|
%I #7 Mar 30 2012 18:35:54
%S 0,3,5,9,13,11,18,23,40,30,27,60,35,45,91,69,98,63,119
%N a(n) = k is the smallest number such that n is the number of distinct primes dividing k^k + 1.
%e a(6) = 11 because the 6 distinct primes dividing 11^11 + 1 = 285311670612 are
%e {2, 3, 23, 89, 199, 58367}.
%p with(numtheory):for n from 1 to 8 do:ind:=0:for k from 1 to 40 while(ind=0)
%p do: x:=k^k+1:y:=nops(factorset(x)):if y=n then ind:=1:printf(`%d, `,k):else
%p fi:od: od:
%Y Cf. A014566.
%K nonn,hard
%O 1,2
%A _Michel Lagneau_, Jan 18 2011
|
