An unusual binomial coefficient identity

Question

I derived the following family of binomial coefficient identities rather indirectly for natural $t \ge 2$ (though it seems to hold more generally). I was hoping someone out there might know if this result exists in the literature, or perhaps come up with a more straightforward proof than mine. I paged through the H. W. Gould volumes of identities but didn't see it, though there were a few that seemed similar.

$$ \binom {n}{k}= (t+1 + t n )\sum_{i=0}^k (-1)^i \, \dfrac{ \displaystyle \binom{t + (t-1) n + k }{i} \binom{n+1}{k-i} } {(i+1) \displaystyle \binom{ t+1 + t n + k }{i+1}} $$

Just to check that I've transcribed it OK, let's work an example, how about $\binom{5}{2}$ with $t=3$? That's $n=5, k=2$.

\begin{align*} %% & %% (t+1 + t n )\sum_{i=0}^k (-1)^i \, %% \dfrac{ %% \displaystyle %% \binom{t + (t-1) n + k }{i} %% \binom{n+1}{k-i} %%} %%{(i+1) \displaystyle %%\binom{ t+1 + t n + k }{i+1}} %%\\ \binom{5}{2} &= {} (3+1+3(5) ) \sum_{i=0}^2 \\ & \quad (-1)^i \, \dfrac{ \displaystyle \binom{3 + (3-1) (5) + 2 }{i} \binom{5+1}{2-i} } {(i+1) \displaystyle \binom{ 3+1 + 3 (5) + 2 }{i+1} } \\ %%% &= %%% 19 \sum_{i=0}^2 (-1)^i \, %%% \dfrac{ %%% \displaystyle %%% \binom{15}{i} %%% \binom{6}{2-i} %%% } %%% {(i+1) \displaystyle %%% \binom{ 21 }{i+1} } \\ &= 19 \left( \dfrac{ \displaystyle \binom{15}{0} \binom{6}{2} } {\displaystyle 1\binom{ 21 }{1} } - \dfrac{ \displaystyle \binom{15}{1} \binom{6}{1} } {2 \displaystyle \binom{ 21 }{2} } + \dfrac{ \displaystyle \binom{15}{2} \binom{6}{0} } {3 \displaystyle \binom{ 21 }{3} } \right) \\ &= 19 \left( \dfrac{ 1 \cdot 15 }{21 } - \dfrac{ 15 \cdot 6} {2 \cdot 210 } + \dfrac{ 105 \cdot 1 }{3 \cdot 1330 } \right) \\&= 19 \cdot \left( \dfrac{5}{7} - \dfrac{3}{14} + \dfrac{1}{38} \right) = 19 \cdot \dfrac{10}{19} \quad\checkmark \end{align*}

Thanks.

Answer 1

Introducing $q=t+1+tn$ the queried identity becomes

$$\frac{1}{q} {n\choose k} = \sum_{p=0}^k (-1)^p {q-1-n+k\choose p} {n+1\choose k-p} \frac{1}{p+1} {q+k\choose p+1}^{-1}.$$

Working with the RHS we see that the second binomial coefficient enforces the upper range of the sum

$$[z^k] (1+z)^{n+1} \sum_{p\ge 0} (-1)^p {q-1-n+k\choose p} \frac{z^p}{p+1} {q+k\choose p+1}^{-1}.$$

This assumes we assign non-singular values for inverse binomial coefficients ${n\choose k}^{-1}$ with $k\gt n$. Recall from MSE 4316307 the following identity which was proved there: with $1\le k\le n$

$$\frac{1}{k} {n\choose k}^{-1} = [v^n] \log\frac{1}{1-v} (v-1)^{n-k}.$$

This fits the bill, we get the exact value when $p+1$ is in the range and the other cases make a zero contribution due to the extractor in $z$ and the fact that $p=q+k\gt k$ when $q\ge 1.$ Apply to the sum to get

$$[z^k] (1+z)^{n+1} [v^{q+k}] \log\frac{1}{1-v} (v-1)^{q+k-1} \sum_{p\ge 0} (-1)^p {q-1-n+k\choose p} z^p (v-1)^{-p} \\ = [z^k] (1+z)^{n+1} [v^{q+k}] \log\frac{1}{1-v} (v-1)^{q+k-1} \left[1-\frac{z}{v-1}\right]^{q-1-n+k} \\ = [z^k] (1+z)^{n+1} [v^{q+k}] \log\frac{1}{1-v} (v-1)^n [v-1-z]^{q-1-n+k}.$$

The contribution from $v$ is

$$\;\underset{v}{\mathrm{res}}\; \frac{1}{v^{q+k+1}} \log\frac{1}{1-v} (v-1)^n [v-1-z]^{q-1-n+k}.$$

Now put $v/(v-1)=w$ so that $v=w/(w-1)$ and $dv = -1/(w-1)^2 \; dw$ to get

$$\;\underset{w}{\mathrm{res}}\; \frac{1}{w^{q+k+1}} (w-1)^{q+k+1} \log\frac{1}{1-w} \frac{1}{(w-1)^n} [1/(w-1)-z]^{q-1-n+k} \frac{1}{(w-1)^2}.$$

With $q-1-n+k+2+n-q-k-1=0$ this becomes

$$\;\underset{w}{\mathrm{res}}\; \frac{1}{w^{q+k+1}} \log\frac{1}{1-w} [1-z(w-1)]^{q-1-n+k} \\ = \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{q+k+1}} \log\frac{1}{1-w} [1+z-wz]^{q-1-n+k}.$$

Using the generalized binomial theorem and writing

$$[1+z-wz]^{q-1-n+k} = (1+z)^{q-1-n+k} \left[ 1 - \frac{wz}{1+z}\right]^{q-1-n+k}$$

we obtain

$$\;\underset{w}{\mathrm{res}}\; \frac{1}{w^{q+k+1}} \log\frac{1}{1-w} \sum_{p\ge 0} {q-1-n+k\choose p} (1+z)^{q-1-n+k-p} (-1)^p w^p z^p.$$

With both extractors (upper range enforced by $w$) we get a non-singular fraction:

$$\sum_{p=0}^{q+k-1} {q-1-n+k\choose p} {q+k-p\choose k-p} (-1)^p \frac{1}{q+k-p} \\ = \frac{1}{q} \sum_{p=0}^{q+k-1} {q-1-n+k\choose p} {q+k-p-1\choose q-1} (-1)^p \\ = \frac{1}{q} [z^k] (1+z)^{q+k-1} \sum_{p\ge 0} {q-1-n+k\choose p} \frac{(-1)^p z^p}{(1+z)^p}.$$

Here we have again extended to infinity as the extractor truncates at $p=k$ and $k\le q+k-1.$ Continuing,

$$\frac{1}{q} [z^k] (1+z)^{q+k-1} \left[1-\frac{z}{1+z}\right]^{q-1-n+k} = \frac{1}{q} [z^k] (1+z)^n = \frac{1}{q} {n\choose k}.$$

This is the claim. The boundary conditions for this are very simple and refer to integers $q\ge 1$ and $n\ge k\ge 0.$

Answer 2

Alternate proof

Starting once more from

$$\sum_{p=0}^k (-1)^p {q-1-n+k\choose p} {n+1\choose k-p} \frac{1}{p+1} {q+k\choose p+1}^{-1}$$

we observe that

$${q-1-n+k\choose p} \frac{1}{p+1} {q+k\choose p+1}^{-1} = \frac{(q-1-n+k)! \times (q+k-1-p)!} {(q-1-n+k-p)! \times (q+k)!} \\ = \frac{1}{n+1} {q+k\choose n+1}^{-1} {q+k-1-p\choose n}$$

which yields for our sum

$$\frac{1}{n+1} {q+k\choose n+1}^{-1} \sum_{p=0}^k (-1)^p {n+1\choose k-p} {q+k-1-p\choose n}.$$

We point out here that this only works when $q+k\ge n+1$ for example with $q=n+1.$ The first binomial coefficient enforces the range of the sum and we get for the sum only

$$[z^k] (1+z)^{n+1} \sum_{p\ge 0} (-1)^p z^p {q+k-1-p\choose q+k-1-n-p} \\ = [z^k] (1+z)^{n+1} [w^{q+k-1-n}] (1+w)^{q+k-1} \sum_{p\ge 0} (-1)^p z^p \frac{w^p}{(1+w)^p} \\ = [z^k] (1+z)^{n+1} [w^{q+k-1-n}] (1+w)^{q+k-1} \frac{1}{1+zw/(1+w)}.$$

The contribution from $w$ is

$$\;\underset{w}{\mathrm{res}}\; \frac{1}{w^{q+k-n}} (1+w)^{q+k-1} \frac{1}{1+zw/(1+w)}.$$

Now put $w/(1+w) = u$ so that $w=u/(1-u)$ and $dw = 1/(1-u)^2 \; du$ to obtain

$$\;\underset{w}{\mathrm{res}}\; \frac{1}{u^{q+k-n}} \frac{1}{(1-u)^{n-1}} \frac{1}{1+zu} \frac{1}{(1-u)^2} \\ = \frac{1}{z} \;\underset{w}{\mathrm{res}}\; \frac{1}{u^{q+k-n}} \frac{1}{(1-u)^{n+1}} \frac{1}{u+1/z}.$$

Here the residue at infinity is zero and we may evaluate using minus the residues at $u=1$ and at $u=-1/z$. We get from the latter

$$[z^k] (1+z)^{n+1} \frac{1}{z} (-1)^{q+k-n} z^{q+k-n} \frac{1}{(1+1/z)^{n+1}} \\ = [z^k] z^{q+k} (-1)^{q+k-n} = 0.$$

The residue at $u=1$ requires the Leibniz rule:

$$(-1)^n \frac{1}{n!} \left(\frac{1}{u^{q+k-n}} \frac{1}{(1+zu)^1}\right)^{(n)} \\ = (-1)^n \frac{1}{n!} \sum_{p=0}^n {n\choose p} \frac{(-1)^p (q+k-n)^{\overline{p}}}{u^{q+k-n+p}} \frac{(-1)^{n-p} 1^{\overline{n-p}} z^{n-p}}{(1+zu)^{1+n-p}}.$$

Set $u=1$ to get

$$[z^k] (1+z)^{n+1} \sum_{p=0}^n {q+k-1-n+p\choose p} \frac{z^{n-p}}{(1+z)^{1+n-p}} \\ = \sum_{p=0}^n {q+k-1-n+p\choose p} {p\choose k-n+p} \\ = [z^{n-k}] \sum_{p=0}^n {q+k-1-n+p\choose p} (1+z)^p \\ = [z^{n-k}] [w^n] \frac{1}{1-w} \sum_{p\ge 0} {q+k-1-n+p\choose p} w^p (1+z)^p \\ = [z^{n-k}] [w^n] \frac{1}{1-w} \frac{1}{(1-w(1+z))^{q+k-n}}.$$

Here we are making use of the boundary condition $n\ge k.$ Continuing,

$$[z^{n-k}] [w^n] \frac{1}{(1-w)^{q+k-n+1}} \frac{1}{(1-wz/(1-w)))^{q+k-n}} \\ = [w^n] \frac{1}{(1-w)^{q+k-n+1}} {q-1\choose n-k} \frac{w^{n-k}}{(1-w)^{n-k}} = {q-1\choose n-k} {k+q\choose k}.$$

We conclude by collecting everything,

$$\frac{1}{n+1} {q+k\choose n+1}^{-1} {q-1\choose n-k} {k+q\choose k}.$$

Note that

$$\frac{1}{n+1} {q+k\choose n+1}^{-1} {q-1\choose n-k} = \frac{1}{n+1} \frac{(n+1)! \times (q-1)!}{(q+k)! \times (n-k)!} = {n\choose k} \frac{1}{q} {q+k\choose q}^{-1}$$

Multiply by ${k+q\choose k}$ to get the desired result.