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	\(C_2j\)
3	\(C_2^2j^2+(C_3-C_2^2)j\)

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

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

Injecting it, we now have

\[ \sum_{k=0}^\infty j^k\sum_{n=0}^{i-1} \begin{pmatrix}n\\k\end{pmatrix}A_{i,n}-\sum_{k=0}^{i-1}A_{i,k}j^k=\sum_{k=1}^{i-1}\sum_{n=0}^{k-1}A_{k,n}j^n\hat B_{i,k} \]

\[ \sum_{n=0}^{i-1}\begin{pmatrix}n\\k\end{pmatrix}A_{i,n}-A_i,k=\sum_{n=0} ^{i-1}A_{n,k}\hat B_{i,n} \]

\[ \sum_{n=k+1}^{i-1}\begin{pmatrix}n\\k\end{pmatrix}A_{i,n}=\sum_{n=k+1} ^{i-1}A_{n,k}\hat B_{i,n} \]

Let \(n\) such that \(k+1=i-N\)

\[ \sum_{n=i-N}^{i-1}\begin{pmatrix}n\\i-N-1\end{pmatrix}A_{i,n}=\sum_{n=i-N} ^{i-1}A_{n,i-N-1}\hat B_{i,n} \]

\[ (i-N)A_{i,i-N}-(i-1)C_2A_{i-1,j-1-N}=\sum_{n=0}^{N-2}A_{n+i-N,i-N-1}\hat B_{i,n+i-N}-\begin{pmatrix}n+i-N+1\\i-N-1\end{pmatrix}A_{i,n+i-N+1} \]

We can do a change of variable such that

\[ A_{i,i-N}=\frac{(i-1)!}{(i-N)!}C_2^iV_N(i) \]

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 \(H_n = \sum_{i=1}^n \frac 1i\), 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

\[ Q_N(n)=\sum_{\sum_i(i-1)j_i}U(j_i,n)\prod_{i\ge2}C_i^{j_i} \]

Injecting it gives

\[ \displaylines{ \sum_{\sum_{i-1}j_i=N}U(j_r,n)C_2\prod_{i\ge 2}C_i^{j_i}=\\ \sum_{i=N+1}^n\sum_{k=0}^{N-2}\left(\sum_{\begin{matrix}\sum(j-1)a_j=N-k\\\sum(j-1)b_j=k+1\end{matrix}}\frac{(k+i-N-1)!(k+i-N)!}{(i-1)!\prod_{r=1}^{N-k+1}a_r!}U(b_i,k+i-N)\prod_{r\ge2}C_{r,a_r+b_r}\\ -\sum_{\sum(j-1)d_j=k+1}\frac{U(d_r,i)}{(N-k)!}C_2^{N-k}\prod_{r\ge2}C_r^{d_r} \right) } \]

Identifying terms, we get

\[ U(j_i,n)=\sum_{m=N+1}^n\sum_{k=0}^{N-2}\left(-\frac{U(\{j_2+1+k-N;j_3;j_4;\ldots\},m)}{(N-k)!}+\sum_{\begin{matrix}\sum(i-1)a_1=N-k\\\sum(i-1)b_i=k+1\\\left\{ \begin{array}{ll} a_i+b_i=j_i, \forall i>2\\ a_2+b_2=j_2+1 \end{array} \right.\end{matrix}} \left(\frac{(k+m-N)!(k+m-N-1)!}{(m-1)!(m+k-N-\sum_{r\ge2}a_r)!}\frac{U(b_r,k+m-N)}{\prod_{r\ge2}a_r!}\right)\right) \]

To simplify formulas, let shift \(j\)'s by 1 such that \(j_2\) is now noted as \(j_1\). The fomula then becomes

\[ U(j_i,n)=\sum_{m=N+1}^n\sum_{k=0}^{N-2}\left(-\frac{U(\{j_1+1+k-N;j_2;\ldots\},m)}{(N-k)!}+\sum_{\begin{matrix}\sum ib_i=k+1\\a_i+b_i=j_i,\forall i:1<i<N\\a_N=j_N\\a_1+b_1=j_1+1\\a_1\ge0,\forall i\\b_i\ge0, \forall i>1\end{matrix}} \left(\frac{(k+m-N)!(k+m-N-1)!}{(m-1)!(m+k-N-\sum a_r)!}\frac{U(b_r,k+m-N)}{\prod a_r!}\right)\right) \]

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

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

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

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

From this last functions, we will define

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

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}	\(F_3(n)-\frac 32 \frac 1{n-1}+\frac 34\)
{1,1}	\(\frac 52 \frac 1{n-1} -\frac 54 -2 F_3(n)\)
{4}	\(\frac{29}{72}-\frac{7}{12}\frac 1{n-2}-\frac{25}{12}\frac{1}{n-1}+\frac 32 \left(\frac 1{n-1}-\frac 12\right) F_2(n-2)-F_4(n)\)
{1,0,1}	\(\frac 16 -\frac 2{n-1} +\frac 1{n-2}+F_2(n-2)\left(\frac 1{n-1}- \frac 12\right)\)
{-1;1;1}	\(-\frac{5}{12}+\frac 12 \left(\frac{1}{n+1}+\frac{1}{n+2}\right)+F_2(n-2)\left(\frac12 -\frac 1{n-1}\right)\)
{2;1}	\(3F_4(n)+F_2(n-2)4\left(\frac 12 -\frac{1}{n-1}\right)+\frac 72 \frac{1}{n-2}+\frac 12 \frac{1}{n-1}-\frac{83}{72}\)
{0;2}	\(\frac{11}{12}-\frac 2{n-1}-\frac 12\frac{1}{n-2}+\frac 52 \left(\frac 1{n-1}-\frac 12\right)F_2(n-2)-3F_4(n)\)

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

\[ K_0(j_1,j_2)=(-1)^{j_1+j_2+1}\begin{pmatrix}j_1+2j_2-1\\j_2\end{pmatrix} \]

\[ K_1(j_1,j_2)=0 \]

\[ K_2(j_1,j_2)=\left(\frac 12 -\frac 1{n-1}\right)(-1)^{j_1+j_2}\left(\begin{pmatrix}j_1+2j_2-3\\j_2\end{pmatrix}-\frac 52 \begin{pmatrix}j_1+2j_2-2\\j_2\end{pmatrix}\right) \]

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.