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A342223
Product_{n>=1} 1 + a(n)*x^n = Sum_{n=-oo..oo} x^(n^2) = theta_3(x).
0
2, 0, 0, 2, -4, 8, -16, 32, -54, 108, -184, 368, -628, 1296, -2160, 4610, -7708, 15848, -27592, 58316, -98496, 207576, -364720, 756872, -1341970, 2778300, -4918536, 10443152, -18512788, 37698416, -69273664, 145105952, -258224544, 534996900, -981494752, 2020011290, -3714566308, 7614288360
OFFSET
1,1
COMMENTS
Coefficients in the power product expansion for theta_3(x), the third Jacobi theta function, described in A000122, also denoted theta_3(0, x) or theta_00(1, x).
See A147541 for additional references.
a(9) = -54 is the first term whose absolute value is not a power of 2.
REFERENCES
Bill Gosper and Joerg Arndt, Discussions in Math-Fun Mailing List, circa Feb 25 2021 - Mar 2 2021.
LINKS
H. Gingold, H. W. Gould, and Michael E. Mays, Power Product Expansions, Utilitas Mathematica 34 (1988), 143-161.
EXAMPLE
This gives 1 + 2x + 2x^4 + 2x^9 + ... = (1+2x)*(1+2x^4)*(1-4x^5)*(1+8x^6)*...
To compute this sequence's terms, start with the series expansion 1 + 2x + 2x^4 + 2x^9 + ...; this gives a(1) = 2, then divide by 1 + a(1)*x to get 1 + 2x^4 - 4x^5 + 8x^6 - 16x^7 ...; this gives a(2) = a(3) = 0 and a(4) = 2, then divide by 1 + a(4)*x to get 1 - 4x^5 + 8x^6 - 16x^7 ...; this gives a(5) = -4, then divide by 1 + a(5)*x to get 1 + 8x^6 - 16x^7 ...
MATHEMATICA
FoldPairList[{Coefficient[#1, q^#2], #1/(1 + q^#2 Coefficient[#1, q^#2])} &, #, Range[#[[5]] - 1]] &[Series[EllipticTheta[3, 0, q], {q, 0, 100}]] (* based on code from Bill Gosper, Feb 25 2021 *)
CROSSREFS
KEYWORD
sign
AUTHOR
Neil Bickford, Mar 06 2021
STATUS
approved