Solving $ F_{n} = \sum_{i=1}^{n-1} (F_{i}\cdot F_{n-i}) $?

Question

I need to find $F_{n}$ in : $$ F_{n} = \sum_{i=1}^{n-1} (F_{i}\cdot F_{n-i}) , F_0 = 0 , n>=2 $$

This equation screams convolution , I think , but I find it as a quite long solution sometimes.

Here, I first tried to play with the indexes :

$$ F_{n-1} = \sum_{i=1}^{n-2} (F_{i}\cdot F_{n-1-i}) $$

But this doesn't seem to do anything productive .

So convolution maybe :

Let $$ A(x) = \sum_{n=0}^{∞} F_{i}\cdot x^n $$

Let $$ B(x) = \sum_{n=0}^{∞} F_{n-i}\cdot x^n $$

And $$C(x) = A(x)\cdot B(x) = \left(\sum_{n=0}^{∞} F_{i}\cdot x^n \right)\cdot \left(\sum_{n=0}^{∞} F_{n-i}\cdot x^n\right)$$

But I can't see how this helps me here , any hints and/or ideas?

Thanks

It's exactly the usual convolution if $F_0 = 0$. Try looking at $A(x)^2$. — Jair Taylor, Jul 21 '13 at 06:14
The only solution is $F_n=0$ for all $n$. Try calculating $F_1$, then $F_2$, then $F_3$, and see what happens. — Gerry Myerson, Jul 21 '13 at 06:22
$$(\sum_{n=1}^\infty a_nx^n)^2=\sum_{n=2}^\infty(\sum_{k=1}^{n-1}a_ka_{n-k})x^n$$ — Ethan Splaver, Jul 21 '13 at 06:23
It appears OP has edited in that the recurrence holds only for $n\ge2$, so $F_1$ is arbitrary. So my earlier comment no longer applies. — Gerry Myerson, Jul 21 '13 at 06:26

score 5 · Accepted Answer · answered Jul 21 '13 at 06:33

5

For convenience set $F_0=0$, and let

$$f(x)=\sum_{n\ge 0}F_nx^n=\sum_{n\ge 1}F_nx^n\;.$$

Then

$$f(x)^2=\sum_{n\ge 2}\sum_{k=0}^nF_kF_{n-k}x^n=\sum_{n\ge 2}\sum_{k=1}^{n-1}F_kF_{n-k}x^n=\sum_{n\ge 2}F_nx^n=f(x)-F_1x\;,$$

and we have $f(x)^2-f(x)+F_1x=0$. Solve this as a quadratic in $f(x)$:

$$f(x)=\frac{1\pm\sqrt{1-4F_1x}}2\;,$$

and since $f(0)=F_0=0$, we must choose

$$f(x)=\frac{1-\sqrt{1-4F_1x}}2\;.$$

The generating function for the Catalan numbers $C_n$ is

$$\frac{1-\sqrt{1-4x}}{2x}=\sum_{n\ge 0}C_nx^n\;,$$

so

$$\begin{align*} f(x)&=F_1x\left(\frac{1-\sqrt{1-4F_1x}}{2F_1x}\right)\\\\ &=F_1x\sum_{n\ge 0}C_nF_1^nx^n\\\\ &=\sum_{n\ge 0}\frac{F_1^{n+1}}{n+1}\binom{2n}nx^{n+1}\\\\ &=\sum_{n\ge 1}\frac{F_1^n}n\binom{2n-2}{n-1}x^n\;, \end{align*}$$

and $$F_n=\frac{F_1^n}n\binom{2n-2}{n-1}\;.$$

answered Jul 21 '13 at 06:33

Brian M. Scott

616,228

+1 But you spoiled the fun! =D Note the OP has set $F_0=0$. – Pedro Jul 21 '13 at 06:34
@Peter: I did get carried away a bit: I got interested in the problem for its own sake! (The setting of $F_0$ to $0$ must have occurred while I was working on the problem.) – Brian M. Scott Jul 21 '13 at 06:35
Of course I am kidding. It is a nice problem. In fact, the evaluation of the coefficients is not something that simple without knowledge of Catalan's GF. I guess I was simply pointing out my answer has been rendered useless =P – Pedro Jul 21 '13 at 06:36
Would you care checking the proof of the second proposition here? – Pedro Jul 21 '13 at 06:44
@Peter: Everything looks okay to me. – Brian M. Scott Jul 21 '13 at 06:51

Solving $ F_{n} = \sum_{i=1}^{n-1} (F_{i}\cdot F_{n-i}) $?

1 Answers1