There have been a couple of questions on Ramanujan's $\tau$ function.
Lehmer's conjecture for Ramanujan's tau function
The Vanishing of Ramanujan's Function tau(n)

A $q$-analogue is given by $$ \tau_q(n) = \sum_{|\lambda|=n}\prod_{(i,j)} \frac{[5-h(i,j)][5+h(i,j)]}{[h(i,j)]^2} $$

Here $[k]=\frac{q^k-q^{-k}}{q-q^{-1}}$. The sum is over all partitions of $n$. The product is over all boxes in the diagram of $\lambda$ and $h(i,j)$ is the hook length of the box $(i,j)$.

This is a $q$-analogue because (despite appearances) it is a Laurent polynomial in $q$ and substituting $q=1$ gives $\tau(n)$.

It would be ridiculous to ask if this is $q$-holonomic.

Is this $q$-analogue known? and does it have any significance in number theory?

  • $\begingroup$ Bruce, wouldn't q-holonomicity imply holonomicity of the $\tau$ sequence? $\endgroup$ Jul 22, 2010 at 8:37
  • $\begingroup$ Wadim, I guess so. Is that known to fail? $\endgroup$ Jul 22, 2010 at 8:42

1 Answer 1


I am not very well familiar with the notion "$q$-holonomic" but learned it quite recently from this question. If the sequence $\tau_q(n)$ were $q$-holonomic, then $\tau(n)$ would be holonomic. The latter means that $$\Delta(x)=x\prod_{n=1}^\infty(1-x^n)^{24}=\sum_{k=1}^\infty\tau(k)x^k$$ satisfies a linear differential equation with rational function coefficients. However, this is known to be false: the function $\Delta(x)$ has singularities at all roots of unity.

Note that the function $\Delta(x)$ satisfies an algebraic differential equation with constant coefficients; the result due to Halphen (and independently, although later, by Ramanujan).

  • $\begingroup$ That's interesting to know. It's what I expected. I was really asking if the $q$-analogue had any significance in number theory. $\endgroup$ Jul 22, 2010 at 9:25
  • $\begingroup$ Sorry. My question didn't make that clear and has been changed. $\endgroup$ Jul 22, 2010 at 9:37
  • $\begingroup$ Bruce, I learn on the $q$-analogue from your question! I was always convinced that modular forms, like $\Delta(x)$, have no reasonable $q$-analogues, as they are themselves "from the $q$-world" (this has to be understood as they come as special cases of basic hypergeometric series). Your $\tau_q$ beats me. Thanks for this example! $\endgroup$ Jul 22, 2010 at 9:40
  • $\begingroup$ As for the change, I've never seen $q$-analogues of sequences which come as coefficients of modular forms. $\endgroup$ Jul 22, 2010 at 9:42

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