# determining the formula of equations similar to ${2}^{x-1}$

the formula for
$$1,2,3,4,5,6,7\space\text{is}\space{x}\\ 1,2,4,7,11,16,22,29\space\text{is}\space\frac{x\left(x-1\right)+2}{2}\\ 1,2,4,8,15,26,42,64,93\space\text{is}\space\frac{x\left(x\left(x-3\right)+8\right)}{6}\\ 1,2,4,8,16,31,57,99,163,256\space\text{is}\space\frac{x\left(x\left(x\left(x-6\right)+23\right)-18\right)+24}{24}\\ 1,2,4,8,16,32,63,120,219,382,638\space\text{is}\space\frac{x\left(x\left(x\left(x\left(x-10\right)+55\right)-110\right)+184\right)}{120}\\ 1,2,4,8,16,32,64,127,247,466,848,1486\space\text{is}\space\frac{x\left(x\left(x\left(x\left(x\left(x-15\right)+115\right)-405\right)+964\right)-660\right)+720}{720}\\ \text{is there a way to calculate the formula of these sequences easily?}$$

You haven't said what these sequences are, but it looks to me like they are (up to some reindexing) the unique polynomials $$P_n(x)$$ of degree $$n$$ with the property that $$P_n(k) = 2^k$$ for the integers $$0 \le k \le n$$. These have closed form

$$P_n(x) = \sum_{i=0}^n {x \choose i}$$

which is not hard to prove using the calculus of finite differences, or just by computing that $$P_n(k) = 2^k$$ for $$0 \le k \le n$$ and arguing that this (together with the degree condition) uniquely determines $$P_n(x)$$.

To any finite sequence of integers $$(a_1,\cdots,a_n)$$, you can associate a polynomial $$P(x)$$ such that $$P(i) = a_i$$. This has nothing to do with the fact that your first few values are of the form $$f(x) = 2^{x-1}$$. The formula is $$P(x) = \sum_{j=1}^n a_j \left( \underset{i \neq i}{\prod_{j=1}^n} \frac{x-i}{j-i} \right).$$ This is the Lagrange polynomial associated to the sequence of pairs $$((1,a_1), \cdots, (n,a_n))$$. It's easy to see that $$P(k) = a_k$$; you just have to notice that in this sum from $$1$$ to $$n$$, if $$1 \le k \le n$$ satisfies $$k \neq j$$, then one of the linear terms on the top of the form $$x-i$$ will vanish when $$k=i$$, so the whole term will disappear. The only term left will be $$\underset{i \neq i}{\prod_{j=1}^n} a_j\frac{x-i}{j-i}$$ evaluated at $$x=j=k$$, and in this case the whole term reduces to $$a_j$$.

A simpler way to write this is to express it in terms of the Lagrange basis polynomials: $$\ell_j(x) = \underset{j \neq i}{\prod_{i=1}^n} \frac{x-i}{j-i}$$ and write $$P(x) = \sum_{j=1}^n a_j \ell_j(x)$$. These polynomial functions satisfy $$\ell_j(i) = \delta_{ij} = \left\{ \begin{matrix} 1 & \text{ if } i = j \\ 0 & \text{ if not} \end{matrix} \right.$$ which is what makes them special. We can then see that $$P(i) = \sum_{j=1}^n a_j \delta_{ij} = a_i$$.

There is not even anything special about the choice of $$1,\cdots,n$$ for the initial sequence. You could have chosen $$n$$ arbitrary pairs $$((x_1,a_1), \cdots, (x_n,a_n))$$ and consider again the Lagrange basis polynomials $$\ell_j(x) = \underset{j \neq i}{\prod_{i=1}^n} \frac{x-x_i}{x_j-x_i}$$ and then the polynomial $$P(x) = \sum_{j=1}^n a_j \ell_j(x)$$ satisfies $$P(x_i) = a_i$$ for $$i=1,\cdots,n$$; this follows from the fact that $$\ell_j(x_i) = \delta_{ij}$$, so $$P(x_i) = \sum_{j=1}^n a_j \delta_{ij} = a_i$$.

This entire process is called polynomial interpolation.

Hope that helps,