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Number of nonnegative k such that k^2 + 2 divides n^2 + 2.
1

%I #17 Jun 12 2024 06:55:47

%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} (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);

%o (PARI) a(n) = my(m=n^2+2); sum(k=0, n, !(m % (k^2+2))); \\ _Michel Marcus_, Jun 12 2024

%Y Cf. A059100, A067201.

%K nonn

%O 0,3

%A _Robert Israel_, Aug 15 2014