login
A135211
Expansion of psi(-x) / psi(-x^3) in powers of x where psi() is a Ramanujan theta function.
2
1, -1, 0, 0, -1, 0, 1, -1, 0, 2, -1, 0, 2, -2, 0, 2, -3, 0, 3, -3, 0, 4, -4, 0, 5, -6, 0, 6, -7, 0, 7, -8, 0, 10, -10, 0, 13, -13, 0, 14, -16, 0, 17, -18, 0, 22, -22, 0, 26, -28, 0, 30, -33, 0, 36, -38, 0, 44, -45, 0, 52, -55, 0, 60, -65, 0, 70, -74, 0, 84, -87, 0, 99, -104, 0, 112, -121, 0, 131, -138, 0, 156, -160
OFFSET
0,10
COMMENTS
Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).
LINKS
Eric Weisstein's World of Mathematics, Ramanujan Theta Functions
FORMULA
Expansion of f(-x^2, -x^4) / f(x, x^5) in powers of x where f() is Ramanujan's two-variable theta function. - Michael Somos, Apr 05 2015
Expansion of q^(1/4) * eta(q) * eta(q^4) * eta(q^6) / ( eta(q^2) * eta(q^3) * eta(q^12) ) in powers of q.
Euler transform of period 12 sequence [ -1, 0, 0, -1, -1, 0, -1, -1, 0, 0, -1, 0, ...].
Given g.f. A(x), then B(q) = A(q^4) / q satisfies 0 = f(B(q), B(q^3)) where f(u, v) = (1 + v^4) - (1 + u*v)^3.
G.f. is a period 1 Fourier series which satisfies f(-1 / (192 t)) = 3^(1/2) g(t) where q = exp(2 Pi i t) and g() is the g.f. for A036018.
a(n) = (-1)^n * A256626(n). a(3*n + 1) = - A036018(n). a(3*n + 2) = 0. - Michael Somos, Apr 05 2015
Convolution inverse is A036018. Convolution square is A062243. Convolution 4th power is A187147. - Michael Somos, Apr 05 2015
EXAMPLE
G.f. = 1 - x -x^4 + x^6 - x^7 + 2*x69 - x^10 + 2*x^12 - 2*x^13 + 2*x^15 + ...
G.f. = 1/q - q^3 - q^15 + q^23 - q^27 + 2*q^35 - q^39 + 2*q^47 - 2*q^51 + ...
MATHEMATICA
a[ n_] := SeriesCoefficient[ q^(1/4) EllipticTheta[ 2, Pi/4, q^(1/2)] / EllipticTheta[ 2, Pi/4, q^(3/2)], {q, 0, n}]; (* Michael Somos, Apr 05 2015 *)
PROG
(PARI) {a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x + A) * eta(x^4 + A) * eta(x^6 + A) / (eta(x^2 + A) * eta(x^3 + A) * eta(x^12 + A)), n))};
CROSSREFS
KEYWORD
sign
AUTHOR
Michael Somos, Nov 22 2007, Nov 23 2007
STATUS
approved