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).

2, 3, 4, 5, 6, 7, 8, 9, 50, 298, 130004, 484950, 3242940, 4264064, 5560625, 36550290, 47746195, 111971979, 129833998, 9865843497, 46793077740, 767609367921, 4432743262896, 42744572298532, 77186414790914, 99320211963544, 99335229415136, 456385296642870

Offset

1,1

Comments

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).

Links

Chai Wah Wu, Table of n, a(n) for n = 1..54 (terms < 10^32, n = 1..53 from Giovanni Resta)

Example

50 = 2*(5^2+0^2);
484950 = 5*(4^5+8^5+4^5+9^5+5^5+0^5).

Prog

(Python)
def mod(n, a):
....kk = 0
....while n > 0:
........kk= kk+(n%10)**a
........n =int(n//10)
....return kk
for a in range (1, 10):
....for c in range (1, 10**7):
........if c==a*mod(c, a) and a<mod(c, a):
............print (a, c, mod(c, a))
(PARI) sdk(d, k) = sum(j=1, #d, d[j]^k);
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

Crossrefs

Cf. A005188, A023052, A257768.

Sequence in context: A349190 A252781 A024660 * A327455 A327735 A133287

Adjacent sequences: A257811 A257812 A257813 * A257815 A257816 A257817

Keyword

nonn,base

Author

Pieter Post, May 10 2015

Extensions

a(16)-a(28) from Giovanni Resta, May 10 2015

Status

approved