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

Author

Pieter Post, May 10 2015

Keywords

• nonn
• base

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).

Examples

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

Links

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

Crossrefs

Cf. A005188, A023052, A257768.

Programs

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

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
				

Extensions

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