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
KEYWORD
nonn,base
AUTHOR
Pieter Post, May 10 2015
EXTENSIONS
a(16)-a(28) from Giovanni Resta, May 10 2015
STATUS
approved