<s>[INST] An expression for $U_{h,0}$ given $U_{n,k}=\frac{c^n}{c^n-1}(U_{n-1,k+1})-\frac{1}{c^n-1}(U_{n-1,k})$ Let $c\in\mathbb{R}\setminus\{ 1\}$, $c>0$.
Let $U_i = \left\lbrace U_{i, 0}, U_{i, 1}, \dots \right\rbrace$, $U_i\in\mathbb{R}^\mathbb{N}$.
We know that $U_{n+1,k}=\frac{c^{n+1}}{c^{n+1}-1}U_{n,k+1}-\frac{1}{c^{n+1}-1}U_{n,k}$.
(As @TedShifrin pointed out, it can also be written $U_{n+1,k}=U_{n,k+1}+\frac{1}{c^{n+1}-1}\left(U_{n,k+1}-U_{n,k}\right)$)
(obviously it implies that if $\lvert U_k \rvert=n$ then $\lvert U_{k+1}\rvert=n-1$ etc)
Here is what I conjectured:
$$\forall h\in\mathbb{N}, 
U_{h,0}=\sum\limits_{p=0}^h\frac{c^{\frac{p^2+p}{2}}}{\left(\prod\limits_{i=1}^p\left(c^i-1\right)\right)\prod\limits_{i=1}^{h-p}\left(1-c^i\right)}U_{0,p}$$
I tested it for some values ($0,1,3,4$) and it seemed to work.
Mathematica also verified it up to at least 10.
What do you think? If it IS true, how can I prove it?

Example with $c=4$, $h=2$ :
$U_{1,0}=\frac{4}{3}U_{0,1}-\frac{1}{3}U_{0,0}$, $U_{1,1}=\frac{4}{3}U_{0,2}-\frac{1}{3}U_{0,1}$
$U_{2,0}=\frac{16}{15}U_{1,1}-\frac{1}{15}U_{1,0}$
Therefore $U_{2,0}=\frac{16}{15}\left(\frac{4}{3}U_{0,2}-\frac{1}{3}U_{0,1}\right)-\frac{1}{15}\left(\frac{4}{3}U_{0,1}-\frac{1}{3}U_{0,0}\right)$
$U_{2,0}=U_{0,0}\left(\frac{1}{15\times 3}\right)-U_{0,1}\left(\frac{16}{15\times 3}+\frac{4}{15\times 3}\right)+U_{0,2}\left(\frac{16\times 4}{15\times 3}\right)$
$U_{2,0}=U_{0,0}\left(\frac{1}{45}\right)-U_{0,1}\left(\frac{4}{9}\right)+U_{0,2}\left(\frac{64}{45}\right)$
The formula gives us :
$U_{2,0}=U_{0,0}\left(\frac{4^0}{(1-4)(1-16)}\right)+U_{0,1}\left(\frac{4^1}{(4-1)(1-4)}\right)+U_{0,2}\left(\frac{4^3}{(16-1)(4-1)}\right)$
$U_{2,0}=U_{0,0}\left(\frac{1}{45}\right)+U_{0,1}\left(\frac{-4}{9}\right)+U_{0,2}\left(\frac{64}{45}\right)$
 [/INST] Here’s what I hope is a good start. I’m stuck at the very end, but maybe you can finish it up. (It is homework, after all.) The special functions experts might have a quicker way to answer this, I imagine.
For $h=0$, the desired result is exactly what “we know.”
Use induction, and assume the result for $h<M$.
$$
\begin{align}
U_{M,0}&=\frac{c^M}{c^M-1}(U_{M-1,1})-\frac{1}{c^M-1}(U_{M-1,0})\\
&=\frac{c^M}{c^M-1}(U_{M-1,1})-
\frac{1}{c^M-1}\sum\limits_{p=0}^{M-1}\frac{c^{\frac{p^2+p}{2}}}{\left(\prod\limits_{i=1}^p\left(c^i-1\right)\right)\prod\limits_{i=1}^{{M-1}-p}\left(1-c^i\right)}U_{0,p}\\
&=\frac{c^M}{c^M-1}(U_{M-1,1})+\frac{1}{1-c^M}\sum\limits_{p=0}^{M-1}\frac{c^{\frac{p^2+p}{2}}}{\left(\prod\limits_{i=1}^p\left(c^i-1\right)\right)\dfrac{\prod\limits_{i=1}^{{M}-p}\left(1-c^i\right)}{1-c^{M{\color{red}{-p}}}}}U_{0,p}
\\
&\mbox{*** Earlier answer omitted the part in red ***}\\
&=\frac{c^M}{c^M-1}(U_{M-1,1})+\frac{1}{1-c^M}\sum\limits_{p=0}^{M-1}\frac{c^{\frac{p^2+p}{2}}}{\left(\prod\limits_{i=1}^p\left(c^i-1\right)\right)\dfrac{\prod\limits_{i=1}^{{M}-p}\left(1-c^i\right)}{1-c^{M{-p}}}}U_{0,p}
\\
&=\frac{c^M}{c^M-1}(U_{M-1,1})+\frac{1}{\prod\limits_{i=1}^M (1-c^i)}+\frac{1}{1-c^M}\sum\limits_{p=1}^{M-1}\frac{c^{\frac{p^2+p}{2}}}{\prod\limits_{i=1}^p\left(c^i-1\right)\prod\limits_{i=1}^{{M}-1-p}\left(1-c^i\right)}U_{0,p}.
\end{align}$$
The proof will be complete if you can show that this equals
$$\sum\limits_{p=0}^M\frac{c^{\frac{p^2+p}{2}}}{\left(\prod\limits_{i=1}^p\left(c^i-1\right)\right)\prod\limits_{i=1}^{{M}-p}\left(1-c^i\right)}U_{0,p},$$
which after correcting my earlier error, I have no idea how to do. Another approach seems needed.
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