About $a_0=0, a_1=1$, $a_n=3 \frac{a_{n-1}}{n-1} + 2a_{n-2}$ for $n > 1$.

Question

Let $\{a_n\}$ be defined by as the following: $a_0=0, a_1=1$

$$a_{n}=3\frac{a_{n-1}}{n-1}+2a_{n-2}, \forall n > 1$$ For example $a_2=3, a_3=\frac{13}{2}$. Is its generating function equal to $$\frac{x}{\left(1-x\sqrt{2}\right)^2}\cdot \left(\frac{1+x\sqrt{2}}{1-x\sqrt{2}}\right)^{\frac{3}{2\sqrt{2}}-1}$$ ?

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Let $s,t > 0$. Let $\{b_n\}$ be defined by as the following: $b_0=0, b_1=1$

$$b_n=s\cdot \frac{b_{n-1}}{n-1} + t\cdot b_{n-2}, \forall n > 1$$

Is its generating function equal to $$\frac{x}{\left(1-x\sqrt{t}\right)^2}\cdot \left(\frac{1+x\sqrt{t}}{1-x\sqrt{t}}\right)^{\frac{s}{2\sqrt{t}}-1}$$ ?

P.S.

$$\frac{x}{\left(1-x\sqrt{t}\right)^2}\cdot \left(\frac{1+x\sqrt{t}}{1-x\sqrt{t}}\right)^{ \frac{s}{2\sqrt{t}}-1}= \frac{x}{s}\cdot \frac{d}{dx} \left(\frac{1+x\sqrt{t}}{1-x\sqrt{t}}\right)^{ \frac{s}{2\sqrt{t}}}$$

I think this generating function is correct.

(17:03) gp > N=26; x='x+O('x^N); Vec(x/(1-2^(1/2)*x)^2 * ((1+2^(1/2)*x)/(1-2^(1/2)*x))^(3/(2^(3/2))-1))
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(17:04) gp >

Answer 1

Incomplete answer. I presume you mean generating function. If so, then $$f(x)=a_0+a_1x+\sum\limits_{n=2}a_nx^n= x+\sum\limits_{n=2}\left(3\frac{a_{n-1}}{n-1}+2a_{n-2}\right)x^n=\\ x+3\sum\limits_{n=2}\frac{a_{n-1}}{n-1}x^n+2\sum\limits_{n=2}a_{n-2}x^n=\\ x+3\sum\limits_{n=2}\frac{a_{n-1}}{n-1}x^n+2x^2\sum\limits_{n=2}a_{n-2}x^{n-2}=\\ x+3x\sum\limits_{n=2}\frac{a_{n-1}}{n-1}x^{n-1}+2x^2f(x)$$ or $$f(x)\left(\frac{1-2x^2}{x}\right)=1+3\sum\limits_{n=1}\frac{a_{n}}{n}x^{n}$$ now we derivate $$\left[f(x)\left(\frac{1-2x^2}{x}\right)\right]'=3\sum\limits_{n=1}a_{n}x^{n-1}=\frac{3}{x}\sum\limits_{n=1}a_{n}x^{n}=\frac{3f(x)}{x}$$ or $$f'(x)\left(\frac{1-2x^2}{x}\right)-f(x)\left(2+\frac{1}{x^2}\right)=\frac{3f(x)}{x}$$ or $$f'(x)\left(1-2x^2\right)-f(x)\left(2x+\frac{1}{x}\right)=3f(x)$$ $$f'(x)\left(1-2x^2\right)=f(x)\left(3+2x+\frac{1}{x}\right)$$ $$\frac{f'(x)}{f(x)}=\frac{3+2x+\frac{1}{x}}{1-2x^2}$$ Solve the differential equation and you have an yes to your first question.

Answer 2

Complete answer -rtybase's way- $$f(x)=b_0+b_1x+\sum\limits_{n=2}b_nx^n= x+\sum\limits_{n=2}\left(s\frac{b_{n-1}}{n-1}+tb_{n-2}\right)x^n=\\ x+sx\sum\limits_{n=2}\frac{b_{n-1}}{n-1}x^{n-1}+tx^2f(x).$$

So $$f(x)\left(\frac{1-tx^2}{x}\right)=1+s\sum\limits_{n=1}\frac{b_{n}}{n}x^{n}.$$ Now we derivate $$\left[f(x)\left(\frac{1-tx^2}{x}\right)\right]'=\frac{s}{x}\sum\limits_{n=1}b_{n}x^{n}=\frac{sf(x)}{x}.$$ So $$f'(x)\left(\frac{1-tx^2}{x}\right)-f(x)\left(t+\frac{1}{x^2}\right)=\frac{sf(x)}{x}.$$ Hence $$\frac{f'(x)}{f(x)}=\frac{tx+s+\frac{1}{x}}{1-tx^2}.$$ Solve the differential equation and we have

$$f(x)=\frac{x}{1-tx^2} * (\frac{1+x\sqrt t}{1-x\sqrt t})^{\frac{s}{2\sqrt t}}= \frac{x}{\left(1-x\sqrt{t}\right)^2}\cdot \left(\frac{1+x\sqrt{t}}{1-x\sqrt{t}}\right)^{ \frac{s}{2\sqrt{t}}-1}.$$