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Search: id:A109675
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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
1
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| 1, 4, 5, 10, 25, 50, 100, 446, 1000, 9775, 10000, 100000
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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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
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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
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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
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Sequence in context: A054173 A049898 A166577 this_sequence A052508 A074098 A126069
Adjacent sequences: A109672 A109673 A109674 this_sequence A109676 A109677 A109678
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KEYWORD
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base,hard,more,nonn
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AUTHOR
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Ryan Propper (rpropper(AT)stanford.edu), Aug 06 2005
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