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%I #29 Apr 15 2018 01:59:33
%S 2,3,4,5,6,7,8,9,50,298,130004,484950,3242940,4264064,5560625,
%T 36550290,47746195,111971979,129833998,9865843497,46793077740,
%U 767609367921,4432743262896,42744572298532,77186414790914,99320211963544,99335229415136,456385296642870
%N Numbers n such that k times the sum of the digits (d) to the power k equal n, so n=k*sum(d^k), for some positive integer k, where k is smaller than sum(d^k).
%C The first nine terms are trivial, but then the terms become scarce. The exponent k must be less than the "sum of the digits" raised to the k-th power, otherwise there will be infinitely many terms containing 1's and 0's, like 11000= 5500*(1^5500+1^5500+0^5500+0^5500+0^5500). It appears this sequence is finite, because there is a resemblance with the Armstrong numbers (A005188).
%H Chai Wah Wu, <a href="/A257814/b257814.txt">Table of n, a(n) for n = 1..54</a> (terms < 10^32, n = 1..53 from Giovanni Resta)
%e 50 = 2*(5^2+0^2);
%e 484950 = 5*(4^5+8^5+4^5+9^5+5^5+0^5).
%o (Python)
%o def mod(n,a):
%o ....kk = 0
%o ....while n > 0:
%o ........kk= kk+(n%10)**a
%o ........n =int(n//10)
%o ....return kk
%o for a in range (1, 10):
%o ....for c in range (1, 10**7):
%o ........if c==a*mod(c,a) and a<mod(c,a):
%o ............print (a,c, mod(c,a))
%o (PARI) sdk(d, k) = sum(j=1, #d, d[j]^k);
%o isok(n) = {d = digits(n); k = 1; while ((val=k*sdk(d,k)) != n, k++; if (val > n, return (0))); k < sdk(d,k);} \\ _Michel Marcus_, May 30 2015
%Y Cf. A005188, A023052, A257768.
%K nonn,base
%O 1,1
%A _Pieter Post_, May 10 2015
%E a(16)-a(28) from _Giovanni Resta_, May 10 2015
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