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a(n) = [x^(n^2)] (theta_3(x) - 1)^n/(2^n*(1 - x)), where theta_3() is the Jacobi theta function.
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%I #5 Apr 17 2018 18:30:57

%S 1,1,1,7,32,177,1269,9263,74452,652710,6078048,60447082,631870024,

%T 6915613084,79113376037,941759419159,11630647314564,148799595377384,

%U 1966441829785081,26793749867965515,375812005722920406,5416574818546042067,80123280319100908258,1214860029446181979357

%N a(n) = [x^(n^2)] (theta_3(x) - 1)^n/(2^n*(1 - x)), where theta_3() is the Jacobi theta function.

%C a(n) = number of positive solutions to (x_1)^2 + (x_2)^2 + ... + (x_n)^2 <= n^2.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/JacobiThetaFunctions.html">Jacobi Theta Functions</a>

%H <a href="/index/Su#ssq">Index entries for sequences related to sums of squares</a>

%t Table[SeriesCoefficient[(EllipticTheta[3, 0, x] - 1)^n/(2^n (1 - x)), {x, 0, n^2}], {n, 0, 23}]

%t Join[{1}, Table[SeriesCoefficient[1/(1 - x) Sum[x^k^2, {k, 1, n}]^n, {x, 0, n^2}], {n, 23}]]

%Y Cf. A000122, A001182, A253663, A298330, A302861, A302863.

%K nonn

%O 0,4

%A _Ilya Gutkovskiy_, Apr 17 2018