OFFSET
1,5
COMMENTS
Is there a pattern to the signs of the terms?
Conjectures:
(C.1) a(4*n) == 0 (mod 2) for n >= 0.
(C.2) a(4*n+1) == a(4*n+2) == a(4*n+3) (mod 2) for n >= 0.
(C.3) a(4*n+1) == a(4*n+3) (mod 4) for n >= 0.
LINKS
Paul D. Hanna, Table of n, a(n) for n = 1..625
Eric Weisstein's World of Mathematics, Jacobi Theta Functions
FORMULA
G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies:
(1) (1 - x)/(1 + x) = Sum_{n=-oo..+oo} (-1)^n * A(x)^(n^2).
(2) (1 - x)/(1 + x) = Product_{n>=1} (1 - A(x)^(2*n)) * (1 - A(x)^(2*n-1))^2, by the Jacobi triple product identity.
(3) (1 - x)^2/(1 + x)^2 = 1 + 4*Sum_{n>=1} (-1)^n * A(x)^(2*n-1)/(1 + A(x)^(2*n-1)), by a q-series identity for the Jacobi theta_4 function.
(4) (1 - x)^4/(1 + x)^4 = 1 + 8*Sum_{n>=1} (-1)^n * n * A(x)^n/(1 + A(x)^n), by a q-series identity for the Jacobi theta_4 function.
(5) A( (1 - theta_4(x))/(1 + theta_4(x)) ) = x, where theta_4(x) = 1 + 2*Sum_{n>=1} (-x)^(n^2) is the Jacobi theta_4 function.
SPECIFIC VALUES.
(V.1) A(x) = -exp(-Pi) at x = (1-y)/(1+y) where y = Pi^(1/4)/gamma(3/4).
(V.2) A(x) = -exp(-2*Pi) at x = (1-y)/(1+y) where y = Pi^(1/4)/gamma(3/4) * (6 + 4*sqrt(2))^(1/4)/2.
(V.3) A(x) = -exp(-3*Pi) at x = (1-y)/(1+y) where y = Pi^(1/4)/gamma(3/4) * (27 + 18*sqrt(3))^(1/4)/3.
(V.4) A(x) = -exp(-4*Pi) at x = (1-y)/(1+y) where y = Pi^(1/4)/gamma(3/4) * (8^(1/4) + 2)/4.
(V.5) A(x) = -exp(-sqrt(3)*Pi) at x = (1-y)/(1+y) where y = gamma(4/3)^(3/2)*3^(13/8)/(Pi*2^(2/3)).
EXAMPLE
G.f. A(x) = x - x^2 + x^3 - 3*x^5 + 9*x^6 - 15*x^7 + 6*x^8 + 56*x^9 - 222*x^10 + 456*x^11 - 322*x^12 - 1562*x^13 + 7604*x^14 - 18062*x^15 + ...
such that
(1 - x)/(1 + x) = 1 - 2*A(x) + 2*A(x)^4 - 2*A(x)^9 + 2*A(x)^16 - 2*A(x)^25 + 2*A(x)^36 - 2*A(x)^49 +- ...
By the Jacobi triple product
(1 - x)/(1 + x) = (1 - A(x)^2)*(1 - A(x))^2 * (1 - A(x)^4) * (1 - A(x)^3)^2 * (1 - A(x)^6) * (1 - A(x)^5)^2 * (1 - A(x)^8) * (1 - A(x)^7)^2 * ...
PROG
(PARI) {a(n) = my(A=[0, 1], t); for(i=1, n, A = concat(A, 0); t = ceil(sqrt(#A+4));
A[#A] = polcoeff( -(1-x)/(1+x) + 1 + 2*sum(n=1, t, (-1)^n * Ser(A)^(n^2)), #A-1)/2); H=A; A[n+1]}
for(n=1, 40, print1(a(n), ", "))
CROSSREFS
KEYWORD
sign
AUTHOR
Paul D. Hanna, Jun 21 2022
STATUS
approved