

A109675


Numbers n such that the sum of the digits of (n^n  1) is divisible by n.


0



1, 4, 5, 10, 25, 50, 100, 446, 1000, 9775, 10000, 100000
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OFFSET

1,2


COMMENTS

n = 10^k is a member of the sequence, for all k >= 0. Proof: Let n = 10^k for some nonnegative integer k. Then n^n  1 has k*10^k 9's and no other digits, so its digits sum to 9*k*10^k = 9*k*n, a multiple of n.


LINKS

Table of n, a(n) for n=1..12.


EXAMPLE

The digits of 9775^9775  1 sum to 175950 and 175950 is divisible by 9775, so 9775 is in the sequence.


MAPLE

sumdigs:= n > convert(convert(n, base, 10), `+`);
select(n > sumdigs(n^n1) mod n = 0, [$1..10^5]); # Robert Israel, Dec 03 2014


MATHEMATICA

Do[k = n^n  1; s = Plus @@ IntegerDigits[k]; If[Mod[s, n] == 0, Print[n]], {n, 1, 10^5}]


PROG

(Python)
A109675_list = [n for n in range(1, 10**4) if not sum([int(d) for d in str(n**n1)]) % n]
# Chai Wah Wu, Dec 03 2014


CROSSREFS

Sequence in context: A242960 A288141 A203853 * A052508 A305887 A074098
Adjacent sequences: A109672 A109673 A109674 * A109676 A109677 A109678


KEYWORD

base,hard,more,nonn


AUTHOR

Ryan Propper, Aug 06 2005


STATUS

approved



