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How to solve a recurrence equation with non-constant coefficients? The equation is $$ 120(3k+1)(18k^3-21k-2)(k+2)a_{3k-6}=120(54k^4-117k^2+4)(k-1)a_{3k-5}+(k-1)^2(3k^2-2)^2k^3(k+2)(k+1)(6k^2+6k-1). $$ Here $a_0=7, a_1=42, a_2=189, a_3=708, a_4=2121$. This equation has non-constant coefficient. Are there some general method to solve this equation?

Thank you very much.

Edit: the other two equations are $$ (3k+2)(18k^3+54k^2+33k-1)a_{3k-5} = (3k+1)(18k^3-21k-2)a_{3k-4}+(1/120)k^2(3k^2+6k+1)(k+1)^2(3k^2-2)(k-1)(k+2)(6k^2+6k-1). $$

$$ (120(k+2))(54k^4+216k^3+207k^2-18k-59)a_{3k-4} = (120(k-1))(3k+2)(18k^3+54k^2+33k-1)a_{3k-3}+k^2(k+3)(k+2)(k+1)^2(6k^2+18k+11)(3k^2+6k+1)(k-1)(3k^2-2). $$

I think one method can be as follows: assume that $a_{3k}=\sum_{i=1}^{8} c_ik^i, a_{3k+1}=\sum_{i=1}^{8} d_ik^i, a_{3k+2}=\sum_{i=1}^{8} e_ik^i$ and compute $a_1, \ldots, a_{24}$. Then plugin $k=1, \ldots, 24$ and solve $c_1, \ldots, c_8, d_1, \ldots, d_8, e_1, \ldots, e_8$. But it is complicated. Are there some simpler method?

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Probably the best "general method" is to generate quite a few values, then go to the Encyclopedia of Integer Sequences oeis.org to see if you get a match. –  GEdgar Jun 15 '12 at 15:46
    
@GEdgar, thank you. But there is no such integer sequence on Encyclopedia of Integer sequences. –  LJR Jun 15 '12 at 15:52
    
So, before we work on it ... is it true that the equation only relates two consecutive terms? And gives no information on terms congruent to 2 mod 3? –  GEdgar Jun 15 '12 at 16:05
    
@GEdgar, the other equation is added. –  LJR Jun 15 '12 at 16:13
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OK, now knowing $a_4$ we can get $a_5$, but there is no way to get $a_6$ –  GEdgar Jun 15 '12 at 16:15
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$$ a_{3k}=7+(41202419/960)*k^4-(395482/15)*k+(265589/96)*k^6-(18229213/240)*k^3+(3429283/48)*k^2-(5861/20)*k^7+(6357/320)*k^8-(3320263/240)*k^5 $$

$$ a_{3k+1}=42+(43899229/960)*k^4-(554191/20)*k+(1428773/480)*k^6-(159973/2)*k^3+(18215401/240)*k^2-306*k^7+(7047/320)*k^8-(290869/20)*k^5 $$

$$ a_{3k+2}=189+(15627897/320)*k^4-(286389/10)*k+(513099/160)*k^6-(6676233/80)*k^3+(6480093/80)*k^2-(6333/20)*k^7+(7773/320)*k^8-(1212243/80)*k^5 $$

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