A393567 Coefficients of the formal power series quotient of the '5th-order' mock theta function f_0(q) by the 3rd-order mock theta function f(q).

1, 0, 1, -3, 8, -20, 52, -134, 342, -877, 2254, -5783, 14836, -38076, 97712, -250738, 643436, -1651170, 4237168, -10873260, 27902576, -71602589, 183743960, -471517110, 1209990193, -3105033096, 7968023710, -20447254552, 52471005784, -134649198974

Offset

0,4

Comments

This is the coefficient sequence a(n) defined by A(q) = f_0(q)/f(q), where f_0(q) = Sum_{n>=0} q^(n^2)/(-q;q)_n and f(q) = Sum_{n>=0} q^(n^2)/(-q;q)_n^2.

Here (-q;q)_n = Product_{k=1..n} (1 + q^k). Since f(q) has constant term 1, the quotient is well-defined as a formal power series over the rationals.

References

George E. Andrews and Frank G. Garvan, Ramanujan's "lost" notebook VI: The mock theta conjectures, Adv. Math. 73 (1989), 242-255.

Srinivasa Ramanujan, Collected Papers, Chelsea, New York, 1962, pp. 354-355.

George N. Watson, The final problem: an account of the mock theta functions, J. London Math. Soc. 11 (1936), 55-80.

Links

Table of n, a(n) for n=0..29.

Formula

G.f.: Sum_{n>=0} a(n)*q^n = (Sum_{n>=0} q^(n^2)/(-q;q)_n) / (Sum_{n>=0} q^(n^2)/(-q;q)_n^2).

a(0)=1 and for n>=1, a(n) = A053256(n) - Sum_{k=1..n} A000025(k)*a(n-k).

Example

A(q) = 1 + q^2 - 3*q^3 + 8*q^4 - 20*q^5 + 52*q^6 - 134*q^7 + ...
Hence a(0)=1, a(1)=0, a(2)=1, a(3)=-3, a(4)=8.

Prog

(Python)
from fractions import Fraction
def conv(a, b, N):
    c = [Fraction(0)] * (N + 1)
    for i, ai in enumerate(a):
        if ai == 0:
            continue
        for j, bj in enumerate(b[:N + 1 - i]):
            c[i + j] += ai * bj
    return c
def invfac(k, p, N):
    c = [Fraction(0)] * (N + 1)
    j = 0
    while j * k <= N:
        c[j * k] = Fraction(((-1) ** j) * (j + 1 if p == 2 else 1))
        j += 1
    return c
def mock(N, p):
    s = [Fraction(0)] * (N + 1)
    d = [Fraction(1)] + [Fraction(0)] * N
    n = 0
    while n * n <= N:
        for i, u in enumerate(d[:N + 1 - n * n]):
            s[n * n + i] += u
        n += 1
        d = conv(d, invfac(n, p, N), N)
    return s
def div(num, den, N):
    a = [Fraction(0)] * (N + 1)
    for n in range(N + 1):
        t = num[n] - sum(den[k] * a[n - k] for k in range(1, n + 1))
        a[n] = t / den[0]
    return a
N = 29
print([int(x) for x in div(mock(N, 1), mock(N, 2), N)])

Crossrefs

Cf. A000025, A053256.

Sequence in context: A295346 A385641 A027220 * A305823 A333679 A230953

Adjacent sequences: A393564 A393565 A393566 * A393568 A393569 A393570

Keyword

sign

Author

Joesph Daniel Burke III, Mar 27 2026

Status

approved