Guess (or upper bound) the general formula for a double sequence Let $t,s \geq 0$ be integers. We have the following recursive formula:
$$f(t+1,s) = f(t,s) + f(t,s-1) + \sum_{0\leq a,b,c \leq h(t):\\a+b+c = s-1}f(t,a)f(t,b)f(t,c),$$ where
$$h(t) = \frac{1}{2}3^t -\frac{1}{2}.$$
The boundary conditions are given by $f(t,0) = 1$ for all $t \geq 0$ and $f(0,s) = 0$ for all $s \geq 1$.
It is not hard to see that 
$$
f(t,1) = 2t
$$
and 
$$
f(t,2) = 4t(t-1).
$$
I wonder what is the general formula for $f(t,s)$. It this is difficult to obtain, then what is a good upper bound for $f(t,s)$?
 A: First notice that $h(t)\geq t$ and $h(t+1)=3h(t)+1$ for all $t\geq 0$. Now, for $s>h(t+1)$, the recurrent formula reduces to
$$f(t+1,s) = f(t,s) + f(t,s-1),$$
which further implies that for all $t\geq 0$ and $s>h(t)$ we have $f(t,s) = 0$.
It follows that the restriction $a,b,c\leq h(t)$ in the recurrent formula is inessential and can be dropped. The simplified formula implies a recurrent formula for the generating function:
$$F_t(x) := \sum_{s=0}^{\infty} f(t,s)x^s.$$
Namely, we have $F_0(x) = 1$, and for $t\geq 0$ 
$$F_{t+1}(x) = (1+x)F_t(x) + xF_t(x)^3.$$
In particular,
$$F_1(x) = (1+x)1+x1^3=1+2x,$$
$$F_2(x) = (1+x)(1+2x) + x(1+2x)^3 = 1+  4x+   8x^2+   12x^3+     8x^4,$$
etc.
UPDATED. I doubt there is a simple formula for $F_t(x)$, although we can get those for some particular values of $x$ such as
$$F_t(0) = 1,$$
$$F_t(-1) = (-1)^{t},$$
and
\begin{split}
F_t(2) &= \sqrt{2} \sinh(3^t\mathrm{arcsinh}(1/\sqrt{2}))\\
& = \frac{1+\sqrt{3}}2 \alpha^{h(t)} + \frac{1-\sqrt{3}}2 \alpha^{-h(t)},
\end{split}
where $\alpha:=2+\sqrt{3}\approx 3.732$. The values $F_t(2)$ are given by the sequence A238799 in the OEIS.
From the last formula it follows that $f(t,s)< \frac{1+\sqrt{3}}{2^{s+1}} \alpha^{h(t)}$.
