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A108825
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Numbers j such that j divides the sum of the digits of j!.
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5
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1, 2, 3, 9, 15, 18, 21, 27, 72, 81, 234, 462, 502, 522, 1314, 1323, 3789, 3897, 6462, 10470, 17532, 17820, 28503, 48248, 48254, 48303, 48644, 48856, 223551, 226149, 227406, 625986, 4838918, 4848004
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OFFSET
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1,2
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COMMENTS
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Especially for larger values, terms can be expected to occur near values of k such that 4.5 times the number of digits in k!, not counting the digits in the string of trailing zeros, is approximately an integer multiple (m) of k. As m increases, such values of k approach k = e * 100^(m/9 + 1/8). - Jon E. Schoenfield, Jun 08 2007
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LINKS
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EXAMPLE
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3! = 6; 6 mod 3 = 0.
9! = 362880; 3 + 6 + 2 + 8 + 8 = 27; 27 mod 9 = 0.
522 is a term because the digit sum of 522!, 4698, is divisible by 522.
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MATHEMATICA
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Do[If[Mod[Plus @@ IntegerDigits[n! ], n] == 0, Print[n]], {n, 1, 10000}]
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CROSSREFS
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KEYWORD
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more,nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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