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

A partition of $n$ is a weakly decreasing tuple of numbers $(\lambda_1,\lambda_2,\lambda_3,....\lambda_k)$ whose sum is $n.$ A natural problem studied is counting partitions whose summands $\lambda_i$ all like in a set $S\subset \mathbb{N}$ and we denote this number $p_S(n)$.

If $S$ is ``dense" enough, one can usually calculate an asymptotic formula that looks like $$p_S(n) = Kn^\beta\exp\left(c\sqrt{n}\right)\left(1+ \frac{c_1}{\sqrt{n}}+ \frac{c_2}{n}+\frac{c_3}{n^\frac{3}{2}} \dots \right).$$

My question is when $S=\mathbb{N}\setminus [0,1,2,\dots c]$ does such a formula already exist in the literature and where can I find it?

share|cite|improve this question
The tuple notation generally implies an ordering, whereas in partition counting one generally considers $1+2$ and $2+1$ to be the same partition of 3. Partitions into parts not exceeding $c$ is the same as partitions into at most $c$ parts, and I think asymptotically that's $Cn^{c-1}$, which is negligible compared to the full number of partitions. – Gerry Myerson Jul 15 '13 at 3:29
I forgot the decreasing part of the definition and its parts greater than c rather than less than c. My fault and I apologize. – Daniel Parry Jul 15 '13 at 3:36
up vote 1 down vote accepted

Parts greater than $c$ is complementary to parts smaller than $c.$ (as pointed out in the comments, parts greater than $c$ is the same as "at least $c$ parts") for the latter, there is a theorem of Szekeres from 1951, see the math review. (of. Course this doesn't answer the question, but will leave answer up for the seekers reference) 2 3rd

share|cite|improve this answer
Thanks, this does help a bit. – Daniel Parry Jul 24 '13 at 17:39

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.