A212326 G.f. satisfies: A(x) = theta_3( x*A(x) )^2, where theta_3(x) is Jacobi's theta_3 function.
1, 4, 20, 112, 676, 4312, 28704, 197600, 1397060, 10090676, 74152456, 552666448, 4167528000, 31736182776, 243698432960, 1884809367456, 14668777816708, 114789815231560, 902661488046900, 7129068237647408, 56524456978032904, 449752267499647104
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + 4*x + 20*x^2 + 112*x^3 + 676*x^4 + 4312*x^5 + 28704*x^6 + ... Given g.f. A(x), let q = x*A(x), then by a q-series identity: A(x) = 1 + 4*q/(1+q^2) + 4*q^2/(1+q^4) + 4*q^3/(1+q^6) + 4*q^4/(1+q^8) + ... A(x) = (1 + 2*q + 2*q^4 + 2*q^9 + 2*q^16 + 2*q^25 + ...)^2. ... Illustrate a(n) = [x^n] theta_3(x)^(2*n+2) / (n+1) by the following table of coefficients in powers theta_3(x)^(2*n+2) for n>=0: n=0: [(1), 4, 4, 0, 4, 8, 0, 0, 4, 4, 8, 0, 0, 8, 0, 0, ...]; n=1: [1, (8), 24, 32, 24, 48, 96, 64, 24, 104, 144, 96, 96, 112, ...]; n=2: [1, 12, (60), 160, 252, 312, 544, 960, 1020, 876, 1560, ...]; n=3: [1, 16, 112, (448), 1136, 2016, 3136, 5504, 9328, 12112, ...]; n=4: [1, 20, 180, 960, (3380), 8424, 16320, 28800, 52020, 88660, ...]; n=5: [1, 24, 264, 1760, 7944, (25872), 64416, 133056, 253704, ...]; n=6: [1, 28, 364, 2912, 16044, 64792, (200928), 503360, ...]; n=7: [1, 32, 480, 4480, 29152, 140736, 525952, (1580800), ...]; ... where the coefficients in parenthesis form the initial terms of this sequence: A = [1/1, 8/2, 60/3, 448/4, 3380/5, 25872/6, 200928/7, 1580800/8, ...]. SPECIFIC VALUES. A(t) = Pi^(1/2)/gamma(3/4)^2 = 1.180340599016096226045337940558488... at t = exp(-Pi)*Pi^(-1/2)*gamma(3/4)^2 = 0.03661139699828536055676055675... A(t) = 2 at t = 0.102664339082522765394312666828624696651144948974... A(t) = 3/2 at t = 0.074809250381600864041128779542209664527032... A(1/9) = 2.33377643668206735588... A(1/10) = 1.926962708064002748766926777581932826845759... A(1/11) = 1.732429109249288812362796164718509396134409... A(1/12) = 1.610547777446005767621845977342631412384684...
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..625
- Eric Weisstein's World of Mathematics, Jacobi Theta Functions
Crossrefs
Cf. A166952.
Programs
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Mathematica
CoefficientList[1/x * InverseSeries[Series[x/EllipticTheta[3, 0, x]^2, {x, 0, 25}], x], x] (* Vaclav Kotesovec, Nov 16 2023 *) (* Calculation of constants {d,c}: *) {1/r, s/Sqrt[Pi*(1 + 4 * r^2 * s^(3/2) * Derivative[0, 0, 2][EllipticTheta][3, 0, r*s])]} /. FindRoot[{s == EllipticTheta[3, 0, r*s]^2, 2*r*Sqrt[s]*Derivative[0, 0, 1][EllipticTheta][3, 0, r*s] == 1}, {r, 1/8}, {s, 3/2}, WorkingPrecision -> 70] (* Vaclav Kotesovec, Nov 16 2023 *) -
PARI
{a(n) = my(A=1+x); for(i=1,n, A = 1 + 4*sum(m=1,n,(x*A)^m/(1+(x*A+x*O(x^n))^(2*m)))); polcoef(A,n)} -
PARI
{a(n) = my(A=1+x); for(i=1,n, A = (1 + 2*sum(m=1,sqrtint(n+1),(x*A+x*O(x^n))^(m^2)))^2); polcoef(A,n)} -
PARI
{a(n) = my(A=1+x); for(i=1,n, A = prod(m=1,n,(1 - (-x)^m*A^m)/(1 + (-x)^m*A^m +x*O(x^n)))^2); polcoef(A,n)} for(n=0,30,print1(a(n),", "))
Formula
G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) A(x) = (1 + 2*Sum_{n>=1} (x*A(x))^(n^2) )^2.
(2) A(x) = 1 + 4*Sum_{n>=1} (x*A(x))^n / (1 + (x*A(x))^(2*n)).
(3) A(x) = Product_{n>=1} (1 - (-x)^n*A(x)^n)^2 / (1 + (-x)^n*A(x)^n)^2.
(4) A( x/theta_3(x)^2 ) = theta_3(x)^2.
(5) A(x) = (1/x)*Series_Reversion(x/theta_3(x)^2), where theta_3(x) = 1 + 2*Sum_{n>=1} x^(n^2).
(6) A(x) = 1 + 4*Sum_{n>=1} (-1)^(n-1) * (x*A(x))^(2*n-1) / (1 - (x*A(x))^(2*n-1)). See formula 57 in the Weisstein link. - Paul D. Hanna, Nov 02 2024
a(n) = [x^n] theta_3(x)^(2*n+2) / (n+1).
a(n) ~ c * d^n / n^(3/2), where d = 8.54148362320612002563896433934021488424489314523756456892173912667254... and c = 1.2437677914754786190190604348779334425700766084860016245397106832001... - Vaclav Kotesovec, Nov 16 2023