

A269665


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.


1



2, 5, 18, 17, 1600, 984, 2888, 460747, 99271723, 792174, 32917926
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OFFSET

0,1


COMMENTS

a(n) is the smallest k such that (k^n + 1)  k! if it exists, otherwise a(n) = 0.


LINKS



EXAMPLE

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).


MATHEMATICA

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 *)


PROG

(PARI) a(n) = {my(k = 1); while (k! % (k^n+1), k++); k; } \\ Michel Marcus, Mar 03 2016


CROSSREFS



KEYWORD

nonn,more


AUTHOR



EXTENSIONS



STATUS

approved



