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Glorfindel
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I'm not entirely sure of what you are asking, but note that Erdos and Lehner proved herehere that $$p(n,m)\sim \frac{n^{m-1}}{m!(m-1)!}$$ holds for $m=o(n^{1/3})$. In generality for any finite set $A$, with $|A|=m$ and $p(n,A)$ denoting the number of partitions of $n$ with parts from $A$, one has $$p(n,A)=\frac{1}{\prod_{a\in A}a}\frac{n^{m-1}}{(m-1)!}+O(n^{m-2}).$$

Such estimations can be deduced from the generating function of $p$ by using methods that are described in many books, for example "Analytic Combinatorics" by Flajolet and Sedgewick.

I'm not entirely sure of what you are asking, but note that Erdos and Lehner proved here that $$p(n,m)\sim \frac{n^{m-1}}{m!(m-1)!}$$ holds for $m=o(n^{1/3})$. In generality for any finite set $A$, with $|A|=m$ and $p(n,A)$ denoting the number of partitions of $n$ with parts from $A$, one has $$p(n,A)=\frac{1}{\prod_{a\in A}a}\frac{n^{m-1}}{(m-1)!}+O(n^{m-2}).$$

Such estimations can be deduced from the generating function of $p$ by using methods that are described in many books, for example "Analytic Combinatorics" by Flajolet and Sedgewick.

I'm not entirely sure of what you are asking, but note that Erdos and Lehner proved here that $$p(n,m)\sim \frac{n^{m-1}}{m!(m-1)!}$$ holds for $m=o(n^{1/3})$. In generality for any finite set $A$, with $|A|=m$ and $p(n,A)$ denoting the number of partitions of $n$ with parts from $A$, one has $$p(n,A)=\frac{1}{\prod_{a\in A}a}\frac{n^{m-1}}{(m-1)!}+O(n^{m-2}).$$

Such estimations can be deduced from the generating function of $p$ by using methods that are described in many books, for example "Analytic Combinatorics" by Flajolet and Sedgewick.

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Gjergji Zaimi
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I'm not entirely sure of what you are asking, but notnote that Erdos and Lehner proved here that $$p(n,m)\sim \frac{n^{m-1}}{m!(m-1)!}$$ holds for $m=o(n^{1/3})$. In generality for any finite set $A$, with $|A|=m$ and $p(n,A)$ denoting the number of partitions of $n$ with parts from $A$, one has $$p(n,A)=\frac{1}{\prod_{a\in A}a}\frac{n^{m-1}}{(m-1)!}+O(n^{m-2}).$$

Such estimations can be deduced from the generating function of $p$ by using methods that are described in many books, for example "Analytic Combinatorics" by Flajolet and Sedgewick.

I'm not entirely sure of what you are asking, but not that Erdos and Lehner proved that $$p(n,m)\sim \frac{n^{m-1}}{m!(m-1)!}$$ holds for $m=o(n^{1/3})$. In generality for any finite set $A$, with $|A|=m$ and $p(n,A)$ denoting the number of partitions of $n$ with parts from $A$, one has $$p(n,A)=\frac{1}{\prod_{a\in A}a}\frac{n^{m-1}}{(m-1)!}+O(n^{m-2}).$$

Such estimations can be deduced from the generating function of $p$ by using methods that are described in many books, for example "Analytic Combinatorics" by Flajolet and Sedgewick.

I'm not entirely sure of what you are asking, but note that Erdos and Lehner proved here that $$p(n,m)\sim \frac{n^{m-1}}{m!(m-1)!}$$ holds for $m=o(n^{1/3})$. In generality for any finite set $A$, with $|A|=m$ and $p(n,A)$ denoting the number of partitions of $n$ with parts from $A$, one has $$p(n,A)=\frac{1}{\prod_{a\in A}a}\frac{n^{m-1}}{(m-1)!}+O(n^{m-2}).$$

Such estimations can be deduced from the generating function of $p$ by using methods that are described in many books, for example "Analytic Combinatorics" by Flajolet and Sedgewick.

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Gjergji Zaimi
  • 85.6k
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  • 402

I'm not entirely sure of what you are asking, but not that Erdos and Lehner proved that $$p(n,m)\sim \frac{n^{m-1}}{m!(m-1)!}$$ holds for $m=o(n^{1/3})$. In generality for any finite set $A$, with $|A|=m$ and $p(n,A)$ denoting the number of partitions of $n$ with parts from $A$, one has $$p(n,A)=\frac{1}{\prod_{a\in A}a}\frac{n^{m-1}}{(m-1)!}+O(n^{m-2}).$$

Such estimations can be deduced from the generating function of $p$ by using methods that are described in many books, for example "Analytic Combinatorics" by Flajolet and Sedgewick.