What is highest coefficient ofthe polynomial $t(t+1)(t+2) \cdots (t+n)$ for arbitrary $n$?

Question

How to find the coefficient of in the polynomial

$t(t+1)(t+2) \cdots (t+n)$ for arbitrary $n$

?

According to me it should be the sum of the products of the numbers from $1,2,\cdots,n$ taking $k$ at a time i.e. it is a sum of $n \choose k$ products. Now if this is true then how do I find out the highest coefficient in that expansion?I feel real trouble in that case.I take quite a few examples and observe that the coefficient of $t^2$ is the highest among all other if $n \geq 2$.But I can't prove it in general.

Please help me in proving this.

Thank you in advance.

Answer 1

Perhaps a missing index in here, but roughly this is:

$$ a_{k}=t^k \sum_{i=1}^{(n \ n+1-k)}\prod_{j \text{ in } C_i(n,n+1-k)} j $$

$C(n,k)$ is the set of all combinations of the integers from 1 to $n$, taking $k$ at each time, and $C_i$ is its $i$th element. The size of $C(n,k)$ is the binomial coefficient $(n \ k)=\frac{n!}{k!(n-k)!}$

$$ C(n,0)=\{\}\\ C(n,1)=\{\{1\},...,\{n\}\}\\ C(n,2)=\{\{1,2\},\{1,3\},...,\{n-2,n\},\{n-1,n\}\}\\ ...\\ C(n,n-1)=\{\{2,...n\},\{1,3,...,n\},\{1,...,n-2,n\},\{1,...,n-1\}\}\\ C(n,n)=\{\{1,...n\}\}\\ C(n,n+1)=\{\}\\ $$

Answer 2

So we have that $$ t\left( {t + 1} \right) \cdots \left( {t + n} \right) = t^{\,\overline {\,n + 1\,} } $$ and $$ t^{\,\overline {\,n + 1\,} } = \sum\limits_{\,\left( {1\, \leqslant } \right)\,k\,\left( { \leqslant \,n + 1} \right)} {\left[ \begin{gathered} n + 1 \\ k \\ \end{gathered} \right]t^{\,k} } $$

Now $$ P(k;\,q) = {1 \over {q!}}\left[ \matrix{ q \cr k \cr} \right] $$ is a discrete Probability distribution
(see for instance this post) whose raw first and second moment are $$ \eqalign{ & E\left[ k \right] = \sum\limits_{\left( {0\, \le } \right)\,g\,\left( { \le \,q} \right)} {{k \over {q!}}\left[ \matrix{ q \cr k \cr} \right]} = H(q) = \gamma + \psi (q + 1) \cr & E\left[ {k^{\,2} } \right] = \sum\limits_{\left( {0\, \le } \right)\,k} {{{k^{\,2} } \over {q!}}\left[ \matrix{ q \cr k \cr} \right] = } \left( {H_{\,q} ^2 + H_{\,q} - {{\pi ^2 } \over 6} + \psi ^{\left( 1 \right)} \left( {q + 1} \right)} \right) \cr} $$

Your question concerns the mode of $\left[ \matrix{ q \cr k \cr} \right]$ which coincides with that of $P(k;\,q)$, but I could not find sufficient information on line about this distribution and its characteristics.

Plotting some examples of $P(k;\,q)$ it comes out that it has quite a peaked and symmetric shape concentrated in the lower values of $k$ and that the mode remains very close to the mean.
So ${\rm mode}\,P(k;q) \approx H(q)$.

If you need a more analytical result, then you can try and interpolate $P(k;\,q)$ by a continuous function, which has a closed expression for the mode.