Exponential digital sums

Exponential digital sums 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.

Some integers (greater than 1), have the property that the digital sum of that integer raised to some integer power greater than 1, is equal to the original integer.

E.G.

92 == 81

8 + 1 == 9

Some integers have this property using more than one exponent.

183 == 5832	5 + 8 + 3 + 2 == 18
186 == 34012224	3 + 4 + 0 + 1 + 2 + 2 + 2 + 4 == 18
187 == 612220032	6 + 1 + 2 + 2 + 2 + 0 + 0 + 3 + 2 == 18

Note: every integer has an exponential digital sum equal to the original integer when using an exponent of 1. And, 0 and 1 raised to any power will have a digital sum of 0 and 1 respectively.

Task

• Find and show the first twenty integers (with their exponents), that satisfy this condition.
• Find and show at least the first ten integers with their exponents, that satisfy this condition in three or more ways.

C++

Translation of: Wren

Library: GMP

#include <iostream>
#include <vector>

#include <gmpxx.h>

using big_int = mpz_class;

int digit_sum(const big_int& n) {
    int sum = 0;
    for (char c : n.get_str())
        sum += c - '0';
    return sum;
}

void exp_digit_sums(int count, int min_ways, int max_power) {
    for (int i = 2; count > 0; ++i) {
        big_int n = i;
        std::vector<int> powers;
        for (int p = 2; p <= max_power; ++p) {
            n *= i;
            if (digit_sum(n) == i)
                powers.push_back(p);
        }
        if (powers.size() >= min_ways) {
            --count;
            auto it = powers.begin();
            std::cout << i << "^" << *it++;
            for (; it != powers.end(); ++it)
                std::cout << ", " << i << "^" << *it;
            std::cout << '\n';
        }
    }
}

int main() {
    std::cout << "First twenty-five integers that are equal to the digital sum "
                 "of that integer raised to some power:\n";
    exp_digit_sums(25, 1, 100);
    std::cout << "\nFirst thirty that satisfy that condition in three or more "
                 "ways:\n";
    exp_digit_sums(30, 3, 500);
}

Output:

First twenty-five integers that are equal to the digital sum of that integer raised to some power:
7^4
8^3
9^2
17^3
18^3, 18^6, 18^7
20^13
22^4
25^4
26^3
27^3, 27^7
28^4, 28^5
31^7
34^7
35^5
36^4, 36^5
40^13
43^7
45^6
46^5, 46^8
53^7
54^6, 54^8, 54^9
58^7
63^8
64^6
68^7

First thirty that satisfy that condition in three or more ways:
18^3, 18^6, 18^7
54^6, 54^8, 54^9
90^19, 90^20, 90^21, 90^22, 90^28
107^11, 107^13, 107^15
181^16, 181^18, 181^19, 181^20
360^45, 360^46, 360^49, 360^51
370^48, 370^54, 370^57, 370^59
388^32, 388^35, 388^36
523^39, 523^42, 523^44, 523^45
603^44, 603^47, 603^54
667^48, 667^54, 667^58
793^57, 793^60, 793^64
1062^72, 1062^77, 1062^81
1134^78, 1134^80, 1134^82, 1134^86
1359^92, 1359^98, 1359^102
1827^121, 1827^126, 1827^131
1828^123, 1828^127, 1828^132
2116^140, 2116^143, 2116^147
2330^213, 2330^215, 2330^229
2430^217, 2430^222, 2430^223, 2430^229, 2430^230
2557^161, 2557^166, 2557^174
2610^228, 2610^244, 2610^246
2656^170, 2656^172, 2656^176
2700^406, 2700^414, 2700^420, 2700^427
2871^177, 2871^189, 2871^190
2934^191, 2934^193, 2934^195
3077^187, 3077^193, 3077^199
3222^189, 3222^202, 3222^210
3231^203, 3231^207, 3231^209
3448^215, 3448^221, 3448^227

jq

Adapted from Wren

Works with gojq, the Go implementation of jq

gojq has infinite-precision integer arithmetic, but when working on the second task, becomes noticeably slow after producing the first ten lines or so.

# whilst/2 is like while/2 but emits the final term rather than the first one
def whilst(cond; update):
     def _whilst:
         if cond then update | (., _whilst) else empty end;
     _whilst;

def digitSum:
  def add(s): reduce s as $_ (0; . + $_);
  add(tostring | explode[] | . - 48);

def expDigitSums(numNeeded; minWays; maxPower):
  {i: 1, c: 0}
  | whilst(.c < numNeeded;
      .i += 1
      | .n = .i
      | reduce range(2; 1+ maxPower) as $p (.res = [];
          .n *= .i
          | if .i == (.n|digitSum)
            then .res += ["\(.i)^\($p)"]
            end)
      | if .res|length >= minWays
        then .c += 1
        end )
  | select(.res|length >= minWays)
  | .res | join(", ");

"First twenty-five integers that are equal to the digital sum of that integer raised to some power:",
expDigitSums(25; 1; 100),

"\nFirst thirty that satisfy that condition in three or more ways:",
expDigitSums(30; 3; 500)

Output:

First twenty-five integers that are equal to the digital sum of that integer raised to some power:
7^4
8^3
9^2
17^3
18^3, 18^6, 18^7
20^13
22^4
25^4
26^3
27^3, 27^7
28^4, 28^5
31^7
34^7
35^5
36^4, 36^5
40^13
43^7
45^6
46^5, 46^8
53^7
54^6, 54^8, 54^9
58^7
63^8
64^6
68^7

First thirty that satisfy that condition in three or more ways:
18^3, 18^6, 18^7
54^6, 54^8, 54^9
90^19, 90^20, 90^21, 90^22, 90^28
107^11, 107^13, 107^15
181^16, 181^18, 181^19, 181^20
360^45, 360^46, 360^49, 360^51
370^48, 370^54, 370^57, 370^59
388^32, 388^35, 388^36
523^39, 523^42, 523^44, 523^45
603^44, 603^47, 603^54
667^48, 667^54, 667^58
793^57, 793^60, 793^64
1062^72, 1062^77, 1062^81
1134^78, 1134^80, 1134^82, 1134^86
1359^92, 1359^98, 1359^102
1827^121, 1827^126, 1827^131
1828^123, 1828^127, 1828^132
2116^140, 2116^143, 2116^147
2330^213, 2330^215, 2330^229
2430^217, 2430^222, 2430^223, 2430^229, 2430^230
2557^161, 2557^166, 2557^174
2610^228, 2610^244, 2610^246
2656^170, 2656^172, 2656^176
2700^406, 2700^414, 2700^420, 2700^427
2871^177, 2871^189, 2871^190
2934^191, 2934^193, 2934^195
3077^187, 3077^193, 3077^199
3222^189, 3222^202, 3222^210
3231^203, 3231^207, 3231^209
3448^215, 3448^221, 3448^227

Julia

function equal_digitalsum_exponents(n::Integer)
    equalpows = Int[]
    if n > 1
        npow, misses = 2, 0
        while misses < n + 50
            dsum = sum(digits(BigInt(n) ^ npow))
            if npow > 10 && dsum > 2 * n
                break # bail here for time contraints (see Wren example)
            elseif dsum == n
                push!(equalpows, npow)
            else
                misses += 1
            end
            npow += 1
        end
    end
    return equalpows
end

function testdigitalequals(lim, wanted, multiswanted)
    found1, found2, multis = 0, 0, Tuple{Int, Vector{Int}}[]
    println("First $wanted integers that are equal to the digital sum of that integer raised to some power:")
    for i in 1:lim
        a = equal_digitalsum_exponents(i)
        if !isempty(a)
            found1 += 1
            if found1 <= wanted
                print(rpad(i, 6), found1 % 10 == 0 ? "\n" : "")
            end
            if length(a) > 2
                found2 += 1
                push!(multis, (i, a))
                if found2 == multiswanted
                    println("\nFirst $multiswanted that satisfy that condition in three or more ways:")
                    for (n, v) in multis
                        println("$n: powers $v")
                    end
                end
                found1 >= wanted && found2 >= multiswanted && break
            end
        end
    end
end

testdigitalequals(6000, 25, 30)

Output:

First 25 integers that are equal to the digital sum of that integer raised to some power:
7     8     9     17    18    20    22    25    26    27    
28    31    34    35    36    40    43    45    46    53    
54    58    63    64    68
First 30 that satisfy that condition in three or more ways:
18: powers [3, 6, 7]
54: powers [6, 8, 9]
90: powers [19, 20, 21, 22, 28]
107: powers [11, 13, 15]
181: powers [16, 18, 19, 20]
360: powers [45, 46, 49, 51]
370: powers [48, 54, 57, 59]
388: powers [32, 35, 36]
523: powers [39, 42, 44, 45]
603: powers [44, 47, 54]
667: powers [48, 54, 58]
793: powers [57, 60, 64]
1062: powers [72, 77, 81]
1134: powers [78, 80, 82, 86]
1359: powers [92, 98, 102]
1827: powers [121, 126, 131]
1828: powers [123, 127, 132]
2116: powers [140, 143, 147]
2330: powers [213, 215, 229]
2430: powers [217, 222, 223, 229, 230]
2557: powers [161, 166, 174]
2610: powers [228, 244, 246]
2656: powers [170, 172, 176]
2700: powers [406, 414, 420, 427]
2871: powers [177, 189, 190]
2934: powers [191, 193, 195]
3077: powers [187, 193, 199]
3222: powers [189, 202, 210]
3231: powers [203, 207, 209]
3448: powers [215, 221, 227]

Phix

Same approach as Wren, but using digitwise maths and calculating the digital sum at the same time.

with javascript_semantics
procedure exponential_digital_sums(integer showfirst, minways, maxpower)
    integer i = 1, shown = 0
    while shown<showfirst do
        i += 1
        sequence n = {i}, res = {}
        for p=2 to maxpower do
            integer ds = 0, d
            atom carry = 0
            for j=1 to length(n) do
                carry += n[j]*i
                d = remainder(carry,10)
                carry = floor(carry/10)
                n[j] = d
                ds += d
            end for
            while carry do
                d = remainder(carry,10)
                carry = floor(carry/10)
                n &= d
                ds += d
            end while
            if p>10 and ds>2*i then exit end if
            if ds=i then
                res = append(res,sprintf("%d^%d",{i,p}))
            end if
        end for
        if length(res)>=minways then
            printf(1,"%s\n",{join(res,", ")})
            shown += 1
        end if
    end while
end procedure
atom t0 = time()
printf(1,"First twenty-five integers that are equal to the digital sum of that integer raised to some power:\n")
exponential_digital_sums(25, 1, 100)
integer {n,mp} = iff(platform()=JS?{18,150}:{30,500}) -- 14.4s on desktop, up to 2116 in 3.1s on JS
printf(1,"\nFirst %s that satisfy that condition in three or more ways:\n",{ordinal(n,true)})
exponential_digital_sums(n, 3, mp)
?elapsed(time()-t0)

Output same as Wren/Raku, second part stops on the 18th (2116) under JS but would match after a 26.5s blank screen if you used the same limits

Python

""" rosettacode.org/wiki/Exponential_digital_sums """


def equal_digitalsum_exponents(num):
    """ return any equal exponential digital sums in an array """
    equalpows = []
    if num > 1:
        npow, misses = 2, 0
        while misses < num + 20:
            dsum = sum(map(int, str(num**npow)))
            if npow > 10 and dsum > 2 * num:
                break  # bail here for time contraints (see Wren example)
            if dsum == num:
                equalpows.append(npow)
            else:
                misses += 1
            npow += 1

    return equalpows


if __name__ == '__main__':

    found1, found2, multis = 0, 0, []
    print('First 25 integers that are equal to the digital sum of the number raised to some power:')
    for i in range(1, 4000):
        a = equal_digitalsum_exponents(i)
        if a:
            S_EXP = ', '.join(f'{i}^{j}' for j in a)
            found1 += 1
            if found1 <= 25:
                print(S_EXP)
            if len(a) > 2:
                found2 += 1
                multis.append(S_EXP)
                if found2 == 30:
                    print(
                        '\n\nFirst 30 that satisfy that condition in three or more ways:')
                    for grp in multis:
                        print(grp)
                if found1 >= 25 and found2 >= 30:
                    break

Output:

Same as Raku.

Raku

Implement a lazy generator. Made some assumptions about search limits. May be poor assumptions, but haven't been able to find any counterexamples. (Edit: and some were bad.)

my @expsum = lazy (2..*).hyper.map( -> $Int {
    my atomicint $miss = 0;
    (2..$Int).map( -> $exp {
        if (my $sum = ($Int ** $exp).comb.sum) > $Int { last if ++⚛$miss > 20 }
        $sum == $Int ?? "$Int^$exp" !! Empty;
    }) || Empty;
});

say "First twenty-five integers that are equal to the digital sum of that integer raised to some power:";
put .join(', ') for @expsum[^25];
say "\nFirst thirty that satisfy that condition in three or more ways:";
put .join(', ') for @expsum.grep({.elems≥3})[^30];

Output:

First twenty-five integers that are equal to the digital sum of that integer raised to some power:
7^4
8^3
9^2
17^3
18^3, 18^6, 18^7
20^13
22^4
25^4
26^3
27^3, 27^7
28^4, 28^5
31^7
34^7
35^5
36^4, 36^5
40^13
43^7
45^6
46^5, 46^8
53^7
54^6, 54^8, 54^9
58^7
63^8
64^6
68^7

First thirty that satisfy that condition in three or more ways:
18^3, 18^6, 18^7
54^6, 54^8, 54^9
90^19, 90^20, 90^21, 90^22, 90^28
107^11, 107^13, 107^15
181^16, 181^18, 181^19, 181^20
360^45, 360^46, 360^49, 360^51
370^48, 370^54, 370^57, 370^59
388^32, 388^35, 388^36
523^39, 523^42, 523^44, 523^45
603^44, 603^47, 603^54
667^48, 667^54, 667^58
793^57, 793^60, 793^64
1062^72, 1062^77, 1062^81
1134^78, 1134^80, 1134^82, 1134^86
1359^92, 1359^98, 1359^102
1827^121, 1827^126, 1827^131
1828^123, 1828^127, 1828^132
2116^140, 2116^143, 2116^147
2330^213, 2330^215, 2330^229
2430^217, 2430^222, 2430^223, 2430^229, 2430^230
2557^161, 2557^166, 2557^174
2610^228, 2610^244, 2610^246
2656^170, 2656^172, 2656^176
2700^406, 2700^414, 2700^420, 2700^427
2871^177, 2871^189, 2871^190
2934^191, 2934^193, 2934^195
3077^187, 3077^193, 3077^199
3222^189, 3222^202, 3222^210
3231^203, 3231^207, 3231^209
3448^215, 3448^221, 3448^227

Wren

Library: Wren-gmp

Well, as the digital sums are all over the place, it's difficult to know how many powers of each number should be tested to solve the task. We therefore have to make an educated guess.

Although it would be possible to use BigInt for this, GMP has been used instead to quicken up the search but even so still takes 110 seconds to run.

import "./gmp" for Mpz

var digitSum = Fn.new { |bi|
    var sum = 0
    for (d in bi.toString.bytes) {
        sum = sum + d - 48
    }
    return sum
}

var expDigitSums = Fn.new { |numNeeded, minWays, maxPower|
    var i = Mpz.one
    var c = 0
    var n = Mpz.new()
    while (c < numNeeded) {
        i.inc
        n.set(i)
        var res = []
        for (p in 2..maxPower) {
            n.mul(i)
            var ds = digitSum.call(n)
            if (i == ds) {
                res.add("%(i)^%(p)")
            }
        }
        if (res.count >= minWays) {
            System.print(res.join(", "))
            c = c + 1
        }
    }
}

System.print("First twenty-five integers that are equal to the digital sum of that integer raised to some power:")
expDigitSums.call(25, 1, 100)

System.print("\nFirst thirty that satisfy that condition in three or more ways:")
expDigitSums.call(30, 3, 500)

Output:

First twenty-five integers that are equal to the digital sum of that integer raised to some power:
7^4
8^3
9^2
17^3
18^3, 18^6, 18^7
20^13
22^4
25^4
26^3
27^3, 27^7
28^4, 28^5
31^7
34^7
35^5
36^4, 36^5
40^13
43^7
45^6
46^5, 46^8
53^7
54^6, 54^8, 54^9
58^7
63^8
64^6
68^7

First thirty that satisfy that condition in three or more ways:
18^3, 18^6, 18^7
54^6, 54^8, 54^9
90^19, 90^20, 90^21, 90^22, 90^28
107^11, 107^13, 107^15
181^16, 181^18, 181^19, 181^20
360^45, 360^46, 360^49, 360^51
370^48, 370^54, 370^57, 370^59
388^32, 388^35, 388^36
523^39, 523^42, 523^44, 523^45
603^44, 603^47, 603^54
667^48, 667^54, 667^58
793^57, 793^60, 793^64
1062^72, 1062^77, 1062^81
1134^78, 1134^80, 1134^82, 1134^86
1359^92, 1359^98, 1359^102
1827^121, 1827^126, 1827^131
1828^123, 1828^127, 1828^132
2116^140, 2116^143, 2116^147
2330^213, 2330^215, 2330^229
2430^217, 2430^222, 2430^223, 2430^229, 2430^230
2557^161, 2557^166, 2557^174
2610^228, 2610^244, 2610^246
2656^170, 2656^172, 2656^176
2700^406, 2700^414, 2700^420, 2700^427
2871^177, 2871^189, 2871^190
2934^191, 2934^193, 2934^195
3077^187, 3077^193, 3077^199
3222^189, 3222^202, 3222^210
3231^203, 3231^207, 3231^209
3448^215, 3448^221, 3448^227

After studying changes in the digital sum as the power increases, one thing I've noticed is that once you get past the first 10 powers, if the point is reached when the digital sum is more than twice the number then it seems to be safe to bail out at that point.

So, if we change the expDigitSums function to the following:

var expDigitSums = Fn.new { |numNeeded, minWays, maxPower|
    var i = Mpz.one
    var c = 0
    var n = Mpz.new()
    while (c < numNeeded) {
        i.inc
        n.set(i)
        var res = []
        for (p in 2..maxPower) {
            n.mul(i)
            var ds = digitSum.call(n)
            if (i == ds) {
                res.add("%(i)^%(p)")
            }
            if (p > 10 && i * 2 < ds) break  // added this line
        }
        if (res.count >= minWays) {
            System.print(res.join(", "))
            c = c + 1
        }
    }
}

the output is the same but the time taken comes down to 36 seconds.

Not very scientific but neither for that matter is guessing how many powers to test.