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%I #30 Apr 30 2022 09:16:19
%S 2,3,6,9,12,15,18,20,21,24,27,30,33,36,38,39,42,45,48,51,54,56,57,63,
%T 66,69,72,74,75,78,81,84,87,90,92,93,96,99,102,105,108,110,111,114,
%U 117,120,123,126,129,132,135,141,144,146,147,150,153,156,159,162
%N Integers k for which there exists some integer m such that the sum of the digits of m^k is equal to m + k.
%C Letting t = m^k, this sequence consists of the integers k for which there exists some integer m such that s(t) = m + k, where s(t) = A007953(t) represents the sum of digits of t. Rearranging gives m = t^(1/k) = s(t) - k; this allows you to find numbers which follow a common online trick like 64^(1/2) = (6 + 4) - 2 or 216^(1/3) = (2 + 1 + 6) - 3. This online trick was the original motivation for this sequence.
%H Samuel Owen, <a href="/A353159/b353159.txt">Table of n, a(n) for n = 1..366</a>
%H Samuel Owen, <a href="/A353159/a353159.txt">Every value of m, for each integer k for k = 2..999</a>
%F For any given k, the value of m is bounded by 0 < m < x, where x is the maximum solution to the equation x = 10^(1/k)*k*floor(9*log_10(x)-1).
%e s(62^9) = 62 + 9, so 9 is a term.
%e s(2157^156) = 2157 + 156, so 156 is a term.
%e s(18045^999) = 18045 + 999, so 999 is a term.
%Y Cf. A007953, A133509.
%K nonn,base
%O 1,1
%A _Samuel Owen_, Apr 27 2022
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