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A307520 Expansion of Product_{k>=1} ((1 - x)^k - x^k)/((1 - x)^k + x^k).

%I A307520 #20 Jan 01 2025 09:54:04
%S A307520 1,-2,-2,-2,0,6,18,38,68,108,148,148,-2,-552,-2004,-5280,-11960,
%T A307520 -24590,-46990,-84254,-141476,-218654,-295794,-295374,3540,1148110,
%U A307520 4378918,12332318,30327568,68633710,146303026,297349758,580052778,1089382158,1969425038,3413378318,5614375970
%N A307520 Expansion of Product_{k>=1} ((1 - x)^k - x^k)/((1 - x)^k + x^k).
%C A307520 Let A(x) = 1 - 2*x - 2*x^2 - 2*x^3 + 6*x^5 + 18*x^6 + ... denote the g.f. of the sequence. Let u denote the lower triangular matrix with 0's on the main diagonal and 1's in every position below the main diagonal. Then the array Sum_{n in Z} (-1)^n * u^(n^2) is equal to the Riordan array (A(x), x). - _Peter Bala_, Dec 31 2024
%F A307520 G.f.: theta_4(x/(1 - x)), where theta_4() is the Jacobi theta function.
%F A307520 For n >= 1, a(n) = 2 * Sum_{k = 1..floor(sqrt(n))} (-1)^k * binomial(n-1, n-k^2). - _Peter Bala_, Dec 31 2024
%e A307520 From _Peter Bala_, Dec 31 2024: (Start)
%e A307520 With the array u as defined above, the lower triangular array Sum_{n = -2..2} (-1)^n * u^(n^2) = I - 2*u + 2*u^4 begins
%e A307520   1;
%e A307520  -2,   1;
%e A307520  -2,  -2,  1;
%e A307520  -2,  -2, -2,  1;
%e A307520   0,  -2, -2, -2,  1;
%e A307520   6,   0, -2, -2, -2,  1;
%e A307520  18,   6,  0, -2, -2, -2,  1;
%e A307520  38,  18,  6,  0, -2, -2, -2,  1;
%e A307520  68,  38, 18,  6,  0, -2, -2, -2,  1; (End)
%p A307520 a(n) := 2*add( (-1)^k * binomial(n-1, n-k^2), k = 1..floor(sqrt(n))):
%p A307520 print(1, seq(a(n), n = 1..40)); # _Peter Bala_, Dec 31 2024
%t A307520 m = 36; CoefficientList[Series[Product[((1 - x)^k - x^k)/((1 - x)^k + x^k), {k, 1, m}], {x, 0, m}], x] (* _Amiram Eldar_, May 14 2021 *)
%o A307520 (PARI) N=66; x='x+O('x^N); Vec(prod(k=1, N, ((1-x)^k-x^k)/((1-x)^k+x^k)))
%Y A307520 Convolution inverse of A318570.
%Y A307520 Cf. A002448, A307521, A307522.
%K A307520 sign,easy
%O A307520 0,2
%A A307520 _Seiichi Manyama_, Apr 12 2019