Suppose $(a_1,\ldots, a_k)$ is an integer partition of $n$, and $(b_1,\ldots,b_k)$ is a rearrangement of the $a$-sequence. Prove the following identity (preferably combinatorially): $$ \sum_{j_2+\cdots+j_k=l,\atop l\geq 0, \, a_t>j_t\geq 0} \quad\frac{(-1)^l l!}{(a_1+l+1)_{l+1}} {a_2\choose j_2}\cdots {a_k \choose j_k}=\sum_{j_2+\cdots+j_k=l,\atop l\geq 0, \, b_t>j_t\geq 0} \quad\frac{(-1)^l l!}{(b_1+l+1)_{l+1}} {b_2\choose j_2}\cdots {b_k \choose j_k}, $$ where $(x)_k$ is the falling factorial.
1 Answer
I assume you meant $j_t\leq a_t$ (not $j_t<a_t$).
It's sufficient to prove that for any analytic function $f(x)$, the function $$F(a_1,a_2) := \sum_{l\geq 0} \sum_{j=0}^{a_2} \frac{(-1)^ll!}{(a_1+l+1)_{l+1}} \binom{a_2}{j} [x^{l-j}]\ f(x)$$ is symmetric, i.e. $F(a_1,a_2) = F(a_2,a_1)$.
Using the property of beta function, we have
\begin{split} F(a_1,a_2) &=\sum_{l\geq 0} \frac{(-1)^ll!}{(a_1+l+1)_{l+1}} [x^l]\ (1+x)^{a_2}f(x) \\ &=\sum_{l\geq 0} (-1)^l \int_0^1 t^l (1-t)^{a_1} {\rm d}t\ [x^l]\ (1+x)^{a_2}f(x)\\ &=\int_0^1 (1-t)^{a_1} (1-t)^{a_2} f(-t) {\rm d}t, \end{split} which is clearly symmetric.
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$\begingroup$ Thanks Max! Actually, it has to be $j_t < a_t$, otherwise the identity does not hold. This can be checked, for instance, using the integer partition $(2,1)$. $\endgroup$ Apr 4, 2022 at 11:16
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$\begingroup$ @rickychen: I do not quite follow. When $k=2$, the sum over $j_2=l\leq a_2$ gives $\frac{1}{a_1+a_2+1}$, which is symmetric. $\endgroup$ Apr 4, 2022 at 18:49
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$\begingroup$ Let me proceed carefully, assuming $j_2 \leq a_2$. For $(2,1)$, the left-hand side gives $$\frac{(-1)^0 0!}{(2+0+1)_1}{1\choose 0}+ \frac{(-1)^1 1!}{(2+1+1)_2}{1\choose 1}$$, while for $(1,2)$, the right-hand side gives $$\frac{(-1)^0 0!}{(1+0+1)_1} {2\choose 0} +\frac{(-1)^1 1!}{(1+1+1)_2}{2\choose 1} +\frac{(-1)^2 2!}{(1+2+1)_3}{2\choose 2}$$. They are not equal! But, if we assume $j_2<a_2$, then they are equal. $\endgroup$ Apr 5, 2022 at 0:04
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$\begingroup$ please see my detailed computation above, one gives $\frac{1}{3}-\frac{1}{12}$, the other gives $\frac{1}{2}-\frac{1}{6}+\frac{2}{24}$. BTW, why I could not see "@max" that I put at the very beginning. $\endgroup$ Apr 5, 2022 at 0:41
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$\begingroup$ You have a numerical error in the second term of the second sum. It should be $\frac{1}{2} - \frac{2}{6} + \frac{2}{24} = \frac{1}{4}$, which matches the first sum and the formula $\frac1{a_1+a_2+1}$. $\endgroup$ Apr 5, 2022 at 1:03