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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
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 (list; graph; refs; listen; history; text; internal format)
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: A219326 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

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Last modified January 23 07:07 EST 2020. Contains 331168 sequences. (Running on oeis4.)