Recurrence Relationship Help Request

I have been playing with the recurrence relation

pn=npn−1+pn−2 p_n = np_{n-1} + p_{n-2} such that p0=0,p1=1 p_0 = 0 , p_1 = 1 .

I have found a few interesting difference forms for it but no closed form formula. Can anyone find a nice closed form formula, generating function or interpretation for what this might count.

All thoughts appreciated and welcome :)

#Combinatorics

Note by Roberto Nicolaides
4 years, 2 months ago

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Comments

If you want to find the closed-form formula, find its generating function. To start you off, you will have

∑n≥2pnxn=∑n≥2npn−1xn+∑n≥2pn−2xn. \sum_{n \geq 2} p_n x^n = \sum_{n \geq 2} np_{n-1} x^n + \sum_{n \geq 2} p_{n-2} x^n.

If we define

f(x)≡∑n≥0pnxn, f(x) \equiv \sum_{n \geq 0} p_n x^n,

then you will have that

∑n≥2pnxn=f(x)−(p0+p1x)=f(x)−x, \begin{aligned} \sum_{n \geq 2} p_n x^n &= f(x) - (p_0 + p_1x) \\ &= f(x) - x, \end{aligned}

∑n≥2npn−1xn=x∑n≥2npn−1xn−1=x(∑n≥2(n−1)pn−1xn−1+∑n≥2pn−1xn−1)=x(x∑n≥2(n−1)pn−1xn−2+f(x))=x2f′(x)+xf(x) \begin{aligned} \sum_{n \geq 2} np_{n-1} x^n &= x \sum_{n \geq 2} np_{n-1} x^{n-1} \\ &= x \left( \sum_{n \geq 2} (n-1)p_{n-1} x^{n-1} + \sum_{n \geq 2} p_{n-1} x^{n-1} \right) \\ &= x \left( x \sum_{n \geq 2} (n-1)p_{n-1} x^{n-2} + f(x) \right) \\ &= x^2 f'(x) + xf(x) \end{aligned}

and

∑n≥2pn−2xn=x2∑n≥2pn−2xn−2=x2f(x). \begin{aligned} \sum_{n \geq 2} p_{n-2} x^n &= x^2 \sum_{n \geq 2} p_{n-2} x^{n-2} \\ &= x^2 f(x). \end{aligned}

So, you end up with the equation

f(x)−x=x2f′(x)+xf(x)+x2f(x), f(x) - x = x^2 f'(x) + xf(x) + x^2 f(x),

which simplifies to

x2f′(x)+(x2+x−1)f(x)=−x. x^2 f'(x) + (x^2 + x - 1) f(x) = -x.

This is a linear ODE which I will let you solve from here. This is one half of the problem; to recover the sequence, you will need to be a little crafty.

A Former Brilliant Member - 2 years, 10 months ago

Whenever you're investigating a sequence, it's not a bad idea to check the OEIS. This is entry A001053. There doesn't seem to be too much information.

Jon Haussmann - 4 years, 2 months ago

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Thanks Jon, this is a nice idea :) I did have a quick look and agreed that there was not a lot of info! I will be having a look at bessel funtions soon to try get more intuition on the problem :)

Roberto Nicolaides - 4 years, 2 months ago
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