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.

Phix

Same approach as Wren, but using digitwise maths and reduced limits on the second part to keep things sane.

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 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)
printf(1,"\nFirst eighteen that satisfy that condition in three or more ways:\n")
--exponential_digital_sums(30, 3, 500) -- 48s, output matches Wren
exponential_digital_sums(18, 3, 150) -- 2.6s on the desktop, 3.1s in a web browser
?elapsed(time()-t0)

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 eighteen 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
"2.6s"

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.

This agrees with the Raku solution for the first part but finds some other qualifying numbers for the second part.

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.

Arguably both 0 and 1 should qualify as all their powers will have digit sums of 0 and 1 respectively. However, as the Raku solution hasn't included them, neither have I.

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 { |max, minWays, limit|
    var i = Mpz.one
    var c = 0
    var n = Mpz.new()
    while (c < max) {
        i.inc
        n.set(i)
        var p = 1
        var res = []
        while (p < limit) {
            n.mul(i)
            p = p + 1
            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