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%I #18 Dec 01 2023 15:52:42
%S 0,0,1,0,1,0,2,1,1,0,1,1,3,0,1,0,3,2,1,0,3,1,3,0,1,0,3,1,1,1,3,2,3,1,
%T 1,0,3,2,1,0,2,1,5,1,1,1,7,1,1,1,1,1,3,0,3,0,7,1,1,1,1,1,3,1,1,0,3,3,
%U 1,2,1,3,7,0,3,1,3,1,1,1,3,2,7,0,1,1,3,1,3,0,3,1,5,0,1,1,3,3,5
%N Half the number of proper divisors (> 1) of n^2 + 1, i.e., tau(n^2 + 1)/2 - 1.
%C For any n > 0, n^2 + 1 cannot be a square and thus has an even number of divisors which always include 1 and n^2 + 1, therefore a(n) = (half that number minus 1) is always a nonnegative integer.
%H Amiram Eldar, <a href="/A147809/b147809.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A000005(A002522(n))/2 - 1 = A147810(n) - 1.
%F Sum_{k=1..n} a(k) ~ c * n * log(n), where c = 3/(2*Pi) = 0.477464... (A093582). - _Amiram Eldar_, Dec 01 2023
%t DivisorSigma[0,Range[100]^2+1]/2-1 (* _Harvey P. Dale_, Feb 11 2015 *)
%o (PARI) A147809(n)=numdiv(n^2+1)/2-1
%Y Cf. A000005, A002522, A147810, A048691, A063647, A093582.
%K easy,nonn
%O 1,7
%A _M. F. Hasler_, Dec 13 2008
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