3.C0=0 C1=1

Continuing from the equation

C_{i, j + 1} = \sum_{k} C_{k, j} {\hat{B}}_{i, k} (C)

C_{i, j + 1} - C_{1, j} C_{i, j} = \sum_{k = 1}^{i - 1} C_{k, j} {\hat{B}}_{i, k}

Taking $C_{1} = 1$ , it becomes

C_{i, j + 1} - C_{i, j} = \sum_{k = 1}^{i - 1} C_{k, j} {\hat{B}}_{i, k}

We can list some $C_{i, j}$ :

$i$	$C_{i, j}$
1	1
2	$C_{2} j$
3	$C_{2}^{2} j^{2} + (C_{3} - C_{2}^{2}) j$

As we can see, they are polynomials. We can thus guess the model

C_{i, j} = \sum_{k = 0}^{i - 1} A_{i, k} j^{k}

Trivial properties:

$A_{1, 0} = 1$
$\sum_{k = 0}^{i - 1} A_{i, k} = C_{i}$

We also have

C_{i, j + 1} = \sum_{k = 0}^{i - 1} A_{i, k} (j + 1)^{k} = \sum_{k = 0}^{\infty} j^{k} \sum_{n = 0}^{i - 1} (\begin{matrix} n \\ k \end{matrix}) A_{i, n} .

Injecting it, we now have

\sum_{k = 0}^{\infty} j^{k} \sum_{n = 0}^{i - 1} (\begin{matrix} n \\ k \end{matrix}) A_{i, n} - \sum_{k = 0}^{i - 1} A_{i, k} j^{k} = \sum_{k = 1}^{i - 1} \sum_{n = 0}^{k - 1} A_{k, n} j^{n} {\hat{B}}_{i, k}

\sum_{n = 0}^{i - 1} (\begin{matrix} n \\ k \end{matrix}) A_{i, n} - A_{i}, k = \sum_{n = 0}^{i - 1} A_{n, k} {\hat{B}}_{i, n}

\sum_{n = k + 1}^{i - 1} (\begin{matrix} n \\ k \end{matrix}) A_{i, n} = \sum_{n = k + 1}^{i - 1} A_{n, k} {\hat{B}}_{i, n}

Let $n$ such that $k + 1 = i - N$

\sum_{n = i - N}^{i - 1} (\begin{matrix} n \\ i - N - 1 \end{matrix}) A_{i, n} = \sum_{n = i - N}^{i - 1} A_{n, i - N - 1} {\hat{B}}_{i, n}

(i - N) A_{i, i - N} - (i - 1) C_{2} A_{i - 1, j - 1 - N} = \sum_{n = 0}^{N - 2} A_{n + i - N, i - N - 1} {\hat{B}}_{i, n + i - N} - (\begin{matrix} n + i - N + 1 \\ i - N - 1 \end{matrix}) A_{i, n + i - N + 1}

We can do a change of variable such that

A_{i, i - N} = \frac{(i - 1)!}{(i - N)!} C_{2}^{i} V_{N} (i)

Injecting it in, leads, after some computation, to

V_{N} (i) - V_{N} (i - 1) = \sum_{n = 0}^{N - 2} {\hat{B}}_{i, n + i - N} \frac{(n + i - N - 1)!}{(i - 1)!} C_{2}^{n - N} V_{n + 1} (n + i - N) - \frac{V_{N - n - 1} (i)}{(n + 2)!}

V_{N} (n) = \sum_{i = N + 1}^{n} \sum_{k = 0}^{N - 2} {\hat{B}}_{i, k + i - N} \frac{(k + i - N - 1)!}{(i - 1)!} C_{2}^{k - N} V_{k + 1} (k + i - N) - \frac{V_{k + 1} (i)}{(N - k)!}

V_{1} (n) = \frac{1}{C_{2}}

V_{2} (n) = C_{2}^{- 3} (C_{3} - C_{2}^{2}) (H_{n} - 1)

with $H_{n} = \sum_{i = 1}^{n} \frac{1}{i}$ , the harmonic numbers. $V_{3} (n)$ is already quite complicated and long to be shown here. Let $Q_{N} (n) = V_{N} (n) C_{2}^{N + 1}$ ,

Q_{N} (n) = \sum_{i = N + 1}^{n} \sum_{k = 0}^{N - 2} {\hat{B}}_{i, k + i - N} \frac{(k + i - N - 1)!}{(i - 1)!} C_{2}^{- 1} Q_{k + 1} (k + i - N) - \frac{C_{2}^{N - k - 1}}{(N - k)!} Q_{k + 1} (i)

Now let

Q_{N} (n) = \sum_{\sum_{i} (i - 1) j_{i}} U (j_{i}, n) \prod_{i \geq 2} C_{i}^{j_{i}}

Injecting it gives

\begin{matrix} \sum_{\sum_{i - 1} j_{i} = N} U (j_{r}, n) C_{2} \prod_{i \geq 2} C_{i}^{j_{i}} = \\ \sum_{i = N + 1}^{n} \sum_{k = 0}^{N - 2} (\sum_{\begin{matrix} \sum (j - 1) a_{j} = N - k \\ \sum (j - 1) b_{j} = k + 1 \end{matrix}} \frac{(k + i - N - 1)! (k + i - N)!}{(i - 1)! \prod_{r = 1}^{N - k + 1} a_{r}!} U (b_{i}, k + i - N) \prod_{r \geq 2} C_{r, a_{r} + b_{r}} - \sum_{\sum (j - 1) d_{j} = k + 1} \frac{U (d_{r}, i)}{(N - k)!} C_{2}^{N - k} \prod_{r \geq 2} C_{r}^{d_{r}}) \end{matrix}

Identifying terms, we get

U (j_{i}, n) = \sum_{m = N + 1}^{n} \sum_{k = 0}^{N - 2} (- \frac{U ({j_{2} + 1 + k - N; j_{3}; j_{4}; \dots}, m)}{(N - k)!} + \sum_{\begin{matrix} \sum (i - 1) a_{1} = N - k \\ \sum (i - 1) b_{i} = k + 1 \\ {\begin{cases} a_{i} + b_{i} = j_{i}, \forall i > 2 \\ a_{2} + b_{2} = j_{2} + 1 \end{cases} \end{matrix}} (\frac{(k + m - N)! (k + m - N - 1)!}{(m - 1)! (m + k - N - \sum_{r \geq 2} a_{r})!} \frac{U (b_{r}, k + m - N)}{\prod_{r \geq 2} a_{r}!}))

To simplify formulas, let shift $j$ 's by 1 such that $j_{2}$ is now noted as $j_{1}$ . The fomula then becomes

U (j_{i}, n) = \sum_{m = N + 1}^{n} \sum_{k = 0}^{N - 2} (- \frac{U ({j_{1} + 1 + k - N; j_{2}; \dots}, m)}{(N - k)!} + \sum_{\begin{matrix} \sum i b_{i} = k + 1 \\ a_{i} + b_{i} = j_{i}, \forall i : 1 < i < N \\ a_{N} = j_{N} \\ a_{1} + b_{1} = j_{1} + 1 \\ a_{1} \geq 0, \forall i \\ b_{i} \geq 0, \forall i > 1 \end{matrix}} (\frac{(k + m - N)! (k + m - N - 1)!}{(m - 1)! (m + k - N - \sum a_{r})!} \frac{U (b_{r}, k + m - N)}{\prod a_{r}!}))

Some properties: Let $J (N) = {j_{i} = 1 if i = N; j_{i} = 0 else}$ , we have

U (J (N), n) = \frac{1}{N - 2} (\frac{1}{(N - 1)!} - \frac{(n - N + 1)!}{(n - 1)!})

U ({N}, n) = \sum_{m = N + 1}^{n} \sum_{k = 0}^{N - 2} \frac{1}{(N - k)!} (\frac{(k + m - N)! (k + m - N - 1)!}{(m - 1)! (m + 2 k - 2 N)!} U ({k + 1}, k + m - N) - U ({k + 1}, m))

U ({2 - N; N - 1}, n) \sum_{m = 2 N - 2}^{n - 1} \frac{U ({3 - N; N - 2}, m - 1)}{m}

From this last functions, we will define

F_{N} (n) = \sum_{i = 2 N - 2}^{n - 1} \frac{F_{N - 1} (i - 1)}{i}

with initial condition

F_{1} (n) = 1

$J$	$U (J, n)$
{0;1}	$F_{2} (n)$
{2}	$- F_{2} (n)$
{-1;2}	$F_{3} (n)$
{3}	$F_{3} (n) - \frac{3}{2} \frac{1}{n - 1} + \frac{3}{4}$
{1,1}	$\frac{5}{2} \frac{1}{n - 1} - \frac{5}{4} - 2 F_{3} (n)$
{4}	$\frac{29}{72} - \frac{7}{12} \frac{1}{n - 2} - \frac{25}{12} \frac{1}{n - 1} + \frac{3}{2} (\frac{1}{n - 1} - \frac{1}{2}) F_{2} (n - 2) - F_{4} (n)$
{1,0,1}	$\frac{1}{6} - \frac{2}{n - 1} + \frac{1}{n - 2} + F_{2} (n - 2) (\frac{1}{n - 1} - \frac{1}{2})$
{-1;1;1}	$- \frac{5}{12} + \frac{1}{2} (\frac{1}{n + 1} + \frac{1}{n + 2}) + F_{2} (n - 2) (\frac{1}{2} - \frac{1}{n - 1})$
{2;1}	$3 F_{4} (n) + F_{2} (n - 2) 4 (\frac{1}{2} - \frac{1}{n - 1}) + \frac{7}{2} \frac{1}{n - 2} + \frac{1}{2} \frac{1}{n - 1} - \frac{83}{72}$
{0;2}	$\frac{11}{12} - \frac{2}{n - 1} - \frac{1}{2} \frac{1}{n - 2} + \frac{5}{2} (\frac{1}{n - 1} - \frac{1}{2}) F_{2} (n - 2) - 3 F_{4} (n)$

We can guess the form as

U (j_{r}, n) = \sum_{l = 0}^{N - 1} K_{l} ({j_{r}}, n) F_{N - l} (n - l)

Through some shady techniques, we can find

K_{0} (j_{1}, j_{2}) = (- 1)^{j_{1} + j_{2} + 1} (\begin{matrix} j_{1} + 2 j_{2} - 1 \\ j_{2} \end{matrix})

K_{1} (j_{1}, j_{2}) = 0

K_{2} (j_{1}, j_{2}) = (\frac{1}{2} - \frac{1}{n - 1}) (- 1)^{j_{1} + j_{2}} ((\begin{matrix} j_{1} + 2 j_{2} - 3 \\ j_{2} \end{matrix}) - \frac{5}{2} (\begin{matrix} j_{1} + 2 j_{2} - 2 \\ j_{2} \end{matrix}))

This is actually only true if $j_{i} = 0 \forall i > 2$ as all of those case, create "parralel" solutions, which should respect similar recurence equations but with different initial conditions.

The form should be injected in the equation to find general solutions but the computations is very complicated and I was not able to get through with it at the moment.