A215594 Expansion of f(-x, -x^4) / f(x, x^4) in powers of x where f(,) is Ramanujan's two-variable theta function.

1, -2, 2, -2, 0, 2, -4, 6, -4, 0, 6, -12, 14, -10, 0, 14, -26, 30, -22, 0, 28, -52, 60, -42, 0, 54, -100, 112, -78, 0, 100, -180, 202, -140, 0, 174, -314, 350, -240, 0, 296, -532, 588, -402, 0, 492, -876, 966, -658, 0, 794, -1412, 1550, -1050, 0, 1260, -2232

Offset: 0

Author

Michael Somos, Aug 16 2012

Keywords

• sign
• easy

Comments

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

Let A(q) denote the g.f. of this sequence. Let m be a nonzero integer. The simple continued fraction expansions of the real numbers A(1/(2*m)) and A(1/(2*m+1)) may be predictable. For a given positive integer n, the sequence of the n-th partial denominators of the continued fractions are conjecturally polynomial or quasi-polynomial in m for m sufficiently large. An example is given below. Cf. A080054 and A098151. - Peter Bala, Jun 10 2025

Examples

			1 - 2*x + 2*x^2 - 2*x^3 + 2*x^5 - 4*x^6 + 6*x^7 - 4*x^8 + 6*x^10 - 12*x^11 + ...
From _Peter Bala_, Jun 10 2025: (Start)
G.f.: A(q) = f(-q, -q^4) / f(q, q^4).
Simple continued fraction expansions of A(1/(2*m)):
  m = 2 [0; 1   1  2    7  1  1   1  10   1  2   12     82  1  6  1   48 ...]
  m = 3 [0; 1   2  2   17  1  1   2  14   2  2   24    318  1  1  1    1 ...]
  m = 4 [0; 1   3  2   31  1  1   3  18   3  2   40    810  2  7  2  161 ...]
  m = 5 [0; 1   4  2   49  1  1   4  22   4  2   60   1654  2  1  1    1 ...]
  m = 6 [0; 1   5  2   71  1  1   5  26   5  2   84   2946  3  7  1    1 ...]
  m = 7 [0; 1   6  2   97  1  1   6  30   6  2  112   4782  3  1  1    1 ...]
  m = 8 [0; 1   7  2  127  1  1   7  34   7  2  144   7258  4  7  1    2 ...]
  m = 9 [0; 1   8  2  161  1  1   8  38   8  2  180  10470  4  1  1    1 ...]
 m = 10 [0; 1   9  2  199  1  1   9  42   9  2  220  14514  5  7  1    3 ...]
 m = 11 [0; 1  10  2  241  1  1  10  46  10  2  264  19486  5  1  1    1 ...]
 m = 12 [0; 1  11  2  287  1  1  11  50  11  2  312  25482  6  7  1    4 ...]
 ...
The sequence of the 4th partial denominators [7, 17, 31, 49, ...] appears to be given by the polynomial 2*m^2 - 1 for m >= 2.
The sequence of the 11th partial denominators [12, 24, 40, 60, ...] appears to be given by the polynomial 2*(m^2 + m) for m >= 2.
The sequence of the 12th partial denominators [82, 318, 810, 1654, ...] appears to be given by the polynomial 2*(8*m^3 - 8*m^2 + 6*m - 3) for m >= 2.
The sequence of the 16th partial denominators appears to become quasi-polynomial in m for m >= 5. (End)
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Michael Somos, Introduction to Ramanujan theta functions
• Eric Weisstein's World of Mathematics, Ramanujan Theta Functions

Crossrefs

Cf. A080054, A098151.

Programs

Mathematica

f[x_, y_] := QPochhammer[-x, x*y]*QPochhammer[-y, x*y]*QPochhammer[x*y, x*y]; A215594[n_] := SeriesCoefficient[f[-x, -x^4]/f[x, x^4], {x, 0, n}]; Table[A215594[n], {n,0,50}] (* G. C. Greubel, Jun 18 2017 *)

PARI

{a(n) = local(A, s); if( n<0, 0, A = x * O(x^n); s = sqrtint( 40*n + 9); polcoeff( sum( k=(-s + 6)\10, (s - 3)\10, (-1)^k *  x^((5*k + 3)*k/2), A) / sum( k=(-s + 6)\10, (s - 3)\10,  x^((5*k + 3)*k/2), A), n))}

Formula

Euler transform of period 10 sequence [ -2, 1, 0, -2, 0, -2, 0, 1, -2, 0, ...].

a(5*n + 4) = 0.