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I would like to solve the following first order non-homogenous recurrence relation $$ \left(n^2+n-1\right) y_{n+1}-(n-2)\,n\,y_n=\frac{1}{256} \left(25 n^3+150 n^2-173 n+32\right)\,. $$

It is a simple exercise to solve the associated homogenous recurrence relation $$ \left(n^2+n-1\right)\,y^H_{n+1}-(n-2)\,n\,y^H_n=0\,, $$ which yields $$ y^H_n = C_1\,\frac{\Gamma (n-2) \Gamma (n)}{\Gamma \left(n-\frac{\sqrt{5}}{2}+\frac{1}{2}\right) \Gamma \left(n+\frac{\sqrt{5}}{2}+\frac{1}{2}\right)}\,, $$ where $C_1$ is a constant and $\Gamma$ is a gamma function.

I tried to use telescoping to solve for $y_n$. In order to do this I set $$ y_n = y_n^H+y_n^H z_n $$ which gives for $z_n$ $$ z_{n}-z_{n-1}=\frac{\left(25 n^3+75 n^2-398 n+330\right)}{256} \frac{\Gamma \left(n-\frac{\sqrt{5}}{2}-\frac{1}{2}\right) \Gamma \left(n+\frac{\sqrt{5}}{2}-\frac{1}{2}\right)}{\Gamma (n-2) \Gamma (n)}\,. $$ However, I was not able to find a closed form for $$ \sum_{k=1}^n \frac{\left(25 k^3+75 k^2-398 k+330\right)}{256} \frac{\Gamma \left(k-\frac{\sqrt{5}}{2}-\frac{1}{2}\right) \Gamma \left(k+\frac{\sqrt{5}}{2}-\frac{1}{2}\right)}{\Gamma (k-2) \Gamma (k)}. $$

Any other ideas?

user12588
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  • What makes you think this has a neat closed form? Trying to use generating functions gives a truly horrendous ODE:

    $$ x^3 Y'' - 2x^2 Y' + Y + c_0 = \frac{165 x^3 - 271 x^2 + 47 x - 16}{128 (1-x)^4} $$

    (where $c_0$ is a parameter) which sage doesn't know how to solve. Here $Y(x) = \sum_n y_n x^n$. The singularities of $Y$ can give good asymptotic estimates for the $y_n$s, and $Y$ has a particularly nice closed form we can compute a closed form for the $y_n$ from this... Unfortunately, this doesn't.

     – HallaSurvivor Jan 05 '22 at 06:00
  • Wolframalpha can "solve" this ODE (see here) but I'm not sure how to taylor expand it and check my work. Also the solution is hideous (see the link) and I wouldn't know how to manipulate it into a form that lets us control even the asymptotics of $y_n$. To me this is a sign that there may not be a great way to proceed. – HallaSurvivor Jan 05 '22 at 06:03

1 Answers1

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Let $\;y_n=u_n+v_n,\;$ where $$v_n=\dfrac{5n^2+35n-56}{256},\quad (n^2+n-1)v_{n+1}-(n-2)nv_n = \dfrac{25n^2+150n^2-173n+16}{256}.\tag1$$ Then $$(n^2+n-1)u_{n+1}-(n-2)nu_n = \dfrac1{16}.\tag2$$

As it was shown in the OP, solution of the homogeneous recurrence relation is $$u^\,_H=\dfrac{\Gamma(n)\Gamma(n-2)}{\Gamma(n+\varphi)\Gamma(n-\varphi+1)},\tag3$$ where $\;\varphi=\dfrac{\sqrt5+1}2\;$ is the golden ratio.

Substitution $\;u=u^\,_H w\;$ leads to the equation $$w_{n+1}-w_n = \dfrac{\Gamma(n+\varphi)\Gamma(n-\varphi+1)}{16\Gamma(n+1)\Gamma(n-1)}.\tag4$$

Therefore, $$y_n=C\,\dfrac{\Gamma(n)\Gamma(n-2)}{\Gamma(n+\varphi)\Gamma(n-\varphi+1)}+\dfrac{5n^2+35n-56}{256}+\sum\limits_{k=m}^{n-1} \dfrac{\Gamma(k+\varphi)\Gamma(k-\varphi+1)}{16\Gamma(k+1)\Gamma(k-1)},\tag5$$ where the parameters $\;C,m\;$ should be defined from addition conditions.

Presentation $(5)$ is not a closed form, but it looks as the maximum of possible.

Yuri Negometyanov
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