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Let $$A_{n}=\sum_{i=0}^{n-3}(-1)^{n+i-2}\dfrac{13n^2-31n-10ni+9i+i^2+16}{(3n-i-3)(3n-i-4)(2n-i-3)!\cdot i!}$$

I want find the $A_{n}$ recursive relations,such as following form $$A_{n}=B_{n}+C_{n}A_{n-1}+D_{n}A_{n-2}+\cdots$$ I have try sometimes,and can't find it,can you help me?

This question has been asked on MSE here without receiving any answers.

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  • $\begingroup$ By general theory, your sequence has to satisfy a recurrence of the desired form where the coefficients are polynomials in $n$. Let $a_n=(2n+3)!A_{n+3}$. See next comment... $\endgroup$ – Michael Stoll Dec 20 '14 at 12:36
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    $\begingroup$ Then experimentally, $2^4 (n+2) (n+3) (2n+3) (2n+7) (27296942n+96266479) a_n +$ $2^3 (n+3) (1373740004n^4 + 11974782476n^3 + 36773573779n^2 + 46446616891n +$ $19095420600) a_{n+1} - 2^2 (n+4) (1258873802n^4 + 12734573713n^3 +$ $46301208703n^2 + 68411873198n + 30162171666) a_{n+2} + 2 (n+5) (1272271211n^4 + 21664023114n^3 + 138039728785n^2 + 392255730894n +$ $423165299544) a_{n+3} - (n+6) (169792631n^4 + 623043934n^3 - 19045733677n^2 -$ $151948512276n - 311659470468) a_{n+4} -$ $3 (n+7) (3n+19) (3n+20) (3057095n + 14424538) a_{n+5} = 0$ for all $n \ge 0$. $\endgroup$ – Michael Stoll Dec 20 '14 at 12:40
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    $\begingroup$ See the book "A=B" by Petkovsek, Wilf and Zeilberger for background and ways of deriving such recurrences in concrete cases: math.upenn.edu/~wilf/AeqB.html. $\endgroup$ – Michael Stoll Dec 20 '14 at 12:43
  • $\begingroup$ +1 for bringing people's attention to that brilliant (though, alas, so far totally useless for my favorite kind of "formal algebra" questions) text again. $\endgroup$ – fedja Dec 20 '14 at 13:19
  • $\begingroup$ A small correction: the coefficient of $a_{n+5}$ has a factor $(n+7)^2$ instead of $n+7$. $\endgroup$ – Michael Stoll Dec 20 '14 at 15:38

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