Sum of how many numbers should N be partitioned.

Question

Partition of integer:

4 = 4                 p(4,1) = 1
  = 1+3, 2+2          p(4,2) = 2
  = 1+1+2             p(4,3) = 1
  = 1+1+1+1           p(4,4) = 1
  

$max(p(4, k)) = 2$ at $k = 2$

5 = 5                 p(5,1) = 1
  = 1+4, 2+3          p(5,2) = 2
  = 1+1+3, 1+2+2      p(5,3) = 2
  = 1+1+1+2           p(5,4) = 1
  = 1+1+1+1+1         p(5,5) = 1
  

$max(p(5, k)) = 2$ at $k = 2$ and $3$

and $p(n) = Σp(n, k)$ for $∀$ $1≤k≤n$ and $n,k$$\in \mathbb N$

$p(4) = p(4, 1) + p(4, 2) + p(4, 3) + p(4, 4) = 1 + 2 + 1 + 1 = 5 $

$p(5) = p(5, 1) + p(5, 2) + p(5, 3) + p(5, 4) + p(5, 5) = 1 + 2 + 2 + 1 + 1 + 1 = 7 $

for this I used Euler's Identity
$p(n, k) = p(n-1, k-1) + p(n-k, k)$

I want to know for which value of $k$, $p(n, k)$ will be maximum for the given value of $n$

Answer 1

As an estimate.
Take a partition $n=a_1+a_2+\ldots a_k$, with the $a_i$ in increasing order. This can be written as $$ka_1+(k-1)(a_2-a_1)+(k-2)(a_3-a_2)+\ldots+1(a_n-a_{n-1})$$ So it is the number of integer solutions to $$2b_2+3b_3+\ldots+kb_k\le n$$ The volume of this region is $$V(n,k)=\frac{n^{k-1}}{k!(k-1)!}$$ The volume may well be related to the number of integer points within it.
The ratio $$\frac{V(n,k+1)}{V(n,k)}=\frac n{k(k+1)}$$ $V(n,k)$ will be maximized when the ratio is $1$, so the number of partitions will be maximized when $$k\approx \sqrt n$$