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A093950 Expansion of 1 / (chi(-x) * chi(-x^7)) in powers of x where chi() is a Ramanujan theta function.

%I A093950 #40 Feb 16 2025 08:32:53
%S A093950 1,1,1,2,2,3,4,6,7,9,12,14,18,22,28,34,41,50,60,72,86,105,124,146,174,
%T A093950 204,240,282,332,386,450,524,606,703,812,940,1082,1243,1428,1636,1873,
%U A093950 2140,2448,2788,3172,3610,4096,4646,5264,5962,6736,7606,8582,9666,10884
%N A093950 Expansion of 1 / (chi(-x) * chi(-x^7)) in powers of x where chi() is a Ramanujan theta function.
%C A093950 Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).
%C A093950 Given g.f. A(x), the right side of Cayley's identity is 2 * q * A(q^2). - _Michael Somos_, Dec 03 2013
%C A093950 Proof of Cayley's identity, from Silviu Radu, Mar 13 2015: (Start)
%C A093950 Up to issues of convergence I observe that the identity may be rewritten after substituting q=e^{2 Pi Iz} as:
%C A093950 E(28z)^(-1) x E(14z)^2 x E(7z)^(-1) x E(4z)^(-1) x E(2z)^2 x E(z)^(-1) -E(14z)^(-1) x E(7z) x E(2z)^(-1) x E(z) = 2 E(28z) x E(14z)^(-1) x E(4z) x E(2z)^(-1)
%C A093950 where E(z)= exp( Pi I z/12) Product_{n>=1} (1-e^{2 Pi I z n}) is the Dedekind eta function.
%C A093950 One can further rewrite the above identity by dividing the whole identity by the first term. We obtain:
%C A093950 1-E(28z) x E(14z)^(-3) x E(7z)^2 x E(4z) x E(2z)^(-3) x E(z)^2
%C A093950 -2 E(28z)^2 x E(14z)^(-3) x E(7z) x E(4z)^2 x E(2z)^(-3) x E(z)=0
%C A093950 What is interesting now about this expression is that each term is a modular function for the group Gamma_0(28).
%C A093950 Furthermore, all the terms except the constant term have two poles, therefore the whole left hand side has at most two poles (at the points z=1/14 and z=1/2).
%C A093950 However we check that in the q-expansion the first three coefficients are zero, which implies that the left hand side also has a zero of order at least three at the point infinity (note that z=I x infty transforms into q=0, q=e^(2 Pi iz} ).
%C A093950 It is impossible that a nonzero modular function has more zeros than poles, therefore it is the zero function. This finishes the proof. (End)
%D A093950 A. Cayley, An elliptic-transcendant identity, Messenger of Math., 2 (1873), p. 179.
%H A093950 G. C. Greubel, <a href="/A093950/b093950.txt">Table of n, a(n) for n = 0..1000</a>
%H A093950 A. Cayley, <a href="/A093950/a093950.pdf">An elliptic-transcendant identity</a>
%H A093950 Michael Somos, <a href="/A010815/a010815.txt">Introduction to Ramanujan theta functions</a>
%H A093950 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/RamanujanThetaFunctions.html">Ramanujan Theta Functions</a>
%F A093950 Expansion of q^(-1/3) * (eta(q^2) * eta(q^14)) / (eta(q) * eta(q^7)) in powers of q.
%F A093950 Euler transform of period 14 sequence [ 1, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 1, 0, ...].
%F A093950 Given g.f. A(x), then B(q) = q * A(q^3) satisfies 0 = f(B(q), B(q^2)) where f(u, v) = u^2 - v - 2*u*v^2.
%F A093950 G.f. is a period 1 Fourier series which satisfies f(-1 / (126 t)) = 1/2 * g(t) where q = exp(2 Pi i t) and g() is the g.f. of A102314. - _Michael Somos_, Dec 03 2013
%F A093950 G.f.: Product_{k>0} (1 + x^k) * (1 + x^(7*k)).
%F A093950 a(n) = A112212(2*n + 1) = - A102314(2*n + 1). - _Michael Somos_, Dec 03 2013
%F A093950 Convolution inverse of A102314.
%F A093950 a(n) = (-1)^n * A246762(n). - _Michael Somos_, Sep 02 2014
%F A093950 a(n) ~ exp(2*Pi*sqrt(2*n/21)) / (2^(7/4) * 21^(1/4) * n^(3/4)). - _Vaclav Kotesovec_, Sep 07 2015
%e A093950 G.f. = 1 + x + x^2 + 2*x^3 + 2*x^4 + 3*x^5 + 4*x^6 + 6*x^7 + 7*x^8 + ...
%e A093950 G.f. = q + q^4 + q^7 + 2*q^10 + 2*q^13 + 3*q^16 + 4*q^19 + 6*q^22 + ...
%t A093950 a[ n_] := SeriesCoefficient[ Product[ 1 + x^k, {k, n}] Product[ 1 + x^k, {k, 7, n, 7}], {x, 0, n}];
%t A093950 a[ n_] := SeriesCoefficient[ QPochhammer[ -x, x] QPochhammer[ -x^7, x^7], {x, 0, n}];
%o A093950 (PARI) {a(n) = if( n<0, 0, polcoeff( prod( k=1, n, 1 + x^k, 1 + x * O(x^n)) * prod( k=1, n\7, 1 + x^(7*k), 1 + x * O(x^n)), n))};
%o A093950 (PARI) {a(n) = local(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x^2 + A) * eta(x^14 + A) / (eta(x + A) * eta(x^7 + A)), n))};
%Y A093950 Cf. A102314, A112212, A246762.
%K A093950 nonn
%O A093950 0,4
%A A093950 _Michael Somos_, Apr 19 2004
%E A093950 Entry revised by _N. J. A. Sloane_, Mar 15 2015 (with thanks to _Doron Zeilberger_)