Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the following recurrence sequence $$ a_{1} = 0\,,\quad a_{2} = 1\,, \qquad\qquad a_{n} = {2 + 2\left(n - 2\right)\, a_{n - 2} + \left(n - 2\right)\left(n - 1\right)\, a_{n - 1} \over n\left(n - 1\right)} $$

It starts form $\displaystyle{% \left\lbrace% 0,\ 1,\ {2 \over 3},\ {5 \over 6},\ {4 \over 5},\ {37 \over 45},\ {52 \over 63},\ {349 \over 420},\ {338 \over 405},\ {11873 \over 14175} \right\rbrace }$.

Please help me to find generating function

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Hint: Consider $A(x)=\sum\limits_{n=1}^\infty a_nx^n$ the generating function of $(a_n)$. The recursion you are interested in is that, for every $n\geqslant2$, $$ n(n-1)a_n=2+2(n-2)a_{n-2}+(n-2)(n-1)a_{n-1}. $$ Can you identify the generating function of $(b_n)$ when $b_n=n(n-1)a_n$? When $b_n=2$? When $b_n=2(n-2)a_{n-2}$? And when $b_n=(n-2)(n-1)a_{n-1}$? Then see what comes out.

share|improve this answer
    
Found it! ((1-E^(-2x)+((2-2x)Ei[2-2x])/E^2)/(1 - x)) –  Филипп Цветков Oct 25 '13 at 10:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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