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Vishnu Namboothiri K
Vishnu Namboothiri K
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Is there any closed formula for 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\}}}\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.

When we take $n-j=0$ or equivalently $j=n$, by varying $n=1,2,3,\ldots$, we get the sequence of Fubini numbers (please see https://oeis.org/A000670) as the sum. I have verified this for first few natural numbers using SageMath software; and of course I do not have a theoretical justification even for this. For other values of $j$, the sum is not coming out in a nice closed form (or as a known sequence of numbers that could be located from OEIS) and I could not so far relate them to Fubini's numbers in any way.

Is there any closed formula for 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\}}}\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*}

When we take $n-j=0$ or equivalently $j=n$, by varying $n=1,2,3,\ldots$, we get the sequence of Fubini numbers (please see https://oeis.org/A000670) as the sum. I have verified this for first few natural numbers using SageMath software; and of course I do not have a theoretical justification even for this. For other values of $j$, the sum is not coming out in a nice closed form (or as a known sequence of numbers that could be located from OEIS) and I could not so far relate them to Fubini's numbers in any way.

Is there any closed formula for 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\}}}\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.

When we take $n-j=0$ or equivalently $j=n$, by varying $n=1,2,3,\ldots$, we get the sequence of Fubini numbers (please see https://oeis.org/A000670) as the sum. I have verified this for first few natural numbers using SageMath software; and of course I do not have a theoretical justification even for this. For other values of $j$, the sum is not coming out in a nice closed form (or as a known sequence of numbers that could be located from OEIS) and I could not so far relate them to Fubini's numbers in any way.

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Vishnu Namboothiri K
Vishnu Namboothiri K
  • 111
  • 4

Closed form for a binomial product sum

Is there any closed formula for 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\}}}\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*}

When we take $n-j=0$ or equivalently $j=n$, by varying $n=1,2,3,\ldots$, we get the sequence of Fubini numbers (please see https://oeis.org/A000670) as the sum. I have verified this for first few natural numbers using SageMath software; and of course I do not have a theoretical justification even for this. For other values of $j$, the sum is not coming out in a nice closed form (or as a known sequence of numbers that could be located from OEIS) and I could not so far relate them to Fubini's numbers in any way.