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 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 (list; graph; refs; listen; history; text; internal format)
 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 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^n-1) 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**n-1)]) % 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

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Last modified March 29 20:20 EDT 2020. Contains 333117 sequences. (Running on oeis4.)