Exponential digital sums: Difference between revisions
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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. |
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. |
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So, if we change the '' |
So, if we change the ''expDigitSums'' function to the following: |
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<syntaxhighlight lang="ecmascript">var expDigitSums = Fn.new { |numNeeded, minWays, maxPower| |
<syntaxhighlight lang="ecmascript">var expDigitSums = Fn.new { |numNeeded, minWays, maxPower| |
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var i = Mpz.one |
var i = Mpz.one |
Revision as of 09:37, 11 August 2023
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
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.