A355151 - OEIS

%I #14 Jul 22 2022 17:45:12

%S 1,-1,1,0,-3,9,-15,6,56,-222,456,-322,-1562,7604,-18062,18544,46676,

%T -293986,797456,-1066316,-1265854,11945620,-36984046,60359006,

%U 20788743,-493365441,1756737971,-3368998030,835358767,20230816709,-84341891481,185935403556

%N G.f. A(x) satisfies: (1 - x)/(1 + x) = Sum_{n=-oo..+oo} (-1)^n * A(x)^(n^2).

%C Is there a pattern to the signs of the terms?

%C Conjectures:

%C (C.1) a(4*n) == 0 (mod 2) for n >= 0.

%C (C.2) a(4*n+1) == a(4*n+2) == a(4*n+3) (mod 2) for n >= 0.

%C (C.3) a(4*n+1) == a(4*n+3) (mod 4) for n >= 0.

%H Paul D. Hanna, <a href="/A355151/b355151.txt">Table of n, a(n) for n = 1..625</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/JacobiThetaFunctions.html">Jacobi Theta Functions</a>

%F G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies:

%F (1) (1 - x)/(1 + x) = Sum_{n=-oo..+oo} (-1)^n * A(x)^(n^2).

%F (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.

%F (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.

%F (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.

%F (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.

%F SPECIFIC VALUES.

%F (V.1) A(x) = -exp(-Pi) at x = (1-y)/(1+y) where y = Pi^(1/4)/gamma(3/4).

%F (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.

%F (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.

%F (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.

%F (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)).

%e 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 + ...

%e such that

%e (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 +- ...

%e By the Jacobi triple product

%e (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 * ...

%o (PARI) {a(n) = my(A=[0,1],t); for(i=1,n, A = concat(A,0); t = ceil(sqrt(#A+4));

%o 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]}

%o for(n=1,40,print1(a(n),", "))

%Y Cf. A354651.

%K sign

%O 1,5

%A _Paul D. Hanna_, Jun 21 2022