Recursion Puzzle(solution)

[Leroy Quet]

>I posted this recursion puzzle on sci.math and rec.puzzles a couple of 
>days ago. I was debating with myself whether to post it to this group, 
>however. But I decided to go ahead and indeed post it.
>(Perhaps it will be fun for some of you to try and solve it.)
>
>---
>
>The following recursion generates a sequence of positive integers. (The 
>sequence may or may not be in the EIS.) -- The fact that the sequence 
>contains all rationals, let alone positive integers, would not be obvious 
>to ME just from looking at the recursion.
>
>
>Let a(0) = 1;
>a(1) = 2;
>a(2) = 5;
>
>and, for m >= 3,
>
>a(m) = 
>
>(1/4)*( (3*a(m-2)^2 *b(m) -9*a(m-2)^3)/a(m-3)^2
>
> +(13*a(m-1)*a(m-2) -8*a(m-2)^2 +2*a(m-2)*b(m) -3*a(m-1)b(m))/a(m-3)
>
> +8*a(m-1) -2*a(m-2) ),
>
>where b(m) = 
>
>sqrt(9*a(m-2)^2 +4*a(m-3)*(a(m-2) -2*a(m-1))).
>
>Find a closed form, in terms of a partial sum, for the sequence of a(m)'s.
>
>(I'll give my answer in a few days if no one else gets it sooner.)

Solution:

a(m)  = 

sum{k=0 to m}  k! * (m-k)!

(Looking at the recursion, it is not obvious to ME that each term of
the sequence is rational, let alone a positive integer.)

---
I can't remember every step I used to get the recursion, but I started
with the identity:

a(m) = m! + a(m-1) *(m+1)/2.

I used this to get that 
m! = (1/4)(a(m) + sqrt(9*a(m)^2 + 4*a(m-1)*(a(m) -2*a(m+1))).

A BUNCH of Algebra (with the help of Mathematica), and I got my
original recursion.

Thanks,
Leroy Quet