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A093950
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Expansion of 1 / (chi(-x) * chi(-x^7)) in powers of x where chi() is a Ramanujan theta function.
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4
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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, 204, 240, 282, 332, 386, 450, 524, 606, 703, 812, 940, 1082, 1243, 1428, 1636, 1873, 2140, 2448, 2788, 3172, 3610, 4096, 4646, 5264, 5962, 6736, 7606, 8582, 9666, 10884
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OFFSET
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0,4
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COMMENTS
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Given g.f. A(x), the right side of Cayley's identity is 2 * q * A(q^2). - Michael Somos, Dec 03 2013
Proof of Cayley's identity, from Silviu Radu, Mar 13 2015: (Start)
Up to issues of convergence I observe that the identity may be rewritten after substituting q=e^{2 Pi Iz} as:
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)
where E(z)= exp( Pi I z/12) Product_{n>=1} (1-e^{2 Pi I z n}) is the Dedekind eta function.
One can further rewrite the above identity by dividing the whole identity by the first term. We obtain:
1-E(28z) x E(14z)^(-3) x E(7z)^2 x E(4z) x E(2z)^(-3) x E(z)^2
-2 E(28z)^2 x E(14z)^(-3) x E(7z) x E(4z)^2 x E(2z)^(-3) x E(z)=0
What is interesting now about this expression is that each term is a modular function for the group Gamma_0(28).
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).
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} ).
It is impossible that a nonzero modular function has more zeros than poles, therefore it is the zero function. This finishes the proof. (End)
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REFERENCES
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A. Cayley, An elliptic-transcendant identity, Messenger of Math., 2 (1873), p. 179.
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LINKS
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FORMULA
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Expansion of q^(-1/3) * (eta(q^2) * eta(q^14)) / (eta(q) * eta(q^7)) in powers of q.
Euler transform of period 14 sequence [ 1, 0, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 1, 0, ...].
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.
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
G.f.: Product_{k>0} (1 + x^k) * (1 + x^(7*k)).
a(n) ~ exp(2*Pi*sqrt(2*n/21)) / (2^(7/4) * 21^(1/4) * n^(3/4)). - Vaclav Kotesovec, Sep 07 2015
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EXAMPLE
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G.f. = 1 + x + x^2 + 2*x^3 + 2*x^4 + 3*x^5 + 4*x^6 + 6*x^7 + 7*x^8 + ...
G.f. = q + q^4 + q^7 + 2*q^10 + 2*q^13 + 3*q^16 + 4*q^19 + 6*q^22 + ...
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MATHEMATICA
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a[ n_] := SeriesCoefficient[ Product[ 1 + x^k, {k, n}] Product[ 1 + x^k, {k, 7, n, 7}], {x, 0, n}];
a[ n_] := SeriesCoefficient[ QPochhammer[ -x, x] QPochhammer[ -x^7, x^7], {x, 0, n}];
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PROG
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(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))};
(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))};
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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