A053254 Coefficients of the '3rd-order' mock theta function nu(q).

1, -1, 2, -2, 2, -3, 4, -4, 5, -6, 6, -8, 10, -10, 12, -14, 15, -18, 20, -22, 26, -29, 32, -36, 40, -44, 50, -56, 60, -68, 76, -82, 92, -101, 110, -122, 134, -146, 160, -176, 191, -210, 230, -248, 272, -296, 320, -350, 380, -410, 446, -484, 522, -566, 612, -660, 715, -772, 830, -896, 966, -1038

Offset: 0

Author

Dean Hickerson, Dec 19 1999

Keywords

• sign
• easy

Comments

In Watson 1936 the function is denoted by upsilon(q). - Michael Somos, Jul 25 2015

Examples

			G.f. = 1 - x + 2*x^2 - 2*x^3 + 2*x^4 - 3*x^5 + 4*x^6 - 4*x^7 + 5*x^8 + ...
		

References

• George E. Andrews, The Theory of Partitions, Addison-Wesley, 1976, (Example 6, p. 29).
• Srinivasa Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa Publishing House, New Delhi, 1988, p. 31.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Leila A. Dragonette, Some asymptotic formulas for the mock theta series of Ramanujan, Trans. Amer. Math. Soc., 72 (1952) 474-500.
• George N. Watson, The final problem: an account of the mock theta functions, J. London Math. Soc., 11 (1936) 55-80.

Crossrefs

Other '3rd-order' mock theta functions are at A000025, A053250, A053251, A053252, A053253, A053255.

Cf. A058140, A067357, A376628.

Programs

Mathematica

Series[Sum[q^(n(n+1))/Product[1+q^(2k+1), {k, 0, n}], {n, 0, 9}], {q, 0, 100}]

PARI

/* Continued Fraction Expansion: */
{a(n)=local(CF); CF=1+x; for(k=0, n, CF=1/(1 + x^(n-k+1)*(1 - x^(n-k+1))*CF+x*O(x^n))); polcoeff(CF, n)} \\ Paul D. Hanna, Jul 09 2013

Formula

G.f.: nu(q) = Sum_{n >= 0} q^(n*(n+1))/((1+q)*(1+q^3)*...*(1+q^(2*n+1)))

(-1)^n*a(n) = number of partitions of n in which even parts are distinct and if k occurs then so does every positive even number less than k.

G.f.: 1/(1 + x*(1-x)/(1 + x^2*(1-x^2)/(1 + x^3*(1-x^3)/(1 + x^4*(1-x^4)/(1 + x^5*(1-x^5)/(1 + ...)))))), a continued fraction. - Paul D. Hanna, Jul 09 2013

a(2*n) = A085140(n). a(2*n + 1) = - A053253(n). - Michael Somos, Jul 25 2015

a(n) ~ (-1)^n * exp(Pi*sqrt(n/6)) / (2^(3/2)*sqrt(n)). - Vaclav Kotesovec, Jun 15 2019

From Peter Bala, Jan 03 2025: (Start)

a(n) = (-1)^n * A067357(n).

nu(-q) = Sum_{n >= 0} q^n * (1 + q)*(1 + q^3)*...*(1 + q^(2*n-1)) (Andrews, p. 29: in Example 6 take x = q and y = -q).

Conjecture: a(n) = (-1)^n * (A376628(n) + A376628(n+1)), or equivalently, (1 + q * nu(-q))/(1 + q) = Sum_{n >= 0} q^(n*(n+1))/((1 - q)*(1 - q^3)*...*(1 - q^(2*n-1))), the g.f. of A376628. (End)