OFFSET
1,2
COMMENTS
Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).
Given g.f. A(q), Greenhill (1895) denotes -64 * A(q^2) by tau_0 on page 409 equation (43). - Michael Somos, Jul 17 2013
REFERENCES
John H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer-Verlag, p. 102.
Albert Eagle, Elliptic functions as they should be, Galloway and Porter Ltd., Cambridge, pp. 72-73.
LINKS
Seiichi Manyama, Table of n, a(n) for n = 1..10000
Kevin Acres and David Broadhurst, Eta quotients and Rademacher sums, arXiv:1810.07478 [math.NT], 2018. See Table 1 p. 10.
A. G. Greenhill, The Transformation and Division of Elliptic Functions, Proceedings of the London Mathematical Society (1895) 403-486.
R. S. Maier, On Rationally Parametrized Modular Equations, arXiv:math/0611041 [math.NT], 2006-2008, see page 4 equation (4).
MathOverflow, Identity for an infinite product [question from _T . Amdeberhan_, Oct 01 2024]
Michael Somos, Introduction to Ramanujan theta functions
Eric Weisstein's World of Mathematics, Ramanujan Theta Functions
Eric Weisstein's World of Mathematics, Elliptic Lambda Function
FORMULA
REVERT(A005149).
Euler transform of period 2 sequence [ 24, 0, 24, 0, ... ]. - Michael Somos, Mar 19 2004
Expansion of (lambda(q) / 16)^2 / (1 - lambda(q)) in powers of q = exp(2 Pi i t) where lambda() is the elliptic modular function A115977. - Michael Somos, Nov 19 2005
Expansion of q / chi(-q)^24 in powers of q where chi() is a Ramanujan theta function.
Expansion of (theta_2(q) * theta_3(q) / (2 * theta_4(q)^2))^4 = (theta_2(q^(1/2))^2 / (4*theta_4(q^(1/2)) * theta_3(q^(1/2))))^4 in powers of q.
G.f.: x * Product_{k > 0} (1 + x^k)^24 = x / Product_{k > 0} (1 - x^(2*k - 1))^24.
G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = u^2 - v - 48*u*v - 4096*u*v^2. - Michael Somos, Mar 19 2004
G.f. is a period 1 Fourier series which satisfies f(-1 / (2 t)) = (1/4096) g(t) where q = exp(2 Pi i t) and g() is the g.f. of A007191. - Michael Somos, Aug 19 2007
j(q) = (f(q) + 16)^3 / f(q), j(q^2) = (f(q) + 256)^3 / f(q)^2 where j(q) is the g.f. for A000521 and f(q) is 4096 times the g.f. for a(n). - Michael Somos, Oct 01 2007
Sum_{n>=1} exp(-2*Pi*n)*a(n) = 1/512. - Simon Plouffe, Feb 20 2011 [Proof: Sum_{n>=0} a(n)/exp(2*Pi*n) = exp(-2*Pi) * (phi(exp(-4*Pi)) / phi(exp(-2*Pi)))^24 = exp(-2*Pi) * A292821^24, where phi(q) is the Euler modular function. - Vaclav Kotesovec, May 13 2023]
a(n) ~ exp(2 * Pi * sqrt(2*n)) / (4096 * 2^(3/4) * n^(3/4)). - Vaclav Kotesovec, Mar 05 2015
a(1) = 1, a(n) = (24/(n-1))*Sum_{k=1..n-1} A000593(k)*a(n-k) for n > 1. - Seiichi Manyama, Apr 01 2017
G.f.: x*exp(24*Sum_{k>=1} (-1)^(k+1)*x^k/(k*(1 - x^k))). - Ilya Gutkovskiy, Feb 06 2018
Expansion of Delta(q^2)/Delta(q) in powers of q where the discriminant Delta(q) is the g.f. of A000594. - Michael Somos, May 27 2022
From Vaclav Kotesovec, May 08 2023, updated May 16 2023: (Start)
Sum_{n>=1} a(n) / exp(Pi*n) = exp(-Pi) * A292820^24 = 1/8.
Sum_{n>=1} a(n) / exp(3*Pi*n) = exp(-3*Pi) * A292887^24 = (2 + sqrt(3) - sqrt(9 + 6*sqrt(3)))^8 / 512.
Sum_{n>=1} a(n) / exp(4*Pi*n) = exp(-4*Pi) * A292822^24 = 99*sqrt(2)/2048 - 35/512.
Sum_{n>=1} a(n) / exp(5*Pi*n) = exp(-5*Pi) * A292904^24 = (2 + sqrt(5) - sqrt((15 + 7*sqrt(5))/2))^12 / 8.
Sum_{n>=1} a(n) / exp(6*Pi*n) = exp(-6*Pi) * (A363020/A363018)^24 = 385/512 + 7*sqrt(3)/16 - sqrt(74664 + 43134*sqrt(3))/256.
Sum_{n>=1} a(n) / exp(7*Pi*n) = exp(-7*Pi) * (A363119/A363117)^24 = (sqrt(7) - 1 - sqrt(22*sqrt(7) - 56))^3 / (2^(15/2) * (2^(1/4)*sqrt(5 + sqrt(7)) + (56 + 23*sqrt(7))^(1/4))^6).
Sum_{n>=1} a(n) / exp(8*Pi*n) = exp(-8*Pi) * (A292864/A259151)^24 = -8963/512 - 99*sqrt(2)/8 + 9*sqrt(126913704 + 89741542*sqrt(2))/4096.
Sum_{n>=1} a(n) / exp(9*Pi*n) = exp(-9*Pi) * (A363120/A363118)^24 = ((6*(3 + sqrt(3)))^(1/3) - 3)^8 / (8*((3*(6 + 7*sqrt(3) + 3*sqrt(14*sqrt(3) - 15)))^(1/3) - 3)^8).
Sum_{n>=1} a(n) / exp(Pi*n/2) = exp(-Pi/2) * A292819^24 = 35 + 99/2^(3/2).
Sum_{n>=1} (-1)^(n+1) * a(n) / exp(Pi*n) = 1/64.
Sum_{n>=1} (-1)^(n+1) * a(n) / exp(2*Pi*n) = 99/2^(3/2) - 35.
Sum_{n>=1} (-1)^(n+1) * a(n) / exp(3*Pi*n) = 97/64 - 7*sqrt(3)/8.
Sum_{n>=1} (-1)^(n+1) * a(n) / exp(4*Pi*n) = -5018696 - 3548754*sqrt(2) + (9/2)*sqrt(2487635528172 + 1759023951091*sqrt(2)).
Sum_{n>=1} (-1)^(n+1) * a(n) / exp(7*Pi*n) = 13880161/64 + 81972*sqrt(7) - 9*sqrt(74328271227 + 28093445864*sqrt(7))/8. (End)
The g.f. A(q) satisfies -(16)^2 * A(q^2) = (lambda(q) + lambda(-q)) = (lambda(q)*lambda(-q)), where lambda(q) = 16*q - 128*q^2 + 704*q^3 - ... is the elliptic modular function in powers of the nome q = exp(i*Pi*t), the g.f. of A115977; lambda(q) = k(q)^2, where k(q) = (theta_2(q) / theta_3(q))^2 is the elliptic modulus. - Peter Bala, Sep 26 2023
From Peter Bala, Sep 26 2023: (Start)
A(q^2) = -A(q)*A(-q).
A(q) = lambda(-q)^2/(16*lambda(q)) = lambda(-q)*(lambda(-q) - 1)/16. (End)
G.f. A(x) satisfies 0 = f(A(x), A(-x)) where f(u, v) = u + v + 48*u*v - 4096*u^2*v^2. - Michael Somos, Oct 07 2024
EXAMPLE
G.f. = q + 24*q^2 + 300*q^3 + 2624*q^4 + 18126*q^5 + 105504*q^6 + 538296*q^7 + ...
MAPLE
q*mul((1+q^m)^24, m=1..30); seq(coeff(series(%, q, n+1), q, n), n=1..25);
MATHEMATICA
a[ n_] := SeriesCoefficient[ q QPochhammer[ q, q^2]^-24, {q, 0, n}]; (* Michael Somos, Jul 11 2011 *)
a[ n_] := SeriesCoefficient[ q / Product[ 1 - q^k, {k, 1, n + 1, 2}]^24, {q, 0, n}]; (* Michael Somos, Jul 11 2011 *)
a[ n_] := With[ {m = ModularLambda[ Log[q]/(Pi I)]}, SeriesCoefficient[ (m/16)^2 / (1 - m), {q, 0, 2 n}]]; (* Michael Somos, Jul 11 2011 *)
a[ n_] := With[ {m = InverseEllipticNomeQ @ q}, SeriesCoefficient[ (m/16)^2 /(1 - m), {q, 0, 2 n}]]; (* Michael Somos, Jul 11 2011 *)
eta[q_]:=q^(1/6) QPochhammer[q]; a[n_]:=SeriesCoefficient[(eta[q^2] / eta[q])^24, {q, 0, n}]; Table[a[n], {n, 4, 25}] (* Vincenzo Librandi, Oct 18 2018 *)
PROG
(PARI) {a(n) = polcoeff( x * prod( k=1, n, 1 + x^k, 1 + x * O(x^n))^24, n)};
(PARI) {a(n) = my(A, A2, m); if( n<0, 0, A = x + O(x^2); m=1; while( m<=n, m*=2; A = subst( A, x, x^2); A2 = A * (1 + 16*A); A = 8 * A2 + (1 + 32*A) * sqrt(A2)); polcoeff( A + 16 * A^2, n))};
(PARI) {a(n) = my(A); if( n<1, 0, n--; A = x * O(x^n); polcoeff( (eta(x^2 + A) / eta(x + A))^24, n))};
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
EXTENSIONS
More terms from Michael Somos, Nov 24 2001
STATUS
approved