I have a question regarding a 2D sequence matrix equation defined as:
- $v_{i,j} = 2(i-j)(2(i-j)-1)v_{i,j-1} - 4(i-j-1)^2 v_{i-1,j-1}$
- $v_{0,0} = -1$
- and $\forall k \neq 0, v_{0,k} = 0$
- $v_{i,j} = 0$ if $i < 0$ or $j<0$
I solved this numerically a couple of times using algorithms, and I tried to guess a form.
I noticed $v_{i,j}/ \Gamma(2(i-j))$ seems to be a polynomial (at a fixed row) in terms of i but really high order.
What I want to do is to find $v_{i,j}$ analytically as a closed formula. So far, I have reached the following conclusions using recursion, $v_{i,0} = −(2i + 1)!$ and $v_{i,1} = \dfrac{(2i + 1)!}{3} (2i(i+1)(2i+1))$ and $v_{i,2}$ is a really high-order polynomial.
Matrix $M$:
Scheme of the recursion:
Also the diagonal is $-(-4)^k$. It looks a bit like Pascal's triangle, it might be involved? Mind helping me, please?
Lemma: If $\{c_i\}$ are distinct constants and $\{u(i, j)\}$ satisfy\begin{gather*} u(i, 0) = 1\ (i \geqslant 0),\ u(-1, j) = 0\ (j \geqslant 0),\\ u(i, j) = u(i - 1, j) + c_i u(i, j - 1)\ (i \geqslant 0,\ j \geqslant 1), \end{gather*} then\begin{gather*} u(i, j) = \sum_{k = 0}^i \frac{c_k^{i + j}}{\prod_{\substack{0 \le l \le i\\l ≠ k}} (c_k - c_l)}. \quad \forall i, j \geqslant 0 \tag{1} \end{gather*}
Proof: It can be calculated by the boundary condition that $u(0, j) = c_0^j$ ($j \geqslant 0$). For $i \geqslant 0$, using Lagrange polynomials to interpolate the function $y = x^i$ yields$$ \sum_{k = 0}^i c_k^i \cdot \frac{\prod_{\substack{0 \le l \le i\\l ≠ k}} (x - c_l)}{\prod_{\substack{0 \le l \le i\\l ≠ k}} (c_k - c_l)} \equiv x^i, $$ and comparing the coefficient of $x^i$ yields$$ \sum_{k = 0}^i \frac{c_k^i}{\prod_{\substack{0 \le l \le i\\l ≠ k}} (c_k - c_l)} = 1 = u(i, 0). $$ Then (1) can be proved by induction on $i$ and $j$.
Now back to the question. The recurrence relation to be solved is\begin{gather*} v(0, 0) = -1,\ v(i, 0) = 0\ (i \geqslant 1),\ v(-1, j) = 0\ (j \geqslant 0),\\ v(i, j) = 2(i - j)(2(i - j) - 1) v(i, j - 1) - 4(i - j - 1)^2 v(i - 1, j - 1)\ (i \geqslant 0,\ j \geqslant 1). \end{gather*} (This is the most likely what you mean based on your expressions for “$v(i, 0)$” and “$v(i, 1)$” since there seems to be a few typos in the indices.) It is easy to prove that $v(i, j) = 0$ for $i > j$. Define\begin{gather*} u(-1, j) = 0\ (j \geqslant 0),\\ u(i, j) = \frac{(-1)^{j + 1}}{4^j \cdot (2i + 1)!} \cdot v(j, i + j)\ (i, j \geqslant 0), \end{gather*} then it is not hard to prove that $\{u(i, j)\}$ satisfy the conditions in the lemma with $c_i = (i + 1)^2$. By the lemma,$$ u(i, j) = \sum_{k = 0}^i \frac{(k + 1)^{2(i + j)}}{\prod_{\substack{0 \le l \le i\\l ≠ k}} ((k + 1)^2 - (l + 1)^2)}, \quad \forall i, j \geqslant 0 $$ thus for $i \leqslant j$,\begin{align*} v(i, j) &= (-1)^{i + 1} 4^i \cdot (2(j - i) + 1)! \cdot u(j - i, i)\\ &= (-1)^{i + 1} 4^i \cdot (2(j - i) + 1)! \sum_{k = 0}^{j - i} \frac{(k + 1)^{2j}}{\prod_{\substack{0 \le l \le j - i\\l ≠ k}} ((k + 1)^2 - (l + 1)^2)}\\ &= (-1)^{i + 1} 4^i \cdot (2(j - i) + 1)! \sum_{k = 0}^{j - i} \frac{(k + 1)^{2j}}{\dfrac{(-1)^{j - i - k}}{2(k + 1)^2} \cdot (j - i - k)!\, (j - i + k + 2)!}\\ &= \frac{4^i}{j - i + 1} \sum_{k = 0}^{j - i} (-1)^{j - k + 1} (k + 1)^{2j + 2} C(2(j - i) + 2, j - i - k). \end{align*}