# Finding the pattern in a sequence of polynomials

I have a series that has a weird behavior, as follows

$A(1)=x_1$

$A(2)={3\over4}x_1+{1\over4}x_2$

$A(3)={2\over3}x_1+{1\over4}x_2+{1\over12}x_3$

$A(4)={5\over8}x_1+{1\over4}x_2+{3\over32}x_3+{1\over32}x_4$

$A(5)={3\over5}x_1+{1\over4}x_2+{1\over10}x_3+{3\over80}x_4+{1\over80}x_5$

I want to have a general formula for $A(n)$, which I know looks like

$A(n)={n+1\over2n}x_1+{1\over4}x_2 + \sum_{i=3}^n \alpha_i x_i$

but I just don't know what is $\alpha_i$. Any ideas?

EDIT. Second part of the question here Finding expected distances and sequences

• Could you give some context of where you found this or how the pattern is meant to continue? I can't see any obvious pattern myself, especially since we only have a few terms of $x_3, x_4, x_5$. Edit: I see the last term is $\frac{1}{n(2^{n-1})}x_n$... but the rest? Jul 18, 2017 at 22:18
• I can see a slight pattern with the terms given $${1\over 3}\alpha_{n-1}=\alpha_n$$ not sure how the coefficients are changing as n is though. at A(n) for clarification.
– user451844
Jul 18, 2017 at 22:22
• All coefficients add to one, just a note. Jul 18, 2017 at 22:24

Note that the "common denominator" in $n^{th}$ row is $n 2^{n-1}$ & the numerators then generate the sequence $1,3,8,20,48,\cdots$ which have the formula $(n+2)2^{n-1}$ https://oeis.org/search?q=1%2C3%2C8%2C20%2C48&language=english&go=Search