# For $J=\{1,2,\dots,n \}$ is there an easy way to compute $\prod\limits_{i\in J | i \ne k} (k-i)$?

When I studied calculus at my university there is one question that I hated the most which is given a finite number of terms for some sequence find the $$n-$$th term. I hated this type of question because there is obviously not a single answer to this question for example: what is the nth term of this sequence $$1,2,4,8,16, \dots$$? For any complex number $$z$$ I can make this sequence

$$\frac{-(n-2)(n-3)(n-4)(n-5)(n-6)}{5!} +\frac{2(n-1)(n-3)(n-4)(n-5)(n-6)}{4!} -\frac{4(n-1)(n-2)(n-4)(n-5)(n-6)}{12} +\frac{8(n-1)(n-2)(n-3)(n-5)(n-6)}{12} -\frac{16(n-1)(n-2)(n-3)(n-4)(n-6)}{4!} +\frac{z (n-1)(n-2)(n-3)(n-4)(n-5)}{5!}$$

This would make $$z$$ is what comes next after $$16$$.

This sequence might seem complicated but it is really easy to find any arbitrary sequence that contain $$n$$ terms for a sequence $$x_1,x_2,\dots,x_n$$ let $$J=\{1,2,\dots,n \}$$ the general sequence is just $$\sum_{k=1}^n x_k \frac{\prod\limits_{i\in J | i \ne k} (n-i)}{\prod\limits_{i\in J | i \ne k} (k-i)}$$

the only problem is how to compute $$\prod\limits_{i\in J | i \ne k} (k-i)$$? I tried to find an easy way to find the general but I failed.

• I haven't read the question in detail, but at first glance it does make me think of Lagrange interpolation Polynomials Commented Jan 25 at 20:02
• @AdamRubinson I have never heard of this theorem before.
– pie
Commented Jan 25 at 20:05

Although not relevant to your problem, if $$k>n$$ the case $$i=k$$ will not appear and the product equals $$\prod_{i\in J|i\neq k} (k-i) = k!/(k-n)!$$
If $$k, split the product into two products $$\prod_{i\in J|i\neq k} (k-i) = \prod_{i\in J|i > k} (k-i)\prod_{i\in J|i < k} (k-i) = (-1)^{n-k}\prod_{i\in J|i > k} (i-k)\prod_{i\in J|i < k} (k-i).$$ The remaining products are just factorials again $$\prod_{i\in J|i\neq k} (k-i) = (-1)^{n-k}(n-k)!(k-1)!.$$
In the example you gave in the question, $$n = 6$$. The terms are $$\begin{array}{c|c|c} k & \prod_{i\in J|i\neq k} (k-i) & \text{Simplified}\\ \hline 1 & (-1)^{(6-1)} (6-1)!0! & -5! \\ \hline 2 & (-1)^{(6-2)} (6-2)!1!& 4! \\ \hline 3 & (-1)^{(6-3)} (6-3)!2!& -3!2! \\ \hline 4 & (-1)^{(6-4)} (6-4)!3!& 2!3! \\ \hline 5 & (-1)^{(6-5)} (6-5)!4!& -4! \\ \hline 6 & (-1)^{(6-6)} (6-6)!5!& 5! \end{array}$$