I have a particular triangular array
$$ \begin{matrix} 1 & \\ 1 & 1 \\ 1 & 2 & 3\\ 1 & 3 & 9 & 15\\ 1 & 4 & 18 & 60 & 105\\ 1 & 5 & 30 & 150 & 525 & 945\\ \end{matrix} $$ The $T(i,j)$th entry is found by the following recursive formula:
$$(2k+1)T(n,k)+T(n,k+1)=T(n+1,k+1)$$
To read, $T(2,1)=2$. Here I am taking $n$ as the row and $k$ as the column. The first entry is the "$0$"th row and "$0$"th column. Also note that $T(n,k)=0$ if $n<k$
I tried looking on the OEIS website for combinations of these numbers and didn't have much success. What would be a method for finding a closed form for this type of recursive triangle? I know that the number $T(n,n)=(2n-1)!!$ where $n!!$ is the (in this case odd) double factorial with the trivial case $T(0,0)=1$
Has anyone ever seen this particular triangle or know of any particular sequences of numbers found here?
EDIT:
As per Ross Milikan's suggestion, looking at the number 60. 60 can be obtained by multiplying 5 and 9 because $T(3,2)=9$ and $=2\cdot 2+1$. Then add $45$ and $15$, since $T(3,3)=15$. Thus, $$60 = T(4,3) = (2\cdot 2+1)T(3,2)+T(3,3)=5\cdot 9+15$$
I was staring at the problem last night and I got inspired. Using Pascal's Triangle,
$$ \begin{matrix} 1 \\ 1 & 1\\ 1 & 2 & 1 \\ 1 & 3 & 3 & 1 \\ 1 & 4 & 6 & 4 & 1\\ 1 & 5 & 10 & 10 & 5 & 1\\ \end{matrix} $$
Looking at column 2, for example we have $(3,9,18,30,...)$. Taking out a $3$ yields $(1,3,6,10,...)$ or the 2nd column of Pascal's Triangle.
Looking at column 3 we have $(15, 60, 150, 300,...)$. Taking out a $15$ yields $(1,4,10,20,...)$ or the third column of Pascal's Triangle.
So looking back at the triangle generated by the above recursion, $$(2k+1)T(n,k)+T(n,k+1)=T(n+1,k+1)$$ it appeared that if we multiplied the $T(n,n)=(2n-1)!!$ by the appropriate entry from Pascal's we achieve success. Thus
$$T(n,k)=\binom{n}{k}(2k-1)!!$$
By plugging in this to the LHS we get $$(2k+1)\binom{n}{k}(2k-1)!!+\binom{n}{k+1}[2(k+1)-1]!!$$ $$=\binom{n}{k}(2k+1)!!+\binom{n}{k+1}(2k+1)!!$$ $$=\left[\binom{n}{k}+\binom{n}{k+1}\right](2k+1)!!$$ $$=\binom{n+1}{k+1}(2k+1)!!$$ $$=T(n+1,k+1)$$