cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A116916 Expansion of q^(-1/8) * (eta(q)^3 + 3 * eta(q^9)^3) in powers of q^3.

Original entry on oeis.org

1, 5, -7, 0, 0, -11, 0, 13, 0, 0, 0, 0, 17, 0, 0, -19, 0, 0, 0, 0, 0, 0, -23, 0, 0, 0, 25, 0, 0, 0, 0, 0, 0, 0, 0, 29, 0, 0, 0, 0, -31, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -35, 0, 0, 0, 0, 0, 37, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 41, 0, 0, 0, 0, 0, 0, -43, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -47, 0, 0, 0, 0, 0
Offset: 0

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Author

Michael Somos, Feb 26 2006

Keywords

Comments

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

Examples

			1 + 5*x - 7*x^2 - 11*x^5 + 13*x^7 + 17*x^12 - 19*x^15 - 23*x^22 + 25*x^26 + ...
q + 5*q^25 - 7*q^49 - 11*q^121 + 13*q^169 + 17*q^289 - 19*q^361 +...
		

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[n_] := SeriesCoefficient[QPochhammer[x + x*O[x]^(3n)]^3 + 3x * QPochhammer[x^9 + O[x]^(3n)]^3, 3n]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Nov 06 2015, adapted from PARI *)
    a[ n_] := With[ {m = Sqrt[ 24 n + 1]}, If[ IntegerQ[ m], m KroneckerSymbol[ 3, m] KroneckerSymbol[ -3, m], 0]]; (* Michael Somos, Apr 27 2018 *)
  • PARI
    {a(n) = if( issquare( 24*n + 1, &n), n * kronecker( 3, n) * kronecker( -3, n))};
    
  • PARI
    {a(n) = if( n<1, n==0, n*=3; polcoeff( eta(x + x * O(x^n))^3 + 3 * x * eta(x^9 + x * O(x^n))^3, n))};

Formula

Expansion of f(-x) * a(x) in powers of x where f() is a Ramanujan theta function and a() is a cubic AGM theta function.
Expansion of f(-x)^3 + 3 * x * f(-x^9)^3 in powers of x^3 where f() is a Ramanujan theta function.
G.f. is a period 1 Fourier series which satisfies f(-1 / (576 t)) = 4608^(1/2) (t / i)^(3/2) g(t) where q = exp(2 Pi i t) and g() is g.f. for A202394.
G.f.: Sum_{k in Z} (-1)^k * (6*k + 1) * x^(k * (3*k + 1) / 2).
a(5*n + 3) = a(5*n + 4) = 0. a(25*n + 1) = 5 * a(n).
a(n) = A010816(3*n).