod
}

func powerful(n, k uint64) []uint64 {
set := make(map[uint64]bool)
var f func(m, r uint64) // recursive closure
f = func(m, r uint64) {
if r < k {
set[m] = true
return
}
for v := uint64(1); v <= iroot(n/m, r); v++ {
if r > k {
if !isSquareFree(v) || gcd(m, v) != 1 {
continue
}
}
f(m*ipow(v, r), r-1)
}
}
f(1, (1<<k)-1)
list := make([]uint64, 0, len(set))
for key := range set {
list = append(list, key)
}
sort.Slice(list, func(i, j int) bool {
return list[i] < list[j]
})
return list
}

func main() {
power := uint64(10)
for k := uint64(2); k <= 10; k++ {
power *= 10
a := powerful(power, k)
le := len(a)
h, t := a[0:5], a[le-5:]
fmt.Printf("%d %2d-powerful numbers <= 10^%-2d: %v ... %v\n", le, k, k, h, t)
}
fmt.Println()
for k := uint64(2); k <= 10; k++ {
power := uint64(1)
var counts []int
for j := uint64(0); j < k+10; j++ {
a := powerful(power, k)
counts = append(counts, len(a))
power *= 10
}
j := k + 10
fmt.Printf("Count of %2d-powerful numbers <= 10^j, j in [0, %d): %v\n", k, j, counts)
}
}
Output:
14  2-powerful numbers <= 10^2 : [1 4 8 9 16] ... [49 64 72 81 100]
20  3-powerful numbers <= 10^3 : [1 8 16 27 32] ... [625 648 729 864 1000]
25  4-powerful numbers <= 10^4 : [1 16 32 64 81] ... [5184 6561 7776 8192 10000]
32  5-powerful numbers <= 10^5 : [1 32 64 128 243] ... [65536 69984 78125 93312 100000]
38  6-powerful numbers <= 10^6 : [1 64 128 256 512] ... [559872 746496 823543 839808 1000000]
46  7-powerful numbers <= 10^7 : [1 128 256 512 1024] ... [7558272 8388608 8957952 9765625 10000000]
52  8-powerful numbers <= 10^8 : [1 256 512 1024 2048] ... [60466176 67108864 80621568 90699264 100000000]
59  9-powerful numbers <= 10^9 : [1 512 1024 2048 4096] ... [644972544 725594112 816293376 967458816 1000000000]
68 10-powerful numbers <= 10^10: [1 1024 2048 4096 8192] ... [7739670528 8589934592 8707129344 9795520512 10000000000]

Count of  2-powerful numbers <= 10^j, j in [0, 12): [1 4 14 54 185 619 2027 6553 21044 67231 214122 680330]
Count of  3-powerful numbers <= 10^j, j in [0, 13): [1 2 7 20 51 129 307 713 1645 3721 8348 18589 41136]
Count of  4-powerful numbers <= 10^j, j in [0, 14): [1 1 5 11 25 57 117 235 464 906 1741 3312 6236 11654]
Count of  5-powerful numbers <= 10^j, j in [0, 15): [1 1 3 8 16 32 63 117 211 375 659 1153 2000 3402 5770]
Count of  6-powerful numbers <= 10^j, j in [0, 16): [1 1 2 6 12 21 38 70 121 206 335 551 900 1451 2326 3706]
Count of  7-powerful numbers <= 10^j, j in [0, 17): [1 1 1 4 10 16 26 46 77 129 204 318 495 761 1172 1799 2740]
Count of  8-powerful numbers <= 10^j, j in [0, 18): [1 1 1 3 8 13 19 32 52 85 135 211 315 467 689 1016 1496 2191]
Count of  9-powerful numbers <= 10^j, j in [0, 19): [1 1 1 2 6 11 16 24 38 59 94 145 217 317 453 644 919 1308 1868]
Count of 10-powerful numbers <= 10^j, j in [0, 20): [1 1 1 1 5 9 14 21 28 43 68 104 155 227 322 447 621 858 1192 1651]

## Java

import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class PowerfulNumbers {

public static void main(String[] args) {
System.out.printf("Task: For k = 2..10, generate the set of k-powerful numbers <= 10^k and show the first 5 and the last 5 terms, along with the length of the set%n");
for ( int k = 2 ; k <= 10 ; k++ ) {
BigInteger max = BigInteger.valueOf(10).pow(k);
List<BigInteger> powerfulNumbers = getPowerFulNumbers(max, k);
System.out.printf("There are %d %d-powerful numbers between 1 and %d.  %nList: %s%n", powerfulNumbers.size(), k, max, getList(powerfulNumbers));
}
System.out.printf("%nTask: For k = 2..10, show the number of k-powerful numbers less than or equal to 10^j, for 0 <= j < k+10%n");
for ( int k = 2 ; k <= 10 ; k++ ) {
List<Integer> powCount = new ArrayList<>();
for ( int j = 0 ; j < k+10 ; j++ ) {
BigInteger max = BigInteger.valueOf(10).pow(j);
}
System.out.printf("Count of %2d-powerful numbers <= 10^j, j in [0, %d]: %s%n", k, k+9, powCount);
}

}

private static String getList(List<BigInteger> list) {
StringBuilder sb = new StringBuilder();
sb.append(list.subList(0, 5).toString().replace("]", ""));
sb.append(" ... ");
sb.append(list.subList(list.size()-5, list.size()).toString().replace("[", ""));
return sb.toString();
}

private static int countPowerFulNumbers(BigInteger max, int k) {
return potentialPowerful(max, k).size();
}

private static List<BigInteger> getPowerFulNumbers(BigInteger max, int k) {
List<BigInteger> powerfulNumbers = new ArrayList<>(potentialPowerful(max, k));
Collections.sort(powerfulNumbers);
return powerfulNumbers;
}

private static Set<BigInteger> potentialPowerful(BigInteger max, int k) {
// Setup
int[] indexes = new int[k];
for ( int i = 0 ; i < k ; i++ ) {
indexes[i] = 1;
}

Set<BigInteger> powerful = new HashSet<>();
boolean foundPower = true;
while ( foundPower ) {

boolean genPowerful = false;
for ( int index = 0 ; index < k ; index++ ) {
BigInteger power = BigInteger.ONE;
for ( int i = 0 ; i < k ; i++ ) {
power = power.multiply(BigInteger.valueOf(indexes[i]).pow(k+i));
}
if ( power.compareTo(max) <= 0 ) {
indexes[0] += 1;
genPowerful = true;
break;
}
else {
indexes[index] = 1;
if ( index < k-1 ) {
indexes[index+1] += 1;
}
}
}
if ( ! genPowerful ) {
foundPower = false;
}
}

return powerful;
}

}

Output:
Task:  For k = 2..10, generate the set of k-powerful numbers <= 10^k and show the first 5 and the last 5 terms, along with the length of the set
There are 14 2-powerful numbers between 1 and 100.
List: [1, 4, 8, 9, 16 ... 49, 64, 72, 81, 100]
There are 20 3-powerful numbers between 1 and 1000.
List: [1, 8, 16, 27, 32 ... 625, 648, 729, 864, 1000]
There are 25 4-powerful numbers between 1 and 10000.
List: [1, 16, 32, 64, 81 ... 5184, 6561, 7776, 8192, 10000]
There are 32 5-powerful numbers between 1 and 100000.
List: [1, 32, 64, 128, 243 ... 65536, 69984, 78125, 93312, 100000]
There are 38 6-powerful numbers between 1 and 1000000.
List: [1, 64, 128, 256, 512 ... 559872, 746496, 823543, 839808, 1000000]
There are 46 7-powerful numbers between 1 and 10000000.
List: [1, 128, 256, 512, 1024 ... 7558272, 8388608, 8957952, 9765625, 10000000]
There are 52 8-powerful numbers between 1 and 100000000.
List: [1, 256, 512, 1024, 2048 ... 60466176, 67108864, 80621568, 90699264, 100000000]
There are 59 9-powerful numbers between 1 and 1000000000.
List: [1, 512, 1024, 2048, 4096 ... 644972544, 725594112, 816293376, 967458816, 1000000000]
There are 68 10-powerful numbers between 1 and 10000000000.
List: [1, 1024, 2048, 4096, 8192 ... 7739670528, 8589934592, 8707129344, 9795520512, 10000000000]

Task:  For k = 2..10, show the number of k-powerful numbers less than or equal to 10^j, for 0 <= j < k+10
Count of  2-powerful numbers <= 10^j, j in [0, 11]: [1, 4, 14, 54, 185, 619, 2027, 6553, 21044, 67231, 214122, 680330]
Count of  3-powerful numbers <= 10^j, j in [0, 12]: [1, 2, 7, 20, 51, 129, 307, 713, 1645, 3721, 8348, 18589, 41136]
Count of  4-powerful numbers <= 10^j, j in [0, 13]: [1, 1, 5, 11, 25, 57, 117, 235, 464, 906, 1741, 3312, 6236, 11654]
Count of  5-powerful numbers <= 10^j, j in [0, 14]: [1, 1, 3, 8, 16, 32, 63, 117, 211, 375, 659, 1153, 2000, 3402, 5770]
Count of  6-powerful numbers <= 10^j, j in [0, 15]: [1, 1, 2, 6, 12, 21, 38, 70, 121, 206, 335, 551, 900, 1451, 2326, 3706]
Count of  7-powerful numbers <= 10^j, j in [0, 16]: [1, 1, 1, 4, 10, 16, 26, 46, 77, 129, 204, 318, 495, 761, 1172, 1799, 2740]
Count of  8-powerful numbers <= 10^j, j in [0, 17]: [1, 1, 1, 3, 8, 13, 19, 32, 52, 85, 135, 211, 315, 467, 689, 1016, 1496, 2191]
Count of  9-powerful numbers <= 10^j, j in [0, 18]: [1, 1, 1, 2, 6, 11, 16, 24, 38, 59, 94, 145, 217, 317, 453, 644, 919, 1308, 1868]
Count of 10-powerful numbers <= 10^j, j in [0, 19]: [1, 1, 1, 1, 5, 9, 14, 21, 28, 43, 68, 104, 155, 227, 322, 447, 621, 858, 1192, 1651]

## Julia

Translation of: Perl
using Primes

is_kpowerful(n, k) = all(x -> x[2] >= k, factor(n)) # not used here

is_squarefree(n) = all(x -> x[2] == 1, factor(n))
rootdiv(n, m, r) = Int128(floor(div(n, m)^(1/r) + 0.0000001))

function genkpowerful(n, k)
ret = Int128[]
function inner(m, r)
if r < k
push!(ret, m)
else
for a in 1:rootdiv(n, m, r)
if r <= k || (gcd(a, m) == 1 && is_squarefree(a))
inner(m * Int128(a)^r, r - 1)
end
end
end
end
inner(1, 2 * k - 1)
return unique(sort(Int.(ret)))
end

function kpowerfulcount(n, k)
count = Int128(0)
function inner(m, r)
if r <= k
count += rootdiv(n, m, r)
else
for a in 1:rootdiv(n, m, r)
if gcd(a, m) == 1 && is_squarefree(a)
inner(m * a^r, r - 1)
end
end
end
end
inner(1, 2*k - 1)
return Int(count)
end

for k in 2:10
a = genkpowerful(10^k, k)
len = length(a)
print("The set of \$k-powerful numbers between 1 and 10^\$k has \$len members:\n[")
for i in [1:5 ; len-5:len]
print(a[i], i == 5 ? " ... " : i == len ? "]\n" : ", ")
end
end
for k in 2:10
print("The count of \$k-powerful numbers from 1 to 10^j for j in 0:\$(k+9) is: ",
[kpowerfulcount(Int128(10)^j, k) for j in 0:(k+9)], "\n")
end

Output:
The set of 2-powerful numbers between 1 and 10^2 has 14 members:
[1, 4, 8, 9, 16 ... 36, 49, 64, 72, 81, 100]
The set of 3-powerful numbers between 1 and 10^3 has 20 members:
[1, 8, 16, 27, 32 ... 512, 625, 648, 729, 864, 1000]
The set of 4-powerful numbers between 1 and 10^4 has 25 members:
[1, 16, 32, 64, 81 ... 4096, 5184, 6561, 7776, 8192, 10000]
The set of 5-powerful numbers between 1 and 10^5 has 32 members:
[1, 32, 64, 128, 243 ... 62208, 65536, 69984, 78125, 93312, 100000]
The set of 6-powerful numbers between 1 and 10^6 has 38 members:
[1, 64, 128, 256, 512 ... 531441, 559872, 746496, 823543, 839808, 1000000]
The set of 7-powerful numbers between 1 and 10^7 has 46 members:
[1, 128, 256, 512, 1024 ... 6718464, 7558272, 8388608, 8957952, 9765625, 10000000]
The set of 8-powerful numbers between 1 and 10^8 has 52 members:
[1, 256, 512, 1024, 2048 ... 53747712, 60466176, 67108864, 80621568, 90699264, 100000000]
The set of 9-powerful numbers between 1 and 10^9 has 59 members:
[1, 512, 1024, 2048, 4096 ... 544195584, 644972544, 725594112, 816293376, 967458816, 1000000000]
The set of 10-powerful numbers between 1 and 10^10 has 68 members:
[1, 1024, 2048, 4096, 8192 ... 6530347008, 7739670528, 8589934592, 8707129344, 9795520512, 10000000000]
The count of 2-powerful numbers from 1 to 10^j for j in 0:11 is: [1, 4, 14, 54, 185, 619, 2027, 6553, 21044, 67231, 214122, 680330]
The count of 3-powerful numbers from 1 to 10^j for j in 0:12 is: [1, 2, 7, 20, 51, 129, 307, 713, 1645, 3721, 8348, 18589, 41136]
The count of 4-powerful numbers from 1 to 10^j for j in 0:13 is: [1, 1, 5, 11, 25, 57, 117, 235, 464, 906, 1741, 3312, 6236, 11654]
The count of 5-powerful numbers from 1 to 10^j for j in 0:14 is: [1, 1, 3, 8, 16, 32, 63, 117, 211, 375, 659, 1153, 2000, 3402, 5770]
The count of 6-powerful numbers from 1 to 10^j for j in 0:15 is: [1, 1, 2, 6, 12, 21, 38, 70, 121, 206, 335, 551, 900, 1451, 2326, 3706]
The count of 7-powerful numbers from 1 to 10^j for j in 0:16 is: [1, 1, 1, 4, 10, 16, 26, 46, 77, 129, 204, 318, 495, 761, 1172, 1799, 2740]
The count of 8-powerful numbers from 1 to 10^j for j in 0:17 is: [1, 1, 1, 3, 8, 13, 19, 32, 52, 85, 135, 211, 315, 467, 689, 1016, 1496, 2191]
The count of 9-powerful numbers from 1 to 10^j for j in 0:18 is: [1, 1, 1, 2, 6, 11, 16, 24, 38, 59, 94, 145, 217, 317, 453, 644, 919, 1308, 1868]
The count of 10-powerful numbers from 1 to 10^j for j in 0:19 is: [1, 1, 1, 1, 5, 9, 14, 21, 28, 43, 68, 104, 155, 227, 322, 447, 621, 858, 1192, 1651]

## Nim

Translation of: C++
import algorithm, math, strformat, strutils

func isSquareFree(n: uint64): bool =
const Primes = [uint64 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41,
43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97]
for p in Primes:
let p2 = p * p
if p2 > n: break
if n mod p2 == 0: return false
result = true

func iroot(n: uint64; r: Natural): uint64 =
result = uint64(pow(float(n), 1 / r) + Adj)

func powerful(n: uint64; k: Natural): seq[uint64] =
var res: seq[uint64]

func f(m: uint64; r: Natural) =
if r < k:
return
let root = iroot(n div m, r)
for v in 1..root:
if r > k and (not v.isSquareFree or gcd(m, v) != 1):
continue
f(m * v^r, r - 1)

f(1, 2 * k - 1)
res.sort()
return res

func powerfulCount(n: uint64; k: Natural): uint64 =
var count = 0u64

func f(m: uint64; r: Natural) =
let root = iroot(n div m, r)
if r <= k:
count += root
return
for v in 1..root:
if v.isSquareFree and gcd(m, v) == 1:
f(m * v^r, r - 1)

f(1, 2 * k - 1)
return count

var p: uint64 = 10
for k in 2..10:
p *= 10
let result = powerful(p, k)
let tail = result[^5..^1].join(" ")
echo &"{result.len} {k:2}-powerful numbers <= 10^{k}: {head} ... {tail}"
echo()

for k in 2..10:
p = 1
var counts: seq[uint64]
for j in 0..k+9:
p *= 10
echo &"Count of {k:2}-powerful numbers <= 10^j, j in 0 ≤ j < {k+10}: {counts.join(\" \")}"
Output:
14  2-powerful numbers <= 10^2: 1 4 8 9 16 ... 49 64 72 81 100
20  3-powerful numbers <= 10^3: 1 8 16 27 32 ... 625 648 729 864 1000
25  4-powerful numbers <= 10^4: 1 16 32 64 81 ... 5184 6561 7776 8192 10000
32  5-powerful numbers <= 10^5: 1 32 64 128 243 ... 65536 69984 78125 93312 100000
38  6-powerful numbers <= 10^6: 1 64 128 256 512 ... 559872 746496 823543 839808 1000000
46  7-powerful numbers <= 10^7: 1 128 256 512 1024 ... 7558272 8388608 8957952 9765625 10000000
52  8-powerful numbers <= 10^8: 1 256 512 1024 2048 ... 60466176 67108864 80621568 90699264 100000000
59  9-powerful numbers <= 10^9: 1 512 1024 2048 4096 ... 644972544 725594112 816293376 967458816 1000000000
68 10-powerful numbers <= 10^10: 1 1024 2048 4096 8192 ... 7739670528 8589934592 8707129344 9795520512 10000000000

Count of  2-powerful numbers <= 10^j, j in 0 ≤ j < 12: 1 4 14 54 185 619 2027 6553 21044 67231 214122 680330
Count of  3-powerful numbers <= 10^j, j in 0 ≤ j < 13: 1 2 7 20 51 129 307 713 1645 3721 8348 18589 41136
Count of  4-powerful numbers <= 10^j, j in 0 ≤ j < 14: 1 1 5 11 25 57 117 235 464 906 1741 3312 6236 11654
Count of  5-powerful numbers <= 10^j, j in 0 ≤ j < 15: 1 1 3 8 16 32 63 117 211 375 659 1153 2000 3402 5770
Count of  6-powerful numbers <= 10^j, j in 0 ≤ j < 16: 1 1 2 6 12 21 38 70 121 206 335 551 900 1451 2326 3706
Count of  7-powerful numbers <= 10^j, j in 0 ≤ j < 17: 1 1 1 4 10 16 26 46 77 129 204 318 495 761 1172 1799 2740
Count of  8-powerful numbers <= 10^j, j in 0 ≤ j < 18: 1 1 1 3 8 13 19 32 52 85 135 211 315 467 689 1016 1496 2191
Count of  9-powerful numbers <= 10^j, j in 0 ≤ j < 19: 1 1 1 2 6 11 16 24 38 59 94 145 217 317 453 644 919 1308 1868
Count of 10-powerful numbers <= 10^j, j in 0 ≤ j < 20: 1 1 1 1 5 9 14 21 28 43 68 104 155 227 322 447 621 858 1192 1651

## Perl

### Generation

use 5.020;
use ntheory qw(is_square_free);
use experimental qw(signatures);
use Math::AnyNum qw(:overload idiv iroot ipow is_coprime);

sub powerful_numbers (\$n, \$k = 2) {

my @powerful;

sub (\$m, \$r) {
if (\$r < \$k) {
push @powerful, \$m;
return;
}
for my \$v (1 .. iroot(idiv(\$n, \$m), \$r)) {
if (\$r > \$k) {
is_square_free(\$v) || next;
is_coprime(\$m, \$v) || next;
}
__SUB__->(\$m * ipow(\$v, \$r), \$r - 1);
}
}->(1, 2*\$k - 1);

sort { \$a <=> \$b } @powerful;
}

foreach my \$k (2 .. 10) {
my @a = powerful_numbers(10**\$k, \$k);
my \$h = join(', ', @a[0..4]);
my \$t = join(', ', @a[\$#a-4..\$#a]);
printf("For k=%-2d there are %d k-powerful numbers <= 10^k: [%s, ..., %s]\n", \$k, scalar(@a), \$h, \$t);
}
Output:
For k=2  there are 14 k-powerful numbers <= 10^k: [1, 4, 8, 9, 16, ..., 49, 64, 72, 81, 100]
For k=3  there are 20 k-powerful numbers <= 10^k: [1, 8, 16, 27, 32, ..., 625, 648, 729, 864, 1000]
For k=4  there are 25 k-powerful numbers <= 10^k: [1, 16, 32, 64, 81, ..., 5184, 6561, 7776, 8192, 10000]
For k=5  there are 32 k-powerful numbers <= 10^k: [1, 32, 64, 128, 243, ..., 65536, 69984, 78125, 93312, 100000]
For k=6  there are 38 k-powerful numbers <= 10^k: [1, 64, 128, 256, 512, ..., 559872, 746496, 823543, 839808, 1000000]
For k=7  there are 46 k-powerful numbers <= 10^k: [1, 128, 256, 512, 1024, ..., 7558272, 8388608, 8957952, 9765625, 10000000]
For k=8  there are 52 k-powerful numbers <= 10^k: [1, 256, 512, 1024, 2048, ..., 60466176, 67108864, 80621568, 90699264, 100000000]
For k=9  there are 59 k-powerful numbers <= 10^k: [1, 512, 1024, 2048, 4096, ..., 644972544, 725594112, 816293376, 967458816, 1000000000]
For k=10 there are 68 k-powerful numbers <= 10^k: [1, 1024, 2048, 4096, 8192, ..., 7739670528, 8589934592, 8707129344, 9795520512, 10000000000]

### Counting

use 5.020;
use ntheory qw(is_square_free);
use experimental qw(signatures);
use Math::AnyNum qw(:overload idiv iroot ipow is_coprime);

sub powerful_count (\$n, \$k = 2) {

my \$count = 0;

sub (\$m, \$r) {
if (\$r <= \$k) {
\$count += iroot(idiv(\$n, \$m), \$r);
return;
}
for my \$v (1 .. iroot(idiv(\$n, \$m), \$r)) {
is_square_free(\$v) || next;
is_coprime(\$m, \$v) || next;
__SUB__->(\$m * ipow(\$v, \$r), \$r - 1);
}
}->(1, 2*\$k - 1);

return \$count;
}

foreach my \$k (2 .. 10) {
printf("Number of %2d-powerful <= 10^j for 0 <= j < %d: {%s}\n", \$k, \$k+10,
join(', ', map { powerful_count(ipow(10, \$_), \$k) } 0..(\$k+10-1)));
}
Output:
Number of  2-powerful <= 10^j for 0 <= j < 12: {1, 4, 14, 54, 185, 619, 2027, 6553, 21044, 67231, 214122, 680330}
Number of  3-powerful <= 10^j for 0 <= j < 13: {1, 2, 7, 20, 51, 129, 307, 713, 1645, 3721, 8348, 18589, 41136}
Number of  4-powerful <= 10^j for 0 <= j < 14: {1, 1, 5, 11, 25, 57, 117, 235, 464, 906, 1741, 3312, 6236, 11654}
Number of  5-powerful <= 10^j for 0 <= j < 15: {1, 1, 3, 8, 16, 32, 63, 117, 211, 375, 659, 1153, 2000, 3402, 5770}
Number of  6-powerful <= 10^j for 0 <= j < 16: {1, 1, 2, 6, 12, 21, 38, 70, 121, 206, 335, 551, 900, 1451, 2326, 3706}
Number of  7-powerful <= 10^j for 0 <= j < 17: {1, 1, 1, 4, 10, 16, 26, 46, 77, 129, 204, 318, 495, 761, 1172, 1799, 2740}
Number of  8-powerful <= 10^j for 0 <= j < 18: {1, 1, 1, 3, 8, 13, 19, 32, 52, 85, 135, 211, 315, 467, 689, 1016, 1496, 2191}
Number of  9-powerful <= 10^j for 0 <= j < 19: {1, 1, 1, 2, 6, 11, 16, 24, 38, 59, 94, 145, 217, 317, 453, 644, 919, 1308, 1868}
Number of 10-powerful <= 10^j for 0 <= j < 20: {1, 1, 1, 1, 5, 9, 14, 21, 28, 43, 68, 104, 155, 227, 322, 447, 621, 858, 1192, 1651}

## Phix

Library: Phix/mpfr

### priority queue

Tried a slightly different approach, but it gets 7/10 at best on performance, and simply not worth attempting under pwa/p2js

without javascript_semantics
include mpfr.e

integer pq = NULL   -- data is the prime powers, eg {2,2} for 2^2*3^2
-- priority is the mpz of that, eg/ie 36

procedure padd(mpz n, np, sequence s)
if pq=NULL then
pq = pq_new(MIN_HEAP,routine_id("mpz_cmp"))
end if
if mpz_cmp(np,n)<=0 then
end if
end procedure

procedure add_next(integer k, mpz n, p, sequence s)
mpz np = mpz_init()
integer l = length(s)
for i=1 to l do
integer ki = iff(s[i]=0?k:1)
mpz_ui_pow_ui(np,get_prime(i),ki)
mpz_mul(np,np,p)
s[i] += ki
s[i] -= ki
end for
mpz_ui_pow_ui(np,get_prime(l+1),k)
if l>0 and s[l]=k and s[1..l-1]==repeat(0,l-1) then
elsif s={} then
mpz_mul(np,np,p)
end if
end procedure

function powerful(integer k, mpz n)
mpz prevp = mpz_init(1)
sequence res = {prevp}
while pq_size(pq) do
{sequence s, mpz p} = pq_pop(pq)
if mpz_cmp(p,prevp)!=0 then
res &= p
prevp = p
end if
end while
return res
end function

mpz p = mpz_init(10)
for k=2 to 10 do
mpz_mul_si(p,p,10)
sequence s = powerful(k,p)
integer l = length(s)
for i=1 to 5 do
s[i] = mpz_get_str(s[i])
s[-i] = mpz_get_str(s[-i])
end for
s[6..-6] = {"..."}
s = join(s,",")
printf(1,"%d %2d-powerful numbers <= 10^%-2d : %s\n",{l,k,k,s})
end for

puts(1,"\n")
--Not really fast enough, so only doing to k+5..k+8 for the first 4, but k+9 for last 5
for k=2 to 10 do
sequence counts = {}
integer lim = k+min(k+3,9)
mpz_set_si(p,1)
for j=0 to lim do
printf(1,"counting (%d/%d)...\r",{j,lim})
counts &= length(powerful(k,p))
mpz_mul_si(p,p,10)
end for
printf(1,"Counts of %2d-powerful numbers <= 10^(0..%d) : %v\n",{k,lim,counts})
end for
Output:
14  2-powerful numbers <= 10^2  : 1,4,8,9,16,...,49,64,72,81,100
20  3-powerful numbers <= 10^3  : 1,8,16,27,32,...,625,648,729,864,1000
25  4-powerful numbers <= 10^4  : 1,16,32,64,81,...,5184,6561,7776,8192,10000
32  5-powerful numbers <= 10^5  : 1,32,64,128,243,...,65536,69984,78125,93312,100000
38  6-powerful numbers <= 10^6  : 1,64,128,256,512,...,559872,746496,823543,839808,1000000
46  7-powerful numbers <= 10^7  : 1,128,256,512,1024,...,7558272,8388608,8957952,9765625,10000000
52  8-powerful numbers <= 10^8  : 1,256,512,1024,2048,...,60466176,67108864,80621568,90699264,100000000
59  9-powerful numbers <= 10^9  : 1,512,1024,2048,4096,...,644972544,725594112,816293376,967458816,1000000000
68 10-powerful numbers <= 10^10 : 1,1024,2048,4096,8192,...,7739670528,8589934592,8707129344,9795520512,10000000000

Counts of  2-powerful numbers <= 10^(0..7) : {1,4,14,54,185,619,2027,6553}
Counts of  3-powerful numbers <= 10^(0..9) : {1,2,7,20,51,129,307,713,1645,3721}
Counts of  4-powerful numbers <= 10^(0..11) : {1,1,5,11,25,57,117,235,464,906,1741,3312}
Counts of  5-powerful numbers <= 10^(0..13) : {1,1,3,8,16,32,63,117,211,375,659,1153,2000,3402}
Counts of  6-powerful numbers <= 10^(0..15) : {1,1,2,6,12,21,38,70,121,206,335,551,900,1451,2326,3706}
Counts of  7-powerful numbers <= 10^(0..16) : {1,1,1,4,10,16,26,46,77,129,204,318,495,761,1172,1799,2740}
Counts of  8-powerful numbers <= 10^(0..17) : {1,1,1,3,8,13,19,32,52,85,135,211,315,467,689,1016,1496,2191}
Counts of  9-powerful numbers <= 10^(0..18) : {1,1,1,2,6,11,16,24,38,59,94,145,217,317,453,644,919,1308,1868}
Counts of 10-powerful numbers <= 10^(0..19) : {1,1,1,1,5,9,14,21,28,43,68,104,155,227,322,447,621,858,1192,1651}

### translation

Translation of: Go
Translation of: Sidef

Significantly faster. The last few counts were wrong when using native ints/atoms, so I went gmp, which means it also works on 32-bit.

with javascript_semantics
include mpfr.e

function powerful(mpz n, integer k, sequence res={}, mpz m=NULL, integer r=0)
if m=NULL then
m = mpz_init(1)
r = 2*k-1
end if
if r<k then
res = append(res,iff(mpz_fits_atom(m)?mpz_get_atom(m):mpz_get_str(m)))
else
mpz t = mpz_init()
mpz_fdiv_q(t, n, m) -- t := floor(n/m)
mpz_nthroot(t,t,r) -- t = floor(rth root of t)
if not mpz_fits_integer(t) then ?9/0 end if
integer lim = mpz_get_integer(t)
for v=1 to lim do
if r<=k
or (square_free(v) and mpz_gcd_ui(NULL,m,v)=1) then
mpz_ui_pow_ui(t,v,r) -- t:= v^r
mpz_mul(t,t,m)       -- t*= m
res = powerful(n, k, res, t, r-1)
end if
end for
end if
return res
end function

function powercount(mpz n, integer k, mpz m=NULL, integer r=0)
atom res = 0
if m=NULL then
m = mpz_init(1)
r = 2*k-1
end if
mpz t = mpz_init()
mpz_fdiv_q(t, n, m) -- t := floor(n/m)
mpz_nthroot(t,t,r) -- t = floor(rth root of t)
if not mpz_fits_integer(t) then ?9/0 end if
integer lim = mpz_get_integer(t)
if r<=k then
res += lim
else
for v=1 to lim do
if square_free(v) and mpz_gcd_ui(NULL,m,v)=1 then
mpz_ui_pow_ui(t,v,r) -- t:= v^r
mpz_mul(t,t,m)       -- t*= m
res += powercount(n, k, t, r-1)
end if
end for
end if
return res
end function

atom t0 = time()
mpz p = mpz_init(10)
for k=2 to 10 do
mpz_mul_si(p,p,10)
sequence s = sort(powerful(p, k))
integer l = length(s)
s[6..-6] = {"..."}
s = ppf(s,{pp_FltFmt,"%d",pp_StrFmt,-1,pp_IntCh,false,pp_Maxlen,100})
printf(1,"%d %2d-powerful numbers <= 10^%-2d : %s\n",{l,k,k,s})
end for
puts(1,"\n")
for k=2 to 10 do
mpz_set_si(p,1)
sequence counts = {}
for j=0 to k+9 do
counts = append(counts, powercount(p, k))
mpz_mul_si(p,p,10)
end for
printf(1,"Counts of %2d-powerful numbers <= 10^(0..%d) : %V\n",{k,k+9,counts})
end for
?elapsed(time()-t0)
Output:
14  2-powerful numbers <= 10^2  : {1,4,8,9,16, ..., 49,64,72,81,100}
20  3-powerful numbers <= 10^3  : {1,8,16,27,32, ..., 625,648,729,864,1000}
25  4-powerful numbers <= 10^4  : {1,16,32,64,81, ..., 5184,6561,7776,8192,10000}
32  5-powerful numbers <= 10^5  : {1,32,64,128,243, ..., 65536,69984,78125,93312,100000}
38  6-powerful numbers <= 10^6  : {1,64,128,256,512, ..., 559872,746496,823543,839808,1000000}
46  7-powerful numbers <= 10^7  : {1,128,256,512,1024, ..., 7558272,8388608,8957952,9765625,10000000}
52  8-powerful numbers <= 10^8  : {1,256,512,1024,2048, ..., 60466176,67108864,80621568,90699264,100000000}
59  9-powerful numbers <= 10^9  : {1,512,1024,2048,4096, ..., 644972544,725594112,816293376,967458816,1000000000}
68 10-powerful numbers <= 10^10 : {1,1024,2048,4096,8192, ..., 7739670528,8589934592,8707129344,9795520512,10000000000}

Counts of  2-powerful numbers <= 10^(0..11) : {1,4,14,54,185,619,2027,6553,21044,67231,214122,680330}
Counts of  3-powerful numbers <= 10^(0..12) : {1,2,7,20,51,129,307,713,1645,3721,8348,18589,41136}
Counts of  4-powerful numbers <= 10^(0..13) : {1,1,5,11,25,57,117,235,464,906,1741,3312,6236,11654}
Counts of  5-powerful numbers <= 10^(0..14) : {1,1,3,8,16,32,63,117,211,375,659,1153,2000,3402,5770}
Counts of  6-powerful numbers <= 10^(0..15) : {1,1,2,6,12,21,38,70,121,206,335,551,900,1451,2326,3706}
Counts of  7-powerful numbers <= 10^(0..16) : {1,1,1,4,10,16,26,46,77,129,204,318,495,761,1172,1799,2740}
Counts of  8-powerful numbers <= 10^(0..17) : {1,1,1,3,8,13,19,32,52,85,135,211,315,467,689,1016,1496,2191}
Counts of  9-powerful numbers <= 10^(0..18) : {1,1,1,2,6,11,16,24,38,59,94,145,217,317,453,644,919,1308,1868}
Counts of 10-powerful numbers <= 10^(0..19) : {1,1,1,1,5,9,14,21,28,43,68,104,155,227,322,447,621,858,1192,1651}
"0.8s"

## Raku

(formerly Perl 6)

Works with: Rakudo version 2020.01
Translation of: Perl

Raku has no handy pre-made nth integer root routine so has the same problem as Go with needing a slight "fudge factor" in the nth root calculation.

sub super (\x) { x.trans([<0123456789>.comb] => [<⁰¹²³⁴⁵⁶⁷⁸⁹>.comb]) }

sub is-square-free (Int \n) {
constant @p = ^100 .map: { next unless .is-prime; .² };
for @p -> \p { return False if n %% p }
True
}

sub powerfuls (\n, \k, \enumerate = False) {
my @powerful;
p(1, 2*k - 1);
sub p (\m, \r) {
if r < k {
enumerate ?? @powerful.push(m) !! ++@powerful[m - 1 ?? (m - 1).chars !! 0];
return
}
for 1 .. ((n / m) ** (1/r) + .0001).Int -> \v {
if r > k {
next unless is-square-free(v);
next unless m gcd v == 1;
}
p(m * v ** r, r - 1)
}
}
@powerful;
}

put "Count and first and last five enumerated n-powerful numbers in 10ⁿ:";
for 2..10 -> \k {
my @powerful = sort powerfuls(10**k, k, True);
printf "%2d %2s-powerful numbers <= 10%-2s: %s ... %s\n", +@powerful, k, super(k),
}

put "\nCounts in each order of magnitude:";
my \$top = 9;
for 2..10 -> \k {
printf "%2s-powerful numbers <= 10ⁿ (where 0 <= n <= %d): ", k, \$top+k;
quietly say join ', ', [\+] powerfuls(10**(\$top + k), k);
}
Output:
Count and first and last five enumerated n-powerful numbers in 10ⁿ:
14  2-powerful numbers <= 10² : 1, 4, 8, 9, 16 ... 49, 64, 72, 81, 100
20  3-powerful numbers <= 10³ : 1, 8, 16, 27, 32 ... 625, 648, 729, 864, 1000
25  4-powerful numbers <= 10⁴ : 1, 16, 32, 64, 81 ... 5184, 6561, 7776, 8192, 10000
32  5-powerful numbers <= 10⁵ : 1, 32, 64, 128, 243 ... 65536, 69984, 78125, 93312, 100000
38  6-powerful numbers <= 10⁶ : 1, 64, 128, 256, 512 ... 559872, 746496, 823543, 839808, 1000000
46  7-powerful numbers <= 10⁷ : 1, 128, 256, 512, 1024 ... 7558272, 8388608, 8957952, 9765625, 10000000
52  8-powerful numbers <= 10⁸ : 1, 256, 512, 1024, 2048 ... 60466176, 67108864, 80621568, 90699264, 100000000
59  9-powerful numbers <= 10⁹ : 1, 512, 1024, 2048, 4096 ... 644972544, 725594112, 816293376, 967458816, 1000000000
68 10-powerful numbers <= 10¹⁰: 1, 1024, 2048, 4096, 8192 ... 7739670528, 8589934592, 8707129344, 9795520512, 10000000000

Counts in each order of magnitude:
2-powerful numbers <= 10ⁿ (where 0 <= n <= 11): 1, 4, 14, 54, 185, 619, 2027, 6553, 21044, 67231, 214122, 680330
3-powerful numbers <= 10ⁿ (where 0 <= n <= 12): 1, 2, 7, 20, 51, 129, 307, 713, 1645, 3721, 8348, 18589, 41136
4-powerful numbers <= 10ⁿ (where 0 <= n <= 13): 1, 1, 5, 11, 25, 57, 117, 235, 464, 906, 1741, 3312, 6236, 11654
5-powerful numbers <= 10ⁿ (where 0 <= n <= 14): 1, 1, 3, 8, 16, 32, 63, 117, 211, 375, 659, 1153, 2000, 3402, 5770
6-powerful numbers <= 10ⁿ (where 0 <= n <= 15): 1, 1, 2, 6, 12, 21, 38, 70, 121, 206, 335, 551, 900, 1451, 2326, 3706
7-powerful numbers <= 10ⁿ (where 0 <= n <= 16): 1, 1, 1, 4, 10, 16, 26, 46, 77, 129, 204, 318, 495, 761, 1172, 1799, 2740
8-powerful numbers <= 10ⁿ (where 0 <= n <= 17): 1, 1, 1, 3, 8, 13, 19, 32, 52, 85, 135, 211, 315, 467, 689, 1016, 1496, 2191
9-powerful numbers <= 10ⁿ (where 0 <= n <= 18): 1, 1, 1, 2, 6, 11, 16, 24, 38, 59, 94, 145, 217, 317, 453, 644, 919, 1308, 1868
10-powerful numbers <= 10ⁿ (where 0 <= n <= 19): 1, 1, 1, 1, 5, 9, 14, 21, 28, 43, 68, 104, 155, 227, 322, 447, 621, 858, 1192, 1651

## Sidef

### Generation

func powerful(n, k=2) {

var list = []

func (m,r) {
if (r < k) {
list << m
return nil
}
for a in (1 .. iroot(idiv(n,m), r)) {
if (r > k) {
a.is_coprime(m) || next
a.is_squarefree || next
}
__FUNC__(m * a**r, r-1)
}
}(1, 2*k - 1)

return list.sort
}

for k in (2..10) {
var a = powerful(10**k, k)
var t = a.tail(5).join(', ')
printf("For k=%-2d there are %d k-powerful numbers <= 10^k: [%s, ..., %s]\n", k, a.len, h, t)
}
Output:
For k=2  there are 14 k-powerful numbers <= 10^k: [1, 4, 8, 9, 16, ..., 49, 64, 72, 81, 100]
For k=3  there are 20 k-powerful numbers <= 10^k: [1, 8, 16, 27, 32, ..., 625, 648, 729, 864, 1000]
For k=4  there are 25 k-powerful numbers <= 10^k: [1, 16, 32, 64, 81, ..., 5184, 6561, 7776, 8192, 10000]
For k=5  there are 32 k-powerful numbers <= 10^k: [1, 32, 64, 128, 243, ..., 65536, 69984, 78125, 93312, 100000]
For k=6  there are 38 k-powerful numbers <= 10^k: [1, 64, 128, 256, 512, ..., 559872, 746496, 823543, 839808, 1000000]
For k=7  there are 46 k-powerful numbers <= 10^k: [1, 128, 256, 512, 1024, ..., 7558272, 8388608, 8957952, 9765625, 10000000]
For k=8  there are 52 k-powerful numbers <= 10^k: [1, 256, 512, 1024, 2048, ..., 60466176, 67108864, 80621568, 90699264, 100000000]
For k=9  there are 59 k-powerful numbers <= 10^k: [1, 512, 1024, 2048, 4096, ..., 644972544, 725594112, 816293376, 967458816, 1000000000]
For k=10 there are 68 k-powerful numbers <= 10^k: [1, 1024, 2048, 4096, 8192, ..., 7739670528, 8589934592, 8707129344, 9795520512, 10000000000]

### Counting

func powerful_count(n, k=2) {

var count = 0

func (m,r) {
if (r <= k) {
count += iroot(idiv(n,m), r)
return nil
}
for a in (1 .. iroot(idiv(n,m), r)) {
a.is_coprime(m) || next
a.is_squarefree || next
__FUNC__(m * a**r, r-1)
}
}(1, 2*k - 1)

return count
}

for k in (2..10) {
var a = (k+10).of {|j| powerful_count(10**j, k) }
printf("Number of %2d-powerful numbers <= 10^j, for 0 <= j < #{k+10}: %s\n", k, a)
}
Output:
Number of  2-powerful numbers <= 10^j, for 0 <= j < 12: [1, 4, 14, 54, 185, 619, 2027, 6553, 21044, 67231, 214122, 680330]
Number of  3-powerful numbers <= 10^j, for 0 <= j < 13: [1, 2, 7, 20, 51, 129, 307, 713, 1645, 3721, 8348, 18589, 41136]
Number of  4-powerful numbers <= 10^j, for 0 <= j < 14: [1, 1, 5, 11, 25, 57, 117, 235, 464, 906, 1741, 3312, 6236, 11654]
Number of  5-powerful numbers <= 10^j, for 0 <= j < 15: [1, 1, 3, 8, 16, 32, 63, 117, 211, 375, 659, 1153, 2000, 3402, 5770]
Number of  6-powerful numbers <= 10^j, for 0 <= j < 16: [1, 1, 2, 6, 12, 21, 38, 70, 121, 206, 335, 551, 900, 1451, 2326, 3706]
Number of  7-powerful numbers <= 10^j, for 0 <= j < 17: [1, 1, 1, 4, 10, 16, 26, 46, 77, 129, 204, 318, 495, 761, 1172, 1799, 2740]
Number of  8-powerful numbers <= 10^j, for 0 <= j < 18: [1, 1, 1, 3, 8, 13, 19, 32, 52, 85, 135, 211, 315, 467, 689, 1016, 1496, 2191]
Number of  9-powerful numbers <= 10^j, for 0 <= j < 19: [1, 1, 1, 2, 6, 11, 16, 24, 38, 59, 94, 145, 217, 317, 453, 644, 919, 1308, 1868]
Number of 10-powerful numbers <= 10^j, for 0 <= j < 20: [1, 1, 1, 1, 5, 9, 14, 21, 28, 43, 68, 104, 155, 227, 322, 447, 621, 858, 1192, 1651]

## Wren

Translation of: Java
Library: Wren-big
Library: Wren-set
Library: Wren-sort
Library: Wren-fmt
import "/big" for BigInt
import "/set" for Set
import "/sort" for Sort
import "/fmt" for Fmt

var potentialPowerful = Fn.new { |max, k|
var indexes = List.filled(k, 1)
var powerful = Set.new()
var foundPower = true
while (foundPower) {
var genPowerful = false
for (index in 0...k) {
var power = BigInt.one
for (i in 0...k) power = power * BigInt.new(indexes[i]).pow(k+i)
if (power <= max) {
indexes[0] = indexes[0] + 1
genPowerful = true
break
}
indexes[index] = 1
if (index < k - 1) indexes[index+1] = indexes[index+1] + 1
}
if (!genPowerful) foundPower = false
}
return powerful.map { |p| BigInt.new(p) }.toList
}

var countPowerfulNumbers = Fn.new{ |max, k| potentialPowerful.call(max, k).count }

var getPowerfulNumbers = Fn.new { |max, k|
var powerfulNumbers = potentialPowerful.call(max, k)
Sort.quick(powerfulNumbers)
return powerfulNumbers
}

for (k in 2..10) {
var max = BigInt.ten.pow(k)
var powerfulNumbers = getPowerfulNumbers.call(max, k)
var count = powerfulNumbers.count
Fmt.print("There are \$d \$d-powerful numbers between 1 and \$i", count, k, max)
Fmt.print("List: [\$i ... \$i]", powerfulNumbers[0..4], powerfulNumbers[-5..-1])
}
System.print()
for (k in 2..10) {
var powCount = []
for (j in 0...k+10) {
var max = BigInt.ten.pow(j)
}
Fmt.print("Count of \$2d-powerful numbers <= 10^j, j in [0, \$d]: \$n", k, k + 9, powCount)
}
Output:
There are 14 2-powerful numbers between 1 and 100
List: [1 4 8 9 16 ... 49 64 72 81 100]
There are 20 3-powerful numbers between 1 and 1000
List: [1 8 16 27 32 ... 625 648 729 864 1000]
There are 25 4-powerful numbers between 1 and 10000
List: [1 16 32 64 81 ... 5184 6561 7776 8192 10000]
There are 32 5-powerful numbers between 1 and 100000
List: [1 32 64 128 243 ... 65536 69984 78125 93312 100000]
There are 38 6-powerful numbers between 1 and 1000000
List: [1 64 128 256 512 ... 559872 746496 823543 839808 1000000]
There are 46 7-powerful numbers between 1 and 10000000
List: [1 128 256 512 1024 ... 7558272 8388608 8957952 9765625 10000000]
There are 52 8-powerful numbers between 1 and 100000000
List: [1 256 512 1024 2048 ... 60466176 67108864 80621568 90699264 100000000]
There are 59 9-powerful numbers between 1 and 1000000000
List: [1 512 1024 2048 4096 ... 644972544 725594112 816293376 967458816 1000000000]
There are 68 10-powerful numbers between 1 and 10000000000
List: [1 1024 2048 4096 8192 ... 7739670528 8589934592 8707129344 9795520512 10000000000]

Count of  2-powerful numbers <= 10^j, j in [0, 11]: [1, 4, 14, 54, 185, 619, 2027, 6553, 21044, 67231, 214122, 680330]
Count of  3-powerful numbers <= 10^j, j in [0, 12]: [1, 2, 7, 20, 51, 129, 307, 713, 1645, 3721, 8348, 18589, 41136]
Count of  4-powerful numbers <= 10^j, j in [0, 13]: [1, 1, 5, 11, 25, 57, 117, 235, 464, 906, 1741, 3312, 6236, 11654]
Count of  5-powerful numbers <= 10^j, j in [0, 14]: [1, 1, 3, 8, 16, 32, 63, 117, 211, 375, 659, 1153, 2000, 3402, 5770]
Count of  6-powerful numbers <= 10^j, j in [0, 15]: [1, 1, 2, 6, 12, 21, 38, 70, 121, 206, 335, 551, 900, 1451, 2326, 3706]
Count of  7-powerful numbers <= 10^j, j in [0, 16]: [1, 1, 1, 4, 10, 16, 26, 46, 77, 129, 204, 318, 495, 761, 1172, 1799, 2740]
Count of  8-powerful numbers <= 10^j, j in [0, 17]: [1, 1, 1, 3, 8, 13, 19, 32, 52, 85, 135, 211, 315, 467, 689, 1016, 1496, 2191]
Count of  9-powerful numbers <= 10^j, j in [0, 18]: [1, 1, 1, 2, 6, 11, 16, 24, 38, 59, 94, 145, 217, 317, 453, 644, 919, 1308, 1868]
Count of 10-powerful numbers <= 10^j, j in [0, 19]: [1, 1, 1, 1, 5, 9, 14, 21, 28, 43, 68, 104, 155, 227, 322, 447, 621, 858, 1192, 1651]

## zkl

Translation of: Go
fcn isSquareFree(n){
var [const] primes2=T(2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97,) // seems to be enough
.apply("pow",2);
foreach p2 in (primes2){
if(p2>n) break;
if(n%p2 == 0) return(False);
}
True
}

const adj=0.0001; // adjustment to ensure f/r square root exact for perfect integer squares
fcn iroot(n,r){ (n.toFloat().pow(1.0/r) + adj).toInt() }

fcn powerfuls(n,k){
fcn(m,r, n,k,powerful){ // recursive closure
if(r<k){ powerful[m]=True; return(); }
foreach v in ([1..iroot(n/m, r)]){
if(r>k){ if(not isSquareFree(v) or m.gcd(v)!=1) continue; }
self.fcn(m*v.pow(r), r-1, n,k,powerful)
}
}(1,(2).pow(k)-1, n,k, powerful:=Dictionary());
powerful.keys.apply("toInt").sort();
}
println("Count, first and last five enumerated n-powerful numbers in 10\u207f:");
foreach k in ([2..10]){
ps:=powerfuls((10).pow(k),k);
println("%2d: %s ... %s".fmt(ps.len(),ps[0,5].concat(" "),ps.tail(5).concat(" ")));
}
Output:
Count, first and last five enumerated n-powerful numbers in 10ⁿ:
14: 1 4 8 9 16 ... 49 64 72 81 100
20: 1 8 16 27 32 ... 625 648 729 864 1000
25: 1 16 32 64 81 ... 5184 6561 7776 8192 10000
32: 1 32 64 128 243 ... 65536 69984 78125 93312 100000
38: 1 64 128 256 512 ... 559872 746496 823543 839808 1000000
46: 1 128 256 512 1024 ... 7558272 8388608 8957952 9765625 10000000
52: 1 256 512 1024 2048 ... 60466176 67108864 80621568 90699264 100000000
59: 1 512 1024 2048 4096 ... 644972544 725594112 816293376 967458816 1000000000
68: 1 1024 2048 4096 8192 ... 7739670528 8589934592 8707129344 9795520512 10000000000
Translation of: Perl

Overflows a 64 bit int at the very end, which results in a count of zero.

fcn powerfulsN(n,k){	// count
fcn(m,r, n,k,count){ // recursive closure
if(r<=k){ count.incN(iroot(n/m, r)); return() }
foreach v in ([1..iroot(n/m, r)]){
if(not isSquareFree(v) or m.gcd(v)!=1) continue;
self.fcn(m*v.pow(r), r-1, n,k,count)
}
}(1,2*k - 1, n,k, count:=Ref(0));
count.value
}
println("Counts in each order of magnitude:");
foreach k in ([2..10]){
print("%2s-powerful numbers <= 10\u207f (n in [0..%d)): ".fmt(k, 10+k));
ps:=[0 .. k+10 -1].apply('wrap(n){ powerfulsN((10).pow(n), k) });
println(ps.concat(" "));
}
Output:
Counts in each order of magnitude:
2-powerful numbers <= 10ⁿ (n in [0..12)): 1 4 14 54 185 619 2027 6553 21044 67231 214122 680330
3-powerful numbers <= 10ⁿ (n in [0..13)): 1 2 7 20 51 129 307 713 1645 3721 8348 18589 41136
4-powerful numbers <= 10ⁿ (n in [0..14)): 1 1 5 11 25 57 117 235 464 906 1741 3312 6236 11654
5-powerful numbers <= 10ⁿ (n in [0..15)): 1 1 3 8 16 32 63 117 211 375 659 1153 2000 3402 5770
6-powerful numbers <= 10ⁿ (n in [0..16)): 1 1 2 6 12 21 38 70 121 206 335 551 900 1451 2326 3706
7-powerful numbers <= 10ⁿ (n in [0..17)): 1 1 1 4 10 16 26 46 77 129 204 318 495 761 1172 1799 2740
8-powerful numbers <= 10ⁿ (n in [0..18)): 1 1 1 3 8 13 19 32 52 85 135 211 315 467 689 1016 1496 2191
9-powerful numbers <= 10ⁿ (n in [0..19)): 1 1 1 2 6 11 16 24 38 59 94 145 217 317 453 644 919 1308 1868
10-powerful numbers <= 10ⁿ (n in [0..20)): 1 1 1 1 5 9 14 21 28 43 68 104 155 227 322 447 621 858 1192 0