cp's OEIS Frontend

A257814 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).

%I A257814 #31 May 22 2025 10:21:42
%S A257814 2,3,4,5,6,7,8,9,50,298,130004,484950,3242940,4264064,5560625,
%T A257814 36550290,47746195,111971979,129833998,9865843497,46793077740,
%U A257814 767609367921,4432743262896,42744572298532,77186414790914,99320211963544,99335229415136,456385296642870
%N A257814 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 A257814 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 A257814 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 A257814 50 = 2*(5^2+0^2);
%e A257814 484950 = 5*(4^5+8^5+4^5+9^5+5^5+0^5).
%o A257814 (Python)
%o A257814 def mod(n,a):
%o A257814     kk = 0
%o A257814     while n > 0:
%o A257814         kk= kk+(n%10)**a
%o A257814         n =int(n//10)
%o A257814     return kk
%o A257814 for a in range (1, 10):
%o A257814     for c in range (1, 10**7):
%o A257814         if c==a*mod(c,a) and a<mod(c,a):
%o A257814             print (a,c, mod(c,a))
%o A257814 (PARI) sdk(d, k) = sum(j=1, #d, d[j]^k);
%o A257814 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 A257814 Cf. A005188, A023052, A257768.
%K A257814 nonn,base
%O A257814 1,1
%A A257814 _Pieter Post_, May 10 2015
%E A257814 a(16)-a(28) from _Giovanni Resta_, May 10 2015