%I #15 Jul 28 2023 17:13:24
%S 0,0,2,2,6,8,12,16,26,30,38,44,56,60,74,82,96,108,128,138,154,166,188,
%T 196,220,238,262,278,304,324,344,366,398,416,452,468,506,526,562,588,
%U 616,644,686,714,754,780,824,848,894,930,976,1008,1056,1090,1134,1170
%N Number of unit square lattice cells inside half-plane (two adjacent quadrants) of origin centered circle of diameter n.
%H G. C. Greubel, <a href="/A136513/b136513.txt">Table of n, a(n) for n = 1..1000</a>
%F Lim_{n -> oo} a(n)/(n^2) -> Pi/8.
%F a(n) = 2 * Sum_{k=1..floor(n/2)} floor(sqrt((n/2)^2 - k^2)).
%F a(n) = 2 * A136483(n).
%F a(n) = (1/2) * A136485(n).
%F a(n) = [x^(n^2)] (theta_3(x^4) - 1)^2 / (2 * (1 - x)). - _Ilya Gutkovskiy_, Nov 24 2021
%e a(3) = 2 because a circle centered at the origin and of radius 3/2 encloses (-1,1) and (1,1) in the upper half plane.
%t Table[2*Sum[Floor[Sqrt[(n/2)^2 -k^2]], {k,Floor[n/2]}], {n,100}]
%o (Magma)
%o A136513:= func< n | n eq 1 select 0 else 2*(&+[Floor(Sqrt((n/2)^2-j^2)): j in [1..Floor(n/2)]]) >;
%o [A136513(n): n in [1..100]]; // _G. C. Greubel_, Jul 27 2023
%o (SageMath)
%o def A136513(n): return 2*sum(isqrt((n/2)^2-k^2) for k in range(1,(n//2)+1))
%o [A136513(n) for n in range(1,101)] # _G. C. Greubel_, Jul 27 2023
%o (PARI) a(n) = 2*sum(k=1, n\2, sqrtint((n/2)^2-k^2)); \\ _Michel Marcus_, Jul 27 2023
%Y Alternating merge of A136514 and A136515.
%Y Cf. A136483, A136485.
%K easy,nonn
%O 1,3
%A Glenn C. Foster (gfoster(AT)uiuc.edu), Jan 02 2008