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  1. Would anyone know if there any functions that fit this recurrence relation:

$F_n = \frac{1}{n+3/2} \left ( F_{n+2} - c F_{n+1} \right )$

where $c$ is a constant parameter.

or, more general:

$F_n = g(n) \left ( F_{n+2} - c F_{n+1} \right )$

  1. Any reference or books that have a list of functions and their recurrence relation?
uday
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2 Answers2

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The "backward" recurrence relation: $$ F_n = \frac{1}{n+3/2}\left(F_{n+2}-c F_{n+1}\right) $$ is equivalent to the standard recurrence relation: $$ F_{n+2}= c F_{n+1} + (n+3/2) F_n$$ so any choice of $F_0,F_1$ leads to a solution, and the situation is the same for the second recurrence.

Jack D'Aurizio
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  • how do I solve $F_{n+2} = c F_{n+1} + (n+3/2) F_{n}$. I found many info on the net to solve $F_{n+2} = c F_{n+1} + c_2 F_{n}$, where $c_2$ is a constant, not a function of $n$ – uday Jul 09 '15 at 14:05
  • I guess it would look a bit like $a\sqrt{n!}$ – Empy2 Jul 09 '15 at 14:18
  • @uday: the usual method is to define $G(x)=\sum_{n\geq 0}F_n x^n$, translate the recurrence relation into a differential equation for $G(x)$, solve it, then recover the coefficients. – Jack D'Aurizio Jul 09 '15 at 14:28
  • @JackD'Aurizio, can you recommend me some reference which shows how to translate the relation into the differential equation where the coefficients are a function of $n$ ? – uday Jul 09 '15 at 14:41
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Use generating functions: Define $f(z) = \sum_{n \ge 0} F_n z^n$, multiply your recurrence by $z^n$ and sum over $n \ge 0$:

$$ \sum_{n \ge 0} F_{n + 2} z^n = c \sum_{n \ge 0} F_{n + 1} z^n + \sum_{n \ge 0} n F_n z^n + \frac{3}{2} \sum_{n \ge 0} F_n z^n $$

Recognize some of the sums:

$$ \frac{f(z) - F_0 - F_1 z}{z^2} = c \frac{f(z) - F_0}{z} + z f'(z) + \frac{3}{2} f(z) $$

Now you've got a first order, linear differential equation for $f$. Solve that one (yes, chances are you get a mess), and pull the coefficients out of that one.

For more general forms of the equation you'll probably be out of luck.

vonbrand
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