Eureka
September 24, 2019
Somebody’s homework today:
A number is a eureka number if the sum of the powers of its digits, with powers increasing, is equal to the number. For instance, 89 is a eureka number because 8**1 + 9**2 = 89, and 1306 is a eureka number because 1**1 + 3**2 + 6**4 = 1306.
Your task is to write a program to identify eureka numbers, and determine how many eureka numbers exist that are less than a million. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Programming Praxis 
Here is my take on it using Julia 1.1.1: https://pastebin.com/vw8CzNVC
To be honest, I was expecting a few more Eureka numbers in this range. Anyway, good drill to warm up for some more challenging coding. Cheers!
Here is my solution using Klong:
l::1+&{[i]; i::0; x=+/{(1:$"",x)^i::i+1}'$x}'!1000000; .p(l); .p(#l);"Done" [1 2 3 4 5 6 7 8 9 10 90 136 176 519 599 1307 1677 2428] 18 "Done"I really should have proofread before I changed my solution starting with 1, to a solution starting with 0:
Klong version
Right l::1+&{[i]; i::0; x=+/{(1:$"",x)^i::i+1}'$x}'1+!999999; .p(l); .p(#l);"" [1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427] 17 "" --- Wrong l::1+&{[i]; i::0; x=+/{(1:$"",x)^i::i+1}'$x}'!1000000; .p(l); .p(#l);"Done" [1 2 3 4 5 6 7 8 9 10 90 136 176 519 599 1307 1677 2428] 18 "Done" --- Right l::&{[i]; i::0; x=+/{(1:$"",x)^i::i+1}'$x}'!1000000; .p(l); .p(#l); "Done" [0 1 2 3 4 5 6 7 8 9 89 135 175 518 598 1306 1676 2427] 18 "Done"Here’s a solution in C.
#include <stdbool.h> #include <stdio.h> #include <stdlib.h> // exponentiation by squaring int ipow(int x, int y) { int result = 1; while (true) { if (y & 1) result *= x; y >>= 1; if (!y) break; x *= x; } return result; } bool is_eureka(int x) { int n = 0; // number of digits int y = x; while (x) { ++n; x /= 10; } x = y; int sum = 0; for (int i = n; i > 0; --i) { int z = x % 10; sum += ipow(z, i); x /= 10; } return sum == y; } int main(void) { int count = 0; for (int i = 0; i < 1000000; ++i) { if (is_eureka(i)) ++count; } printf("%d\n", count); return EXIT_SUCCESS; }Output:
Here’s a quick and dirty Haskell version.
{-# LANGUAGE ScopedTypeVariables #-} import Data.List (foldl', unfoldr) import Data.Tuple (swap) -- The list of a number's digits. The least significant digits are first. The -- empty list represents 0. toDigits :: Integral a => a -> [a] toDigits n = unfoldr step n where step 0 = Nothing step i = Just $ swap $ i `quotRem` 10 -- True iff n is a eureka number, where n is assumed to be >= 0. isEureka :: Integral a => a -> Bool isEureka n = let m = foldl' step 0 $ zip [1..] $ reverse $ toDigits n in m == n where step s (pow :: a, base :: a) = s + base^pow main :: IO () main = do print $ length $ filter isEureka [0..999999 :: Int]On the OEIS page on this series MAPLE code was given to calculate all these numbers faster. I translated the code into Python. It finds all numbers within 10 minutes an takes about 45 minutes to try all the numbers up to 22 digits (the maximum possible). The code uses a branch and bound strategy. The commented code and the output is on Ideone.
def eureka1(ndigits): b = [[x ** (j+1) - x * 10 ** (ndigits - (j+1)) for x in range(10)] for j in range(ndigits)] smin = [min(bi) for bi in b] # min value for every row of b smax = [max(bi) for bi in b] # max value for every row of b perm = sorted(range(ndigits), key=lambda j: smax[j]-smin[j], reverse=True) scmin = [sum(smin[perm[i]] for i in range(j+1, ndigits)) for j in range(ndigits)] scmax = [sum(smax[perm[i]] for i in range(j+1, ndigits)) for j in range(ndigits)] b = [b[j] for j in perm] def branch(): Q = [(0, 0, [])] while Q: pos, sofar, num = Q.pop() x0 = 1 if perm[pos] == 0 else 0 for x in range(x0, 10): current = sofar + b[pos][x] if scmin[pos] <= -current <= scmax[pos]: if pos == ndigits-1: yield num + [x] else: Q.append((pos+1, current, num + [x])) for num in branch(): decimal = sum(d*10**(ndigits-perm[i]-1) for i, d in enumerate(num)) print(decimal) for ndigits in range(1, 25): eureka1(ndigits)@Paul – thanks for posting that, I’d looked at the OEIS Maple code but couldn’t make head or tail of it, though I suspected it was doing something clever as you have revealed.