Csdrhrt

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

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

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

$$b_n=scdot frac{b_{n-1}}{n-1} + tcdot b_{n-2}, forall n > 1$$

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

P.S.

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

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 >

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

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

So
$$f(x)left(frac{1-tx^2}{x}right)=1+ssumlimits_{n=1}frac{b_{n}}{n}x^{n}.$$
Now we derivate
$$left[f(x)left(frac{1-tx^2}{x}right)right]'=frac{s}{x}sumlimits_{n=1}b_{n}x^{n}=frac{sf(x)}{x}.$$
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

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