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The recurrence relation is

$$a_n = a_{n-1} + a_{n-2} + n,\quad n\ge 2$$

with initial conditions $a_0 = 0$ and $a_1 = 1$.

I know I need to convert the recurrence into series and I have broken it down, but am struggling with getting it into a proper form to do partial fractions.

I have:

$$f(x) = a_0 + a_1x + x\sum_{n\ge 2} a_{n-1}x^{n-1} + x^2\sum_{n\ge 2} a_{n-2}x^{n-2} + \frac{1}{(1-x)^2}$$

Any insight/help is much appreciated!

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  • $\begingroup$ Are you sure about the $n\geq 0$ being the beginning of the summands? You have the first summation beginning with $a_{-1}x^{-1}$ and the second one begins with $a_{-2}x^{-2}. $\endgroup$ Dec 4, 2016 at 4:11
  • $\begingroup$ @DavidSnyder Thanks, you're right, typo on my part. It should be n=2 beginning the summands. $\endgroup$
    – mandib
    Dec 4, 2016 at 4:14

2 Answers 2

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Write $F(x)$ as the generating function, then

$$F(x)-x-0= (x+x^2)F(x)+x\left({1\over (1-x)^2}-1\right).$$

Solving gives:

$$F(x) = {x\over (1-x-x^2)(1-x)^2}$$

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  • $\begingroup$ Oops. Don't forget the last term, $\frac{1}{(1-x)^2}$. $\endgroup$ Dec 4, 2016 at 4:09
  • $\begingroup$ you forgot the $n$ $\endgroup$
    – Phicar
    Dec 4, 2016 at 4:09
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    $\begingroup$ @Phicar yeah already caught it, but thanks! $\endgroup$ Dec 4, 2016 at 4:10
  • $\begingroup$ @DavidSnyder yes, and I forgot to take out the constant term from the first one as well @_@ $\endgroup$ Dec 4, 2016 at 4:22
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    $\begingroup$ @mandib yes, that's the general strategy! :) And sorry for all the edits, keeping track of the little nitty gritty bits is always the devil from the details. $\endgroup$ Dec 4, 2016 at 4:25
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Just manipulate the generating function; \begin{align} f(x)=x+\sum_{n=2}^\infty a_nx^n=&x+x\sum_{n=2}^\infty a_{n-1}x^{n-1}+x^2\sum_{n=2}^\infty a_{n-2}x^{n-2}+x\sum_{n=2}^\infty nx^{n-1}\\ =&x+xf(x)+x^2f(x)+x\,\frac{d}{dx}\sum_{n=2}^\infty x^{n}\\ =&x+xf(x)+x^2f(x)+x\,\frac{d}{dx}\underbrace{\left(\frac{1}{1-x}-x-1\right)}_{x^2/(1-x)}\\ =&x+xf(x)+x^2f(x)+x\,\frac{2x-x^2}{(1-x)^2}, \end{align}

and now solve for $f(x)$.

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