Iterated rascal triangle row sums

Question

In this manuscript the authors propose the following conjecture (1) \begin{align*} \sum_{k=0}^{4i+3} \binom{4i+3}{k}_i &= 2^{4i+2} \end{align*} where $\binom{4i+3}{k}_i$ is iterated rascal number defined as \begin{align*} \binom{n}{k}_i &= \sum_{m=0}^{i} \binom{n-k}{m} \binom{k}{m} \end{align*} So that conjecture (1) can be expressed in terms of binomial coefficients (2) $$ \sum_{k=0}^{4i+3} \sum_{m=0}^{i} \binom{4i+3-k}{m} \binom{k}{m} = 2^{4i+2} $$

My approach

To prove the conjecture (1) I utilize relations in terms of binomial coefficients and iterated rascal numbers. Recall the row sums property of binomial coefficients \begin{equation*} \sum_{k=0}^{4i+3} \binom{4i+3}{k} = 2^{4i+3} \end{equation*} If conjecture (1) is true, then it is also true that \begin{equation*} \sum_{k=0}^{4i+3} \binom{4i+3}{k} - \sum_{k=0}^{4i+3} \binom{4i+3}{k}_i = 2^{4i+2} \end{equation*} because $2^{4i+3} - 2^{4i+2} = 2^{4i+2}$.

Expanding both sums applying Vandermonde convolution $\binom{a+b}{r} = \sum_{m=0}^{r} \binom{a}{m} \binom{b}{r-m}$ we get \begin{align*} 2^{4i+2} &= \sum_{k=0}^{4i+3} \sum_{m=0}^{k} \binom{4i+3-k}{m} \binom{k}{k-m} - \sum_{k=0}^{4i+3} \sum_{m=0}^{i} \binom{4i+3-k}{m} \binom{k}{k-m}\\ 2^{4i+2} &= \sum_{k=0}^{4i+3} \sum_{m=0}^{k} \binom{4i+3-k}{m} \binom{k}{k-m} - \sum_{m=0}^{i} \binom{4i+3-k}{m} \binom{k}{k-m} \end{align*} Note that $\binom{n}{k} \geq \binom{n}{k}_i$ for each $n,k,i$. Now we have three possible relations between $i,k$: $k < i$, $k = i$, $k > i$.

If $k < i$ then inner sums equal zero \begin{equation*} \sum_{m=0}^{k} \binom{4i+3-k}{m} \binom{k}{k-m} - \sum_{m=0}^{i} \binom{4i+3-k}{m} \binom{k}{k-m} = 0 \end{equation*} Because $\binom{k}{k-m}$ in the sum over $i$ is zero for all $m > k$.

If $k = i$ obviously \begin{equation*} \sum_{m=0}^{k} \binom{4i+3-k}{m} \binom{k}{k-m} - \sum_{m=0}^{i} \binom{4i+3-k}{m} \binom{k}{k-m} = 0 \end{equation*}

If $k > i$ then \begin{equation*} \sum_{m=0}^{k} \binom{4i+3-k}{m} \binom{k}{k-m} - \sum_{m=0}^{i} \binom{4i+3-k}{m} \binom{k}{k-m} = \sum_{m=i+1}^{k} \binom{4i+3-k}{m} \binom{k}{k-m} \end{equation*} Thus, we have to prove that \begin{equation*} 2^{4i+2} = \sum_{k} \sum_{m=i+1}^{k} \binom{4i+3-k}{m} \binom{k}{k-m} \end{equation*} Let $m$ to iterate from 0 \begin{equation*} 2^{4i+2} = \sum_{k} \sum_{m=0}^{k} \binom{4i+3-k}{i+1+m} \binom{k}{i+1+m} \end{equation*} Although, above equation almost exactly matches Vandermonde identity, it cannot be applied directly. Even it were applied, the result would disprove the main conjecture giving $2^{4i+3}$ as row sums. My validations show that indeed conjecture true for $i \leq 100$. Therefore, propose the following conjecture:

Conjecture 2.1 For every $i$ \begin{equation*} 2^{4i+2} = \sum_{k} \sum_{m=0}^{k} \binom{4i+3-k}{i+1+m} \binom{k}{i+1+m} \end{equation*} Above conjecture validated up to $i=100$.

Mathematica scripts I used to validate all conjectures can be found on GitHub

Statement of the problem: It can be either a tip, a correction of my proof, or your own perspective

Answer 1

Let $n=i$. We need to find the following sum: \begin{align*} \sum\limits_{m=0}^n \sum\limits_k\binom{4n+3-k}{m}\binom{k}{m} \quad (1.0) \end{align*} Note that we can assume that $k$ goes over all integers without bounds. We can rewrite the inner sum as \begin{align*} [x^{4n+3}] \sum\limits_{i,j} \binom{i}{m} x^i \binom{j}{m} x^j &= [x^{4n+3}] \frac{x^m}{(1-x)^{m+1}} \frac{x^m}{(1-x)^{m+1}} \quad (2.0) \\ &= [x^{4n+3}] \frac{x^{2m}}{(1-x)^{2m+2}} \end{align*} which evaluates to \begin{align*} \binom{4n+4}{2m+1} \quad (3.0) \end{align*} meaning that the sum is same as $S=\sum\limits_{m=0}^n \binom{4n+4}{2m+1}$. Note that \begin{align*} \binom{4n+4}{2m+1} = \binom{4n+4}{4n+4-(2m+1)} = \binom{4n+4}{2(2n+1-m)+1} \end{align*} Therefore, by substitution $m \to 2n+1-m$, we have \begin{align*} S =\sum\limits_{m=0}^n \binom{4n+4}{2(2n+1-m)+1} = \sum\limits_{m=n+1}^{2n+1} \binom{4n+4}{2m+1} \end{align*} But this means that $2S = \sum\limits_{m=0}^{2n+1} \binom{4n+4}{2m+1} =2^{4n+3}$, which implies $S = 2^{4n+2}$. $\square$

P.S. Note that, more generally, it means that $\sum\limits_k \binom{t-k}{m} \binom{k}{m} = \binom{t+1}{2m+1}$, so we have \begin{align*} \sum\limits_{k=0}^t \binom{t}{k}_n = \sum\limits_{m=0}^n \binom{t+1}{2m+1} \end{align*} P.S.S. Also see this question for the identity above.

Answer 2

This maybe a little aside from the problem you propose but my coauthor and I recently uploaded a preprint to Arvix which has a combinatorial proof of this conjecture (Corollary 2) of Gibbs and Miceli, Two Combinatorial Interpretations of Rascal Numbers.

The proof falls out from the fact that $$\sum_{k=0}^{n} \binom{n}{k}_i = \sum_{k=0}^{2i+1} \binom{n}{k}.$$ This identity has a fairly nice combinatorial proof using the interpretation of $\binom{n}{k}_i$ as counting the number of binary words of length $n$ with $k$ ones and at most $i$ ascents. Let $w = w_1 \ldots w_n$ be such a binary word then $$ w \mapsto \begin{cases} \operatorname{Des}(w) \cup \operatorname{Asc}(w) \cup \{ n \} & w_1 = 0 \\ \operatorname{Des}(w) \cup \operatorname{Asc}(w) & w_1=1 \end{cases}. $$ Note that we are not taking the final position to be a descent as some authors do so $\operatorname{Des}(w) \subseteq \{ 1, \ldots, n-1 \}$ and $\operatorname{Asc}(w) \subseteq \{ 1, \ldots, n-1 \}$. Then, since $\operatorname{asc}(w) \le j$, we also have that $\operatorname{des}(w) \le j+1$ with equality only when $1 \in \operatorname{Des}(w)$ but this only occurs when $w_1 = 1$. So, this map always results in subsets of $\{1, \ldots, n \}$ with at most $2i+1$ elements. Furthermore, this map is invertible. Given a set $S \subseteq \{ 1, \ldots, n \}$ with $|S| \le 2i+1$, we know the first letter of the it's corresponding $w$ by checking if $n \in S$. From here, we simply being by placing $1$s or $0$s (depending on whether $n \in S$ or not). If after placing $i$ bits, $i \in S$ then we switch to placing the bit (e.g. if we were placing $1$s, we switch the placing $0$s). After placing $n$ bits, we've reconstructed $w$. Thus, this map is a bijection which gives us our desired identity. Now, apply this result to $$\sum_{k=0}^{4i+3} \binom{4i+3}{k}_i = \sum_{k=0}^{2i+1} \binom{4i+3}{k} = 2^{4i+2}.$$

Hope this provides a new perspective!

Answer 3

A variation: We use the coefficient of operator to denote the coefficient of $z^n$ of a series. This way we can write for instance \begin{align*} \binom{n}{k}=[z^k](1+z)^n\tag{1} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{k=0}^{4i+3}}&\color{blue}{\sum_{m=0}^i\binom{4i+3-k}{m}\binom{k}{m}}\\ &=\sum_{m=0}^i\sum_{k=m}^{\infty}\binom{4i+3-k}{m}\binom{k}{m}\tag{2}\\ &=\sum_{m=0}^i\sum_{k=0}^{\infty}\binom{4i+3-k-m}{m}\binom{k+m}{m}\tag{3}\\ &=\sum_{m=0}^i\sum_{k=0}^{\infty}[z^m](1+z)^{4i+3-k-m}\binom{-m-1}{k}(-1)^k\tag{4}\\ &=\sum_{m=0}^i[z^m](1+z)^{4i+3-m}\sum_{k=0}^{\infty}\binom{-m-1}{k}\frac{(-1)^k}{(1+z)^k}\\ &=\sum_{m=0}^i[z^m](1+z)^{4i+3-m}\left(1-\frac{1}{1+z}\right)^{-m-1}\tag{5}\\ &=\sum_{m=0}^i[z^{2m+1}](1+z)^{4i+4}\tag{6}\\ &=\sum_{m=0}^i\binom{4i+4}{2m+1}=\frac{1}{2}\sum_{m=0}^{2i+1}\binom{4i+4}{2m+1} =\frac{1}{4}\sum_{m=0}^{4i+4}\binom{4i+4}{m}\tag{7}\\ &\,\,\color{blue}{=2^{4i+2}} \end{align*} according to the claim.

Comment:

• In (2) we set the upper limit of the inner sum to $\infty$ which does not change anything, since $\binom{4i+3-k}{m}=0$ when the upper index $4i+3-k<m$. We also set the lower limit of the inner sum to $m$ since $\binom{k}{m}=0$ if $k<m$ and $m>0$.

• In (3) we shift the index of the inner sum to start with $k=0$.

• In (4) we apply (1) accordingly and use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (5) we apply the binomial series expansion.

• In (6) we do some simplifications and apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

• In (7) we select the coefficient of $z^{2m+1}$ and do some transformations.