%I #13 Dec 06 2015 11:35:35
%S 1,6,1,24,64,1,384
%N a(n) = smallest k such that (A115091(n)-k)! == -1 (mod A115091(n)^2).
%C The values of m in A115091.
%C A115091(n) is in A007540 iff a(n) = 1.
%e a(2) = 6, because 6 is the smallest k such that (A115091(2)-k)! == -1 (mod A115091(2)^2), which yields the congruence (11-6)! == -1 (mod 11^2).
%t t = Select[Prime@ Range@ 120, AnyTrue[Range@ #, Function[m, Divisible[m! + 1, #^2]]] &]; Table[k = 1; While[Mod[(t[[n]] - k)!, t[[n]]^2] != t[[n]]^2 - 1, k++]; k, {n, 7}] (* _Michael De Vlieger_, Nov 10 2015, Version 10 *)
%o (PARI) forprime(p=1, , for(k=1, p-1, if(Mod((p-k)!, p^2)==-1, print1(k, ", "); break({1}))))
%Y Cf. A007540, A115091.
%K nonn,hard,more
%O 1,2
%A _Felix Fröhlich_, Nov 08 2015
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