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%I #19 Jul 28 2023 16:52:02
%S 0,0,1,1,3,4,6,8,13,15,19,22,28,30,37,41,48,54,64,69,77,83,94,98,110,
%T 119,131,139,152,162,172,183,199,208,226,234,253,263,281,294,308,322,
%U 343,357,377,390,412,424,447,465,488,504,528,545,567,585,612,628,654
%N Number of unit square lattice cells inside quadrant of origin-centered circle of diameter n.
%H Ivan Panchenko, <a href="/A136483/b136483.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = Sum_{k=1..floor(n/2)} floor(sqrt((n/2)^2 - k^2)).
%F Lim_{n -> oo} a(n)/(n^2) -> Pi/16 (A019683).
%F a(n) = (1/4) * A136485(n) = (1/2) * A136513(n).
%F a(n) = [x^(n^2)] (theta_3(x^4) - 1)^2 / (4 * (1 - x)). - _Ilya Gutkovskiy_, Nov 23 2021
%e a(5) = 3 because a circle of radius 5/2 in the first quadrant encloses (2,1), (1,1), (1,2).
%t Table[Sum[Floor[Sqrt[(n/2)^2 -k^2]], {k,Floor[n/2]}], {n,100}]
%o (Magma)
%o A136483:= func< n | n eq 1 select 0 else (&+[Floor(Sqrt((n/2)^2-j^2)): j in [1..Floor(n/2)]]) >;
%o [A136483(n): n in [1..100]]; // _G. C. Greubel_, Jul 28 2023
%o (SageMath)
%o def A136483(n): return sum(isqrt((n/2)^2-j^2) for j in range(1,(n//2)+1))
%o [A136483(n) for n in range(1,101)] # _G. C. Greubel_, Jul 28 2023
%o (PARI) a(n) = sum(k=1, n\2, sqrtint((n/2)^2 - k^2)); \\ _Michel Marcus_, Jul 28 2023
%Y Alternating merge of A136484 and A001182.
%Y Cf. A019683, A136485, A136513.
%K easy,nonn
%O 1,5
%A Glenn C. Foster (gfoster(AT)uiuc.edu), Jan 02 2008
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