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A217786
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Expansion of (psi(x^3) / psi(x))^2 in powers of x where psi() is a Ramanujan theta function.
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7
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1, -2, 3, -4, 7, -12, 17, -24, 36, -52, 71, -96, 133, -182, 240, -316, 420, -552, 713, -916, 1182, -1516, 1920, -2424, 3063, -3852, 4806, -5976, 7430, -9204, 11336, -13924, 17088, -20908, 25473, -30960, 37586, -45518, 54939, -66172, 79603, -95556, 114399
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OFFSET
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0,2
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COMMENTS
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LINKS
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FORMULA
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Expansion of q^(-1/2) * eta(q)^2 * eta(q^6)^4 / (eta(q^2)^4 * eta(q^3)^2) in powers of q.
Euler transform of period 6 sequence [ -2, 2, 0, 2, -2, 0, ...].
Given g.f. A(x), B(q) = q * A(q^2) satisfies 0 = f(B(q), B(q^2)) where f(u, v) = (1 - v) * (v - u^2) - 4 * u^2 * v.
Given g.f. A(x), B(q) = q * A(q^2) satisfies 0 = f(B(q), B(q^3)) where f(u, v) = u * (u + 3*v)^2 - v * (1 + 3*u*v)^2.
G.f. is a period 1 Fourier series which satisfies f(-1 / (12 t)) = (1/3) g(t) where q = exp(2 Pi i t) and g() is the g.f. of A217771.
G.f.: Product_{k>0} (1 + x^k + x^(2*k))^2 * (1 - x^k + x^(2*k))^4.
a(n) ~ (-1)^n * exp(Pi*sqrt(2*n/3)) / (6^(5/4) * n^(3/4)). - Vaclav Kotesovec, Jun 06 2018
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EXAMPLE
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G.f. = 1 - 2*x + 3*x^2 - 4*x^3 + 7*x^4 - 12*x^5 + 17*x^6 - 24*x^7 + 36*x^8 + ...
G.f. = q - 2*q^3 + 3*q^5 - 4*q^7 + 7*q^9 - 12*q^11 + 17*q^13 - 24*q^15 + 36*q^17 + ...
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MATHEMATICA
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a[ n_] := SeriesCoefficient[ x^(-1/2) (EllipticTheta[ 2, 0, x^(3/2)] / EllipticTheta[ 2, 0, x^(1/2)])^2, {x, 0, n}];
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PROG
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(PARI) {a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x + A)^2 * eta(x^6 + A)^4 / (eta(x^2 + A)^4 * eta(x^3 + A)^2), n))};
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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