Twisted Padé approximants 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.
 A: 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.
