%I #9 Aug 18 2014 02:57:54
%S 1,1,3,1,4,2,2,2,5,1,5,2,2,2,7,1,4,2,2,3,4,1,6,2,2,3,4,1,4,2,3,2,8,1,
%T 4,2,3,2,4,1,5,4,2,2,7,1,4,3,2,3,5,1,8,2,2,2,4,1,11,3,2,3,4,2,4,2,2,2,
%U 5,2,6,3,2,2,7,1,7,2,2,2,7,1,5,2,2,3,6,1,4,2,2,3,7,1,5,4,3,2,4,1,4,2,3,3,5,1,4,3,3,3,4,1,8,4,2,2,4,1,6,2,3,2,5,1,6,2,2,2
%N Number of nonnegative k such that k^2+2 divides n^2+2.
%C a(A067201(n)) = 1, but these are not the only cases where a(k) = 1.
%H Robert Israel, <a href="/A242111/b242111.txt">Table of n, a(n) for n = 0..10000</a>
%F G.f. sum(k=0..infinity, sum(i in S(k), x^i)/(1 - x^(k^2+2))) where S(k) = {i : 0 <= i < k^2 + 2, i^2 + 2 == 0 mod k^2 + 2}.
%e For n=2, n^2+2=6 is divisible by 0^2+2=2, 1^2+2=3 and 2^2+2=6 so a(2)=3.
%p a:= n -> nops(select(t -> issqr(t-2),numtheory:-divisors(n^2+2))):
%p seq(a(n),n=0..100);
%Y Cf. A059100, A067201.
%K nonn
%O 0,3
%A _Robert Israel_, Aug 15 2014
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