A008440 Expansion of Jacobi theta constant theta_2^6 /(64q^(3/2)).
1, 6, 15, 26, 45, 66, 82, 120, 156, 170, 231, 276, 290, 390, 435, 438, 561, 630, 651, 780, 861, 842, 1020, 1170, 1095, 1326, 1431, 1370, 1716, 1740, 1682, 2016, 2145, 2132, 2415, 2550, 2353, 2850, 3120, 2810, 3321, 3486, 3285, 3906, 4005, 3722, 4350
Offset: 0
Examples
G.f. = 1 + 6*x + 15*x^2 + 26*x^3 + 45*x^4 + 66*x^5 + 82*x^6 + ... - _Michael Somos_, Jun 25 2019 G.f. = q^3 + 6*q^7 + 15*q^11 + 26*q^15 + 45*q^19 + 66*q^23 + 82*q^27 + ...
References
- B. C. Berndt, Fragments by Ramanujan on Lambert series, in Number Theory and Its Applications, K. Gyory and S. Kanemitsu, eds., Kluwer, Dordrecht, 1999, pp. 35-49, see Entry 6.
- J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer-Verlag, p. 102.
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 0..1000 from Vincenzo Librandi)
- B. C. Berndt, Fragments by Ramanujan on Lambert series.
- K. Ono, S. Robins and P. T. Wahl, On the representation of integers as sums of triangular numbers, Aequationes mathematicae, August 1995, Volume 50, Issue 1-2, pp 73-94. Theorem 4.
Crossrefs
Programs
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Mathematica
CoefficientList[(QPochhammer[q^2]^2 / QPochhammer[q])^6 + O[q]^50, q] (* Jean-François Alcover, Nov 05 2015 *) a[ n_] := If[ n < 0, 0, -DivisorSum[ 4 n + 3, Re[I^(# - 1)] #^2 &] / 8]; (* Michael Somos, Jun 25 2019 *)
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PARI
{a(n) = if( n<0, 0, polcoeff( sum(k=0, (sqrtint(8*n+1)-1)\2, x^((k^2+k)/2), x * O(x^n))^6, n))}; /* Michael Somos, May 23 2006 */
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PARI
{a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( (eta(x^2 + A)^2 / eta(x + A))^6, n))}; /* Michael Somos, May 23 2006 */
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PARI
{a(n)= -sumdiv(4*n + 3, d, real(I^(d-1))*d^2)/8}; /* Michael Somos, Oct 24 2012 */
Formula
Expansion of Ramanujan phi^6(q) in powers of q.
Expansion of q^(-3/4)(eta(q^2)^2/eta(q))^6 in powers of q.
Euler transform of period 2 sequence [6, -6, ...]. - Michael Somos, May 23 2006
G.f.: (Sum_{n>=0} x^((n^2+n)/2))^6.
a(n) = (-1/8)*Sum_{d divides (4n+3)} Chi_2(4;d)*d^2. - Michel Marcus, Oct 24 2012. See the Ono et al. link. Theorem 4.
a(n) =(-1/8)*A002173(4*n+3). This is the preceding formula. - Wolfdieter Lang, Jan 12 2017
a(0) = 1, a(n) = (6/n)*Sum_{k=1..n} A002129(k)*a(n-k) for n > 0. - Seiichi Manyama, May 06 2017
G.f.: exp(Sum_{k>=1} 6*(x^k/k)/(1 + x^k)). - Ilya Gutkovskiy, Jul 31 2017
Comments