MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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

Let $S_n$ denote the set of partions of $n$ such that every part is greater than 1. Partitions $(x_1,\ldots,x_k), (y_1,\ldots,y_l) \in S_n$ are said to have almost equal product if $$\prod_{i=1}^k (x_i+1) = \prod_{i=1}^l (y_i+1)$$.

For example if $n = 14$ the partitions (3,3,8) and (2,5,7) are almost equal since (3+1)(3+1)(8+1) = (2+1)(5+1)(7+1).

Now if we denote by $S'_n$ the largest subset of $S_n$ not containing almost equal partitons,then I would like to find the asymptotic value of $|S'_n|$. I believe $|S'_n|$ is at least subexponential in $n$ but do not know how to prove this. Is there any way to perhaps find a bound on the number of pairs of almost equal partitions and take it from there?

share|cite|improve this question
The only "hook" I see in this one is that if you have two almost equal partitions A and B for n, then concatenating k to A and B give you almost equal partitions for n+k. If you were able to compute the quantities you desired for small n, as well as something like the number of "minimal" or "reduced" such partitions, then you might build it into a recurrence to compute larger values. Presently I dare not suggest what the recurrence would look like. Richard Stanley might know. Gerhard "Sometimes Won't Handle The Truth" Paseman, 2011.03.14 – Gerhard Paseman Mar 14 '11 at 20:21
Is there a generally agreed definition of "subexponential"? – Gerry Myerson Mar 14 '11 at 23:39
up vote 4 down vote accepted

I don't know what you mean by "at least subexponential." A subexponential upper bound is trivial because the number of partitions of $n$ with no restrictions grows subexponentially.

A little work gives a superpolynomial lower bound. Choose an arbitrary subset $\{p_1,...p_k\}$ of the primes between $5$ and $f(n)$ for some function $f$ to be determined later. There are $2^{\pi(f(n))}$ such subsets, and we will create a partition of $n$, $(2,...,2,3,...,3,p_1-1,p_2-1,...p_k-1)$ such that the primes dividing the product $\prod (x_i+1)$ are $\{2,3,p_1,...p_k\}$ hence the products for partitions constructed from different subsets are distinct.

For an easy lower bound, take $f(n)=\sqrt n$. The sum of the primes up to $\sqrt n$ is smaller than $n$ by enough to create a partition of $n$ by padding with $2,...,2,3,...,3$. The prime number theorem says $\pi(\sqrt n) \sim \frac{\sqrt n}{\log \sqrt n}$ so there are at least $\exp(C \frac {\sqrt n}{\log n})$ partitions of $n$ with parts greater than $1$ and different (augmented) products. I expect that you can get rid of the $\log n$ in the denominator or at least replace it with $\log\log n$ by using a better $f(n)$.

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.