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A269665
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For n>=0, let A_n be the set of natural numbers k such that (k^n + 1) | k!. If A is nonempty, then a(n) is the least element of A_n; otherwise a(n) = 0.
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1
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2, 5, 18, 17, 1600, 984, 2888, 460747, 99271723, 792174, 32917926
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OFFSET
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0,1
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COMMENTS
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a(n) is the smallest k such that (k^n + 1) | k! if it exists, otherwise a(n) = 0.
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LINKS
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EXAMPLE
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For n=2, a(2) is equal to 18 because k=18 is the least natural number k such that (k^2+1)|k! (see A120416).
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MATHEMATICA
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For[k = 0, k < 11, k++, x = 0; r = 0; n = 1; While[x != 1, If[Mod[n!, n^k + 1] != 0, x = 0, x = 1; r = n]; n++]; Print[r]]
Table[SelectFirst[Range[10^4], Divisible[#!, #^n + 1] &], {n, 0, 6}] (* Michael De Vlieger, Mar 04 2016, Version 10 *)
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PROG
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(PARI) a(n) = {my(k = 1); while (k! % (k^n+1), k++); k; } \\ Michel Marcus, Mar 03 2016
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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