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We know that a binomial coefficient is denoted as $$\binom{n}{k}$$ and the sum of binomial coefficients is denoted as \begin{equation} \sum_{k=0}^{n}\binom{n}{k} \tag{1} \end{equation} I think it is easy to understand that \begin{equation} m=\sum_{k=0}^{n}\binom{n}{k}f(k) \tag{2} \end{equation} Note: (2) is just an illustration of how a summation involving binomial coefficients can be formed in a particular way.

Now, let us assume there is a set of numbers that behave like binomial coefficients and let's call them psuedo-binomial coefficients. For example, $1, 3, 8, 9, 8, 3, 1$ and $2, 6, 9, 11, 11, 9, 6, 2$ follow the structure or order of arrangement of binomial coefficients but are actually not. Also, they obey the rule that $\binom{n}{k} = \binom{n}{n-k}$. We can generalize the psuedo-binomial coefficients as $a_1, a_2, a_3\cdots a_i, a_{i-1}, a_{i-2} \cdots a_1$ or $a_1, a_2, a_3\cdots a_i, a_i, a_{i-1}, a_{i-2} \cdots a_1$, where $a_1<a_2<a_3<\cdots<a_i$. Now, I want the psuedo-binomial coefficients to align with $f(k)$ so that it looks exactly like (2), except for not minding that the initial value of $k$ being $0$. So, i came up with the following statement:

If $A_p =$ {$a_1, a_2, a_3\cdots a_i, a_{i-1}, a_{i-2} \cdots a_1$} or $A_p =$ {$a_1, a_2, a_3\cdots a_i, a_i, a_{i-1}, a_{i-2} \cdots a_1$} is a well-ordered set of natural numbers, such that $a_1<a_2<a_3<\cdots<a_i$, and $n$ is the number of elements of $A_p$, $\binom{n}{k}_p$ for $k= 1, 2, 3\cdots n$, is the $k$th element of $A_p$, $f(k)$ is an expression involving $k$, $m$ is some real or complex number depending on $f(k)$, then \begin{equation} m=\sum_{k=1}^{n}\binom{n}{k}_{p} f(k) \tag{3} \end{equation}

Note: $k$ in my statement starts from $1$ so that the position of the first element of $A_p$ is $1$ and not $0$. Also, let us forget that $\binom{n}{k} = \frac{n!}{(n-k)!k!}$ as that won't work for pseudo-binomial coefficients. Instead, let's just focus on $\binom{n}{k} =\binom{n}{n-k}$ which is a general rule for both binomial coefficients and pseudo-binomial coefficients. So, I think this way, it will be valid to say (3) works for binomial coefficients as well. For example, let $1, 3, 3, 1$ be the elements of $A_p$, then we can say $\binom{4}{1}_p =1$, $\binom{4}{2}_p =3$, $\binom{4}{3}_p =3$, and $\binom{4}{4}_p =1$.

Question: is the above statement well defined, such that (3) deals with both binomial coefficients and pseudo-binomial coefficients? If not, can someone come up with a well defined statement?

Shuaib Lateef
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    I'm confused as to what you want here? Do you want to find $A$ such that $\sum_k \binom nk f(k) = \sum_k \binom nk _p f(k) = m$? I.e. they have the same sum? But what is $f$ here? Is it an arbitrary function? Is $A$ allowed to depend on $f$? – Milten Aug 25 '21 at 19:56
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    You seem to ask if the "statement" $m=\sum_k\binom nk _p f(k)$ is well defined. Well sure, it's a valid equation and notation; it just says that the sum is equal to some number. I wouldn't call it a statement unless $f$, $m$ and $A$ are defined first. For example, $\sum_k k\binom nk = n2^{n-1}$ is a statement, which can be proven. As for your notation, it's odd to use both $A$ and $p$, which look independent. I'd either write $\binom nk _A$ or let the set be notated $A_p$ to signify the relationship. – Milten Aug 25 '21 at 20:09
  • @Milten I think I get what you are trying to say. Thanks for your response. – Shuaib Lateef Aug 25 '21 at 20:17

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