Start with [this triangle (OEIS A118981)](https://oeis.org/A118981). This triangle is simple to generate with the following recurrence relation (though $T(0,0)$ ends up different from the OEIS version):
$$
T(0,0) = 2;T(1,0) = 1;T(1,1) = 1;
$$
$$T(n,k) = T(n-1, k) + T(n-1, k-1) + T(n-2, k-2), n>1 $$


The sequence in question is generated by taking the largest element from each row, then dividing it by the row number $n$. The resulting sequence is as follows (listed up to $a\_{20}$):


$1, 1.5, 2, 3, 5, 10, 19, 37, 77.\overline 7, 158, 331, 715, 1505, 3287, 7224.4, 15529, 34839, 77453.8 \overline3, 168950, 386050, ...$


My question is: **Why are the majority of these numbers [convoluted convolved Fibonacci numbers?](https://arxiv.org/pdf/math/0311205.pdf)** Even the ones which are not integers become convoluted convolved Fibonacci numbers when rounded down to the nearest integer. Some examples:


$$floor(a\_{15}) = 7224 = G\_{11}^{(5)}$$
$$a\_{16} = 15529 = G\_{12}^{(5)}$$
$$a\_{17} = 34839 = G\_{12}^{(6)}$$
$$a\_{19} = 168950 = G\_{13}^{(7)}$$
$$a\_{20} = 386050 = G\_{14}^{(7)}$$
$$floor(a\_{21}) = 865264 = G\_{15}^{(7)}$$



Why do convoluted convolved Fibonacci numbers pop up from this triangle?
We have $$T(n,k) = [x^ny^{n-k}]\ \frac{2 - (1+y)x}{1-(1+y)x-x^2}=[y^{n-k}]\ L\_n(1+y),$$
where $L\_n$ is the $n$-th [Lucas polynomial](https://en.wikipedia.org/wiki/Fibonacci_polynomials).


For $k<n$, we have an explicit formula:
\begin{split}
T(n,k) &= \sum\_{i=0}^{\lfloor n/2\rfloor} \frac{n}{n-i}\binom{n-i}i \binom{n-2i}{n-k}\\
&= \frac{n}{n-k}\sum\_{i=0}^k \binom{n-i-1}{k-i} \binom{k-i}i\\
&=\frac{n}{n-k} F\_{k+1}^{(n-k)}.
\end{split}
So, relation to convoluted Fibonacci numbers should come at no surprise.


In the given examples for $G\_{j+1}^{(r)}$ with $3\nmid n$, we have $\gcd(r,j)=1$, implying that $G\_{j+1}^{(r)} = \frac1r F\_{j+1}^{(r)}$.


Numerical evidence suggests that for a fixed $n$, $T(n,k)$ is unimodal and achieves its maximum at $k=n-\left\lfloor \frac{n}3\right\rfloor$, suggesting that
$$\frac1n\max\_k T(n,k) = \frac1{\lfloor n/3\rfloor} F\_{n-\lfloor n/3\rfloor + 1}^{(\lfloor n/3\rfloor)} \stackrel{3\nmid n}{=} G\_{n-\lfloor n/3\rfloor + 1}^{(\lfloor n/3\rfloor)}.$$


The case $3\mid n$ can be addressed similarly.


