MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am trying to do some experimentation with the values of Thompson series, but I have having a hard time finding a table that has these Thompson series with as many terms as I'd like. The tables I've seen only give me the required Head character values up to the $q^{10}$ term.

Can someone point me to a resource that either has more terms, or (even better), a resource that might help me compute them myself in sage?

Thanks a lot for any help.

share|cite|improve this question
up vote 9 down vote accepted

From MathSciNet:

MR1037906 (90m:11065) Reviewed McKay, John(3-CONC); Strauss, Hubertus(3-CONC) The q-series of monstrous moonshine and the decomposition of the head characters. Comm. Algebra 18 (1990), no. 1, 253–278. 11F22 (20C15 20D08 33A99)

The authors tabulate the first fifty coefficients of the q-expansions of the Hauptmoduln which arise in the Monster-modular connection, and provide the decompositions of the corresponding Monster characters into irreducibles. This work to some extent duplicates earlier computations of S. D. Smith [in Finite groups—coming of age (Montreal, PQ, 1982), 303–313, Contemp. Math., 45, Amer. Math. Soc., Providence, RI, 1985; MR0822245 (87h:20037)] but the calculations are based on different methods which are also briefly discussed. The authors observe a surprising periodicity of the signs of the q-coefficients, a phenomenon which as yet remains unexplained.

share|cite|improve this answer
Great, thanks a lot! – user43645 Aug 24 '13 at 22:36
The periodicity should follow from asymptotic formulas for the coefficients of these modular functions analogous to those given by Hardy and Ramanujan for the partition function. There are also various elementary approaches; the simplest example is a proof that the $X_0(2)$ Hauptmodul $$ \left(\frac{\eta(q)}{\eta(q^2)}\right)^{24} = q^{-1} - 24 + 276q - 2048q^2 + 11202q^3 - 49152q^4 + - \cdots $$ has alternating signs by rearranging the product $\eta(q)/\eta(q^2)$ as $(1-q)(1-q^3)(1-q^5)(1-q^7)\cdots$ which clearly becomes nonnegative on substituting $-q$ for $q$. – Noam D. Elkies Aug 25 '13 at 4:53

Michael Somos (you can contact him at has a Pari-GP code which can help you calculate as many terms as you like, while D. Madore has also computed the first 3200 terms.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.