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a(n) = smallest m such that n-th prime divides m^m + 1.
1

%I #10 Jul 21 2018 19:31:37

%S 1,5,2,3,21,6,24,27,11,14,15,6,25,63,23,26,117,30,99,35,9,39,165,11,6,

%T 10,51,213,54,7,63,261,174,23,74,33,78,27,83,86,357,30,95,12,14,11,15,

%U 111,453,18,12,119,90,501,4,131,82,135,117,60,45,138,51,95,54,114,75

%N a(n) = smallest m such that n-th prime divides m^m + 1.

%H Harvey P. Dale, <a href="/A073943/b073943.txt">Table of n, a(n) for n = 1..1000</a>

%e 6th prime is 13 and 13 first divides 6^6 + 1 = 46657, so a(6) = 6.

%t smp[n_]:=Module[{m=1},While[PowerMod[m,m,n]!=n-1,m++];m]; smp/@Prime[ Range[ 70]] (* _Harvey P. Dale_, Jul 21 2018 *)

%Y Cf. A014566.

%K easy,nonn

%O 1,2

%A _Jason Earls_, Nov 13 2002