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).
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)) = (lamda(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)
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