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A109663
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Numbers k such that the sum of the digits of (k^k + k!) is divisible by k.
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1
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1, 2, 3, 9, 15, 18, 27, 36, 51, 81, 93, 169, 181, 348, 444, 504, 528, 1881, 2031, 9843, 16479, 16685, 45435, 129056, 138510, 214008, 358326
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OFFSET
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1,2
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COMMENTS
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The quotients are 2, 3, 2, 6, 6, 5, 8, 7, 6, 9, 9, 10, 10, 12, 12, 12, 12, 15, 15, 18, 19, 19, 21, 23, 22, 24, 25.
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LINKS
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EXAMPLE
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The digits of 1881^1881 + 1881! sum to 28215 and 28215 is divisible by 1881, so 1881 is in the sequence.
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MATHEMATICA
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Do[s = n^n + n!; k = Plus @@ IntegerDigits[s]; If[Mod[k, n] == 0, Print[n]], {n, 1, 10000}]
Select[Range[360000], Divisible[Total[IntegerDigits[#^#+#!]], #]&] (* Harvey P. Dale, Dec 27 2018 *)
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CROSSREFS
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KEYWORD
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base,more,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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