This concerns difference/limit ratio results for special restricted partitions.

Let $r,a, b$ be nonnegative integers; define $p(r,a,b)$ to be the number of partitions of the integer $r$ using at most $b$ positive integers and each of them is at most $a$; if $r = 0$ and $a,b >0$, then $p(0,a,b)$ is taken to be one. Then, assuming $ab > 1$, $p(1,a,b) = 1$, and $p(ab,a,b) = 1$; moreover, $p(r,a,b) \neq 0$ iff $0 \leq r \leq ab$. These are well known and closely related to Gauss polynomials (explicitly, the coefficient of $z^r$ in the $[a+b,b]$ Gaussian polynomial is $p(r,a,b)$) (notation varies considerably in the literature).

It is also well known that if $a, b$ are fixed, then the sequence $(p(0,a,b), p(1,a,b), \dots , p(ab,a,b))$ is unimodal, and symmetric (about $ab/2$).

What I am interested in is the ratios of consecutive $p(r,a,m-a)$ with $r$ varying (note the change in notation, $b \mapsto m-a$) over the interval $2a \leq r \leq a(m-a)/2$, with estimates uniform in $m$ and $a$, with $a$ not too close to $0$ or $m$ (for the latter, we can just restrict to $a \leq m/2$).

To that end, define $$R_{m} = \min_{ 2a \leq r \leq \frac{a(m-a)}{2} {\rm \ and \ }3 \leq a \leq m-4} \frac{p(r-1,a,m-a)}{p(r,a,m-a}.$$ It is important to keep track of the conditions below the $\min$ symbol. Since $p(r, a,m-a)$ is increasing (as $r$ runs over $0,1,2, \dots, a(m-a)/2$) (with $a$, $m$ fixed), obviously $R_{m} < 1$. It is relatively easy to show that $R_m \geq 1/2$ for all but a few values of $m$.

Question Is it true that $R_{m} \to 1 $?

This might require a further restriction on $a$; small values (or those close to $m$) tend to screw it up; alternatively, impose more conditions on $r$, e.g., $r \geq 3a$).

I imagine there are such results in the literature, but I am unfamiliar with it. For $r$ near the middle ($a(m-a)/2$), asymptotic estimates for $p(r,a,m-a)$ given in [RP Stanley & F Zanello, Some asymptotic results on q-binomial coefficients, Ann Comb 20 (2016) 623--34, Thm 2.6]) suggest that the problem probably reduces to relatively small $r$. An exact formula for $p(r,a,b)$ is given in [G Almkvist & GE Andrews, A Hardy-Ramanujan formula for restricted partitions, J Number Theory, 38 (1991) 135--144, Thm 4], but it is difficult to see how to use it. Without the $b$, asymptotic formulas were obtained by Szekeres [An asymptotic formula in the theory of partions I and II, Quart J Math 4 (1951) 85--108 and 4 (1953) 96--11].

Motivation I am looking at very simple random walks on the simplest torsion-free nonabelian group, given by $hg = z gh$ where $z$ is central. Then the coefficient of $z^r g^a h^b$ in $(g + h)^{a+b}$ is $p(r,a,b)$. I want to obtain the limit ratio result above (or some version of it), because it will almost certainly lead to a description of all extremal harmonic functions on the subcone starting at $1$ and proceeding by multiplication by $g+h$. Less cryptically, I have a bunch of such harmonic functions, and want to prove they constitute all of them.

In general, limit ratio results are easier to prove than establishing asymptotics (since the latter usually yields the former).

  • $\begingroup$ In the definition of $R_m$ are you fixing $a$ or are both $a$ and $r$ varying? $\endgroup$
    – Nate
    Mar 1, 2021 at 16:39
  • $\begingroup$ Both $a$ and $r$ are varying, subject to $3 \leq a \leq m-3$ and $2a \leq r \leq a(m-a)$, although symmetry can be applied, and reduce the range accordingly. I replaced inf by min. $\endgroup$ Mar 1, 2021 at 21:16
  • $\begingroup$ $2a\le r\le a(m-a)/2$ (as in OP) yields $a\le m-4$, so $a=m-3$ is not possible $\endgroup$ Mar 1, 2021 at 22:50
  • $\begingroup$ @FedorPetrov Right; fixed. thx $\endgroup$ Mar 1, 2021 at 23:17

1 Answer 1


This is exactly Lemma 1 in the recent paper by Vershik and Malyutin. By the way, motivated by random walks on Heisenberg group.

  • $\begingroup$ It's not quite the result I was hoping for, in that it requires my $a$ to go to infinity (it may be that this is necessary in any event), but it is close. I also see that I wasn't the only one thinking about these problems (there is a lot of other stuff in the paper that I had been working on). What's amazing is that you found the reference. And I didn't know that Anatoly was still active. $\endgroup$ Mar 2, 2021 at 15:23
  • $\begingroup$ I did not exactly find it: I was aware. I work at the same lab (where Anatoly Moiseevich is the head and the most active guy), and the authors discussed this lemma with me. I think, the case of bounded $a$ should be easy, as this is polynomial in $r$, is not it? $\endgroup$ Mar 2, 2021 at 15:31
  • $\begingroup$ Yes, I realize (now) that I should have included a condition guaranteeing $a \to \infty$ (else it's not going to be true), which is exactly what the result is in the Vershik-Malyutin paper. Give my regards to Anatoly. $\endgroup$ Mar 2, 2021 at 19:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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