login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

Expansion of 1 / (chi(-x) * chi(-x^7)) in powers of x where chi() is a Ramanujan theta function.
5

%I #39 Mar 12 2021 22:24:42

%S 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 204,240,282,332,386,450,524,606,703,812,940,1082,1243,1428,1636,1873,

%U 2140,2448,2788,3172,3610,4096,4646,5264,5962,6736,7606,8582,9666,10884

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

%C Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

%C Given g.f. A(x), the right side of Cayley's identity is 2 * q * A(q^2). - _Michael Somos_, Dec 03 2013

%C Proof of Cayley's identity, from Silviu Radu, Mar 13 2015: (Start)

%C Up to issues of convergence I observe that the identity may be rewritten after substituting q=e^{2 Pi Iz} as:

%C 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 where E(z)= exp( Pi I z/12) Product_{n>=1} (1-e^{2 Pi I z n}) is the Dedekind eta function.

%C One can further rewrite the above identity by dividing the whole identity by the first term. We obtain:

%C 1-E(28z) x E(14z)^(-3) x E(7z)^2 x E(4z) x E(2z)^(-3) x E(z)^2

%C -2 E(28z)^2 x E(14z)^(-3) x E(7z) x E(4z)^2 x E(2z)^(-3) x E(z)=0

%C What is interesting now about this expression is that each term is a modular function for the group Gamma_0(28).

%C 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 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 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 A. Cayley, An elliptic-transcendant identity, Messenger of Math., 2 (1873), p. 179.

%H G. C. Greubel, <a href="/A093950/b093950.txt">Table of n, a(n) for n = 0..1000</a>

%H A. Cayley, <a href="/A093950/a093950.pdf">An elliptic-transcendant identity</a>

%H Michael Somos, <a href="/A010815/a010815.txt">Introduction to Ramanujan theta functions</a>

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

%F Expansion of q^(-1/3) * (eta(q^2) * eta(q^14)) / (eta(q) * eta(q^7)) in powers of q.

%F Euler transform of period 14 sequence [ 1, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 1, 0, ...].

%F 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 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 G.f.: Product_{k>0} (1 + x^k) * (1 + x^(7*k)).

%F a(n) = A112212(2*n + 1) = - A102314(2*n + 1). - _Michael Somos_, Dec 03 2013

%F Convolution inverse of A102314.

%F a(n) = (-1)^n * A246762(n). - _Michael Somos_, Sep 02 2014

%F a(n) ~ exp(2*Pi*sqrt(2*n/21)) / (2^(7/4) * 21^(1/4) * n^(3/4)). - _Vaclav Kotesovec_, Sep 07 2015

%e 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 G.f. = q + q^4 + q^7 + 2*q^10 + 2*q^13 + 3*q^16 + 4*q^19 + 6*q^22 + ...

%t a[ n_] := SeriesCoefficient[ Product[ 1 + x^k, {k, n}] Product[ 1 + x^k, {k, 7, n, 7}], {x, 0, n}];

%t a[ n_] := SeriesCoefficient[ QPochhammer[ -x, x] QPochhammer[ -x^7, x^7], {x, 0, n}];

%o (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 (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 Cf. A102314, A112212, A246762.

%K nonn

%O 0,4

%A _Michael Somos_, Apr 19 2004

%E Entry revised by _N. J. A. Sloane_, Mar 15 2015 (with thanks to _Doron Zeilberger_)