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Feb 3, 2016 at 1:25 comment added Robert Israel i.e. multiply Tony's equation by $x^n$, sum for $n=1 \ldots \infty$ (interchanging the order of summations) and add $1$, and you should get $$g(x) = \dfrac{x}{1-4x} + \dfrac{x g(x)}{1-x} + \dfrac{x^2 g(x)}{1-5x+4x^2}$$ where $g(x)$ is the g.f. Solve: $$ g(x) = \dfrac{1-4x+3x^2}{1-6x+7x^2}$$
Feb 2, 2016 at 22:17 comment added Robert Israel You should be able to get the g.f. from Tony Huynh's recursion equation, and it follows from that.
Feb 2, 2016 at 21:53 comment added Gerry Myerson At OEIS, it says $a_n=6a_{n-1}-7a_{n-2}$. I wonder how one gets that?
Feb 2, 2016 at 16:47 comment added Robert Israel Sorry, mistake in my program, corrected. Actually you want $f(2,2)=8$.
Feb 2, 2016 at 16:46 history undeleted Robert Israel
Feb 2, 2016 at 16:46 history edited Robert Israel CC BY-SA 3.0
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Feb 2, 2016 at 16:41 history deleted Robert Israel via Vote
Feb 2, 2016 at 16:36 history edited Robert Israel CC BY-SA 3.0
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Feb 2, 2016 at 16:31 history answered Robert Israel CC BY-SA 3.0