octagonal number theorem $q$-Pochhammer symbol expression

Question

Setting the exponents of this analogue of the series in Euler's Pentagonal Number theorem to be the octagonal numbers: $$U(q)= \sum_{n\in\mathbb{Z}} (-1)^{n}q^{n(6n-4)/2}$$

in mpmath:

>>> def oct(q):
...     return nsum(lambda n: ((-1)**n)* q**((6*n*n+4*n)/2), [-inf,inf])
... 
>>> taylor(lambda q: 1/oct(q), 0,30)
[mpf('1.0'), mpf('1.0'), mpf('1.0'), mpf('1.0'), mpf('1.0'), mpf('2.0'),        
mpf('3.0'), mpf('4.0'), mpf('4.0'), mpf('4.0'), mpf('5.0'), mpf('7.0'),    
mpf('10.0'), mpf('12.0'), mpf('13.0'), mpf('14.0'), mpf('16.0'), mpf('21.0'), 
mpf('27.0'), mpf('32.0'), mpf('35.0'), mpf('38.0'), mpf('44.0'), 
mpf('53.999999999999993'), mpf('67.0'), mpf('78.0'), 
mpf('85.999999999999986'), mpf('94.0'), mpf('107.0'), mpf('128.0'), 
mpf('153.0')]

Okay, that's easily oeis:A195848: Expansion of 1 / (psi(x) * chi(-x)) in powers of x where psi(), chi() are Ramanujan theta functions., so I thought, let me test that:

Using the definitions for the Ramanujan theta functions on mathworld, $\psi(q)=(-q;q)_{\infty}(q^{2};q^{2})_{\infty}$ and $\chi(q)=(-q,q^{2})_{\infty}$

>>> def psi(q):
...     return qp(-q,q)*qp(q*q,q*q)
... 
>>> def chi(q):
...     return qp(-q,q*q)
... 

>>> taylor(lambda q: 1/psi(q)*chi(-q), 0,30)
[mpf('1.0'), mpf('-2.0'), mpf('2.0'), mpf('-4.0'), mpf('7.0'), mpf('-10.0'),     
mpf('14.0'), mpf('-20.0'), mpf('30.0'), mpf('-42.0'), mpf('56.0'), 
mpf('-76.0'), mpf('105.0'), mpf('-140.0'), mpf('182.0'), mpf('-240.0'),  
mpf('317.0'), mpf('-410.0'), mpf('524.0'), mpf('-672.0'), mpf('861.0'), 
mpf('-1092.0'), mpf('1372.0000000000002'), mpf('-1723.9999999999998'),   
mpf('2167.0'), mpf('-2702.0'), mpf('3346.0'), mpf('-4144.0'), mpf('5125.0'), 
mpf('-6302.0'), mpf('7716.0')]

But those absolute values of those correspond to A058630:McKay-Thompson series of class 32B for the Monster group., not to $1/U(q)$!

What's the correct expression for $1/U(q)$ in terms of $q$-Pochhammer symbols?

Answer 1

You should have parenthesized your psi(q)*chi(-q), but that would not yield a match either, as $\psi(q)\,\chi(-q) = (q^2,q^2)_\infty$ whose reciprocal just generates partition numbers (at even index values).

But Michael Somos' formula works: The series can be given as $$\frac{(q^2,q^2)_\infty(q^3,q^3)_\infty}{(q,q)_\infty(q^6,q^6)_\infty^2} = \frac{1}{\psi(q^3)\,\chi(-q)}$$ So you have found a typo in the title of A195848. It should read

Expansion of 1/(psi(x^3) * chi(-x)) in powers of x where psi(), chi() are Ramanujan theta functions.

Feel free to report that typo.

Update: The title of A195848 has been corrected.