Timeline for A binomial product sum that turns out to be 1
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Feb 3, 2022 at 2:19 | comment | added | Ira Gessel | Your sum (for $n>0$) is equal to (or should be equal to) $$\sum_{k=1}^n \sum_{j_1+\cdots+j_k=n}(-1)^{n-k}\frac{n!}{j_1!\, j_2!\,\cdots j_k!}.$$ where $j_1,\dots, j_k$ must be positive. This is $n!$ times the coefficient of $x^n$ in $$\sum_{k=0}^\infty\left(\sum_{j=1}^\infty (-1)^{j-1}\frac{x^j}{j!}\right)^k.$$ | |
| Feb 3, 2022 at 2:11 | comment | added | Vishnu Namboothiri K | There is a $p$ appearing in the above expression in the comment (in the power to -1). It should be actually $n$. I am unable to edit and correct it. | |
| Feb 3, 2022 at 1:59 | history | edited | Ira Gessel | CC BY-SA 4.0 |
added 9 characters in body
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| Feb 3, 2022 at 1:57 | comment | added | Vishnu Namboothiri K | Can you please explain how did you arrive at this exponential function? I am asking this because I need to understand the binomial product sum \begin{align*} \sum\limits_{\substack{i_1> i_2> \cdots > i_k\\i_1, i_2, \cdots, i_k \in \{n-j+1,n-j+ 2, \cdots, n-1\}}}(-1)^{p+k+1}\binom{n}{i_1}\binom{i_1}{i_2}\binom{i_2}{i_3}\cdots\binom{i_{k-1}}{i_k}\binom{i_k}{n-j} \end{align*} with the convention that if the summation is over an empty subset, then only $\binom{n}{n-j}$ occurs in the product. Or otherwise, if I am not asking much, what could be the exponential generating function for this sum? | |
| Feb 2, 2022 at 6:38 | history | answered | Ira Gessel | CC BY-SA 4.0 |