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A109675
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Numbers n such that the sum of the digits of (n^n - 1) is divisible by n.
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0
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1, 4, 5, 10, 25, 50, 100, 446, 1000, 9775, 10000, 100000
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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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.
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EXAMPLE
| The digits of 9775^9775 - 1 sum to 175950 and 175950 is divisible by 9775, so 9775 is in the sequence.
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MATHEMATICA
| Do[k = n^n - 1; s = Plus @@ IntegerDigits[k]; If[Mod[s, n] == 0, Print[n]], {n, 1, 10^5}]
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CROSSREFS
| Sequence in context: A049898 A166577 A203853 * A052508 A074098 A196270
Adjacent sequences: A109672 A109673 A109674 * A109676 A109677 A109678
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KEYWORD
| base,hard,more,nonn
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AUTHOR
| Ryan Propper (rpropper(AT)stanford.edu), Aug 06 2005
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