1.The recursion formula implies even integers only.

2. The 2012 limit causes the candidates for a(2) not to exceed 10.

3. Therefore there are 5 solutions i.e. (a(2),a(5))= ((2,2),(4,32),(6,162),(8,512),(10,1250)).

4. Interesting fact : if a(2)=2 than all members are equal to 2, sinceeach of the members is 2^2/2.

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