A263002 Expansion of (f(-x^5) / f(-x))^2 in powers of x where f() is a Ramanujan theta function.

1, 2, 5, 10, 20, 34, 61, 100, 165, 260, 408, 620, 940, 1390, 2045, 2960, 4257, 6040, 8525, 11900, 16522, 22738, 31130, 42300, 57210, 76872, 102834, 136800, 181230, 238900, 313725, 410160, 534330, 693330, 896655, 1155420, 1484274, 1900420, 2426215, 3088100

Offset: 0

Author

Michael Somos, Oct 07 2015

Keywords

• nonn

Comments

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

Number of 5-regular bipartitions of n. - N. J. A. Sloane, Oct 20 2019

Examples

			G.f. = 1 + 2*x + 5*x^2 + 10*x^3 + 20*x^4 + 34*x^5 + 61*x^6 + 100*x^7 + ...
G.f. = q + 2*q^4 + 5*q^7 + 10*q^10 + 20*q^13 + 34*q^16 + 61*q^19 + 100*q^22 + ...
		

References

• Kathiravan, T., and S. N. Fathima. "On L-regular bipartitions modulo L." The Ramanujan Journal 44.3 (2017): 549-558.

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..1000
• Vaclav Kotesovec, A method of finding the asymptotics of q-series based on the convolution of generating functions, arXiv:1509.08708 [math.CO], Sep 30 2015
• Michael Somos, Introduction to Ramanujan theta functions
• Eric Weisstein's World of Mathematics, Ramanujan Theta Functions

Crossrefs

Cf. A058511.

Number of r-regular bipartitions of n for r = 2,3,4,5,6: A022567, A328547, A001936, A263002, A328548.

Programs

Maple

f:=(k,M) -> mul(1-q^(k*j),j=1..M);
LRBP := (L,M) -> (f(L,M)/f(1,M))^2;
S := L -> seriestolist(series(LRBP(L,80),q,60));
S(5); # N. J. A. Sloane, Oct 20 2019

Mathematica

a[ n_] := SeriesCoefficient[ (QPochhammer[ x^5] / QPochhammer[ x])^2, {x, 0, n}];

PARI

{a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( (eta(x^5 + A) / eta(x + A))^2, n))};

Formula

Expansion of q^(-1/3) * (eta(q^5) / eta(q))^2 in powers of q.

Euler transform of period 5 sequence [ 2, 2, 2, 2, 0, ...].

G.f. is a period 1 Fourier series which satisfies f(-1 / (45 t)) = (1/5) g(t) where q = exp(2 Pi i t) and g() is the g.f. of A058511.

Given g.f. A(x), then B(q) = q * A(q^3) satisfies 0 = f(B(q), B(q^2)) where f(u, v) = (u - v^2) * (v - u^2) - 4*u^2*v^2.

Convolution inverse is A058511.

a(n) ~ exp(4*Pi*sqrt(n/15)) / (sqrt(2) * 3^(1/4) * 5^(5/4) * n^(3/4)). - Vaclav Kotesovec, Oct 14 2015

See Maple code for a simple g.f. - N. J. A. Sloane, Oct 20 2019