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






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

  • 2
    $\begingroup$ 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? $\endgroup$
    – Shuri2060
    Jul 18, 2017 at 22:18
  • $\begingroup$ 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. $\endgroup$
    – user451844
    Jul 18, 2017 at 22:22
  • 2
    $\begingroup$ All coefficients add to one, just a note. $\endgroup$ Jul 18, 2017 at 22:24

1 Answer 1


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

So the formula is \begin{eqnarray*} A(n)=\sum_{i=1}^{n} \frac{(n+2-i)2^{n-i-1}}{n 2^{n-1}} x_i=\sum_{i=1}^{n} \frac{(n+2-i)}{n 2^{i}} x_i. \end{eqnarray*}

  • $\begingroup$ You could simplify the factor of xi in the sum to (n+2-i)/(n*2^i) $\endgroup$
    – Penguino
    Jul 18, 2017 at 22:57
  • $\begingroup$ Good thinking @Penguino ... I will edit. $\endgroup$ Jul 18, 2017 at 22:59
  • $\begingroup$ Thank you all. Here is the reasoning behind this question - which possess an even harder problem math.stackexchange.com/questions/2363295/… $\endgroup$
    – fox
    Jul 18, 2017 at 23:56

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