A066536 Number of ways of writing n as a sum of n+1 squares.
1, 4, 12, 32, 90, 312, 1288, 5504, 22608, 88660, 339064, 1297056, 5043376, 19975256, 80027280, 321692928, 1291650786, 5177295432, 20748447108, 83279292960, 335056780464, 1351064867328, 5456890474248, 22063059606912
Offset: 0
Keywords
Examples
There are a(2)=12 solutions (x,y,z) of 2=x^2+y^2+z^2: 3 permutations of (1,1,0), 3 permutations of (-1,-1,0) and 6 permutations of (1, -1,0).
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000
Crossrefs
Programs
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Mathematica
Join[{1}, Table[SquaresR[n+1, n], {n, 24}]] (* Calculation of constants {d,c}: *) {1/r, Sqrt[s/(2*Pi*r^2*Derivative[0, 0, 2][EllipticTheta][3, 0, r*s])]} /. FindRoot[{s == EllipticTheta[3, 0, r*s], r*Derivative[0, 0, 1][EllipticTheta][3, 0, r*s] == 1}, {r, 1/4}, {s, 5/2}, WorkingPrecision -> 70] (* Vaclav Kotesovec, Nov 16 2023 *)
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PARI
{a(n)=local(THETA3=1+2*sum(k=1,sqrtint(n),x^(k^2))+x*O(x^n));polcoeff(THETA3^(n+1), n)} /* Paul D. Hanna, Oct 26 2009*/
Formula
a(n) equals the coefficient of x^n in the (n+1)-th power of Jacobi theta_3(x) where theta_3(x) = 1 + 2*Sum_{n>=1} x^(n^2). - Paul D. Hanna, Oct 26 2009
a(n) is divisible by n+1: a(n)/(n+1) = A166952(n) for n>=0. - Paul D. Hanna, Oct 26 2009
a(n) ~ c * d^n / sqrt(n), where d = 4.13273137623493996302796465... (= 1/radius of convergence A166952), c = 0.70710538549959357505200... . - Vaclav Kotesovec, Sep 10 2014
Extensions
Edited by Dean Hickerson, Jan 12 2002
a(0) added by Paul D. Hanna, Oct 26 2009
Edited by R. J. Mathar, Oct 29 2009
Comments