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Let $p_k(n)$ denote the number of partitions of $n$ into exactly $k$ parts. It is known that $p_k(n)$ satisfies the recurrence

$p_k(n) = p_{k-1}(n-1) + p_k(n-k)$,

where $p_k(n)=0$ for $k>n$, $p_n(n)=1$, and $p_1(n)=1$. It is also known that there are $\binom{n-1}{k-1}$ compositions of $n$ into $k$ parts. Clearly, $p_k(n) \leq \binom{n-1}{k-1}$. Let $c(n,k)>0$ be a positive constant such that $c(n,k)\cdot p_k(n) \leq \binom{n-1}{k-1}$.

What are known upper bounds for the value of $c(n,k)$? I'm looking for a tight upper bound such that $c(n,k)\cdot p_k(n)$ does not exceed $\binom{n-1}{k-1}$.

I have read about upper bounds for $p_k(n)$. However, some of these are greater than $\binom{n-1}{k-1}$, which is not desirable.

EDIT: early replies of $k!$ led me to edit this question.

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  • $\begingroup$ The ratio is certainly bounded above by $k!$, since the map (from the set of compositions to the set of partitions) given by sorting each composition is at most $k!$-to-$1$. $\endgroup$ – Greg Martin Feb 20 at 2:21
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The upper bound for the ratio is $k!$, which is approached as $n$ increases without limit.

For example:

  • $p_5(10000)= 3475694791250$
  • ${9999 \choose 4}=416250145812501$
  • making the ratio about $119.76$
  • compared with $5!=120$

$k!$ is a an upper bound because you can order the $k$ parts of the partition in no more than $k!$ different ways to make different compositions; exactly $k!$ ways if all the parts are distinct but fewer if some are equal.

For large enough $n$ given $k$, the proportion of partitions with equal parts can be arbitrarily small. See this by considering $q_k(n)$ as the number of partitions of $n$ into $k$ distinct positive parts; then $q_k(n)= p_k\left(n-\frac{k(k-1)}{2}\right)$ so $\frac{q_k(n)}{p_k(n)} \to 1$ as $n$ increases

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  • $\begingroup$ Thanks for the reply. $k!$ would indeed be an upper bound for the ratio. I edited the question since I probably worded it poorly the first time. $\endgroup$ – no_chi Feb 20 at 2:43
  • $\begingroup$ Perhaps I do not understand your edited question but $c(n,k)$ does not look like a constant to me, as it varies with $n$ $\endgroup$ – Henry Feb 20 at 9:02
  • $\begingroup$ I want to establish that $p_k(n) \leq \binom{n-1}{k-1}$ by showing that $\dfrac{\binom{n-1}{k-1}}{p_k(n)}\geq c(n,k)\geq 1$. (And eventually, compute for savings in operations when using only $p_k(n)$ elements as opposed to $\binom{n-1}{k-1}$.) $c(n,k)$ need not be a constant since it may vary with $n$ and $k$.) $\endgroup$ – no_chi Feb 20 at 9:16

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