1
$\begingroup$

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 $$\frac{x}{\left(1-x\sqrt{2}\right)^2}\cdot \left(\frac{1+x\sqrt{2}}{1-x\sqrt{2}}\right)^{\frac{3}{2\sqrt{2}}-1}$$ ?


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))
%1 = [1, 3.0000000000000000000000000000000000000, 6.5000000000000000000000000000000000000, 12.500000000000000000000000000000000000, 22.375000000000000000000000000000000000, 38.425000000000000000000000000000000000, 63.962500000000000000000000000000000001, 104.26250000000000000000000000000000000, 167.02343750000000000000000000000000000, 264.19947916666666666666666666666666667, 413.30671875000000000000000000000000000, 641.11897253787878787878787878787878788, 986.89318063446969696969696969696969697, 1509.9825252221736596736596736596736597, 2297.3540452451194638694638694638694638, 3479.4358594933712121212121212121212120, 5247.1023141452460300116550116550116547, 7884.8309508947270177738927738927738924, 11808.343120106279896318958818958818958, 17634.137131279919282334989571831677093, 26261.806809904547684988166073692389480, 39019.960949689059662525431439905124111, 57844.517385675785323957072798280932247, 85584.858949683656541219176723064282863, 126387.14214006202771556654268694489985]
(17:04) gp >
$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Thank you. The second question may be proved in the same way as you. $\endgroup$ – Manyama Dec 16 '18 at 15:30
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.