Let $f$ be a continuous function defined on $\mathbb Z\_p$. By Mahler theorem, there exists a sequence $(\gamma\_k)\_{k\in\mathbb N}$ of $\mathbb Z\_p$ such that for every $z\in\mathbb Z\_p$
$$f(z)=\sum\_{k\ge0}\gamma\_k\binom{z}{k},$$
where $\binom{z}{k}=\frac1{n!}\prod\_{j=0}^{k-1}(z-j)$. I try to prove (without any success) that for every $n\in\mathbb N$ there exist two polynomials $P\_n,Q\_n\in\mathbb Q\_p[z]$ with degree $\le n$ such that
$$P\_n(z)f(z)+Q\_n(z)=\sum\_{k\ge 2n+1}\beta\_{k,n}\binom{z}{k}\quad(\beta\_{k,n}\in\mathbb Q\_p)$$
Is this true (known to be true??)
If yes, thanks in advance for any reference or hint for the demonstration.



Twisted Padé approximants
This is basically linear algebra (I assume you want $P\_n$ and $Q\_n$ not both zero or equivalently $P\_n$ not zero) with a minor complication coming from expressing the product of two $\binom{z}{k}$'s as a sum of $\binom{z}{k}$'s.


By expressing $\binom{z}{k}$ in terms of $1$, $\frac{z-j}{j+1}$, $\frac{(z-j)(z-j-1)}{(j+1)(j+2)}$, $\dots$, $\prod\_{l=j}^{j+k-1}{\frac{z-l}{l+1}}$, which is a basis of the vector space of all polynomials of degree $\leq k$, we see that $\binom{z}{k}\binom{z}{j}=\sum\_{l=j}^{j+k}{a\_{j,k,l}\binom{z}{l}}$ for certain rational numbers $a\_{j,k,l}$.


Write $P\_n(z)=\sum\_{k=0}^n{p\_k\binom{z}{k}}$, $Q\_n(z)=\sum\_{k=0}^n{q\_k\binom{z}{k}}$.
Then


$$P\_n(z)f(z)+Q\_n(z)=\sum\_{l=0}^{n}{\left(\sum\_{k=0}^{n}\left(\sum\_{j=\max\{0,l-k\}}^{l}{\gamma\_j a\_{j,k,l}}\right)p\_k+q\_l\right)\binom{z}{l}}+\sum\_{l>n}{\left(\sum\_{k=0}^{n}\left(\sum\_{j=\max\{0,l-k\}}^{l}{\gamma\_j a\_{j,k,l}}\right)p\_k\right)\binom{z}{l}}.$$


Equating the first $2n+1$ coefficients on the right-hand side to zero now gives $2n+1$ linear equations in $2(n+1)$ unknowns $p\_k$, $q\_k$ ($k=0,\dots,n$), which have therefore a non-trivial solution.


