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Let $\alpha\in \mathbb{Z}_p$ be an $p$-adic integer and define for $n\in \mathbb{Z}_{\geq 0}$ $${\alpha\choose n} := \frac{\alpha(\alpha-1)\cdot\ldots\cdot(\alpha-n+1)}{n!}.$$

This is again a $p$-adic integer, and we can define $$\alpha_1:=\sum_{n=0}^\infty \overline{{\alpha \choose n}} p^n$$ and $$\alpha_2:=\sum_{n=0}^\infty \overline{{\alpha \choose p^n}} p^n,$$

where $\overline{(\cdot)}$ means reduction modulo $p$. How are $\alpha,\alpha_1,\alpha_2$ related? Are there formulas expressing this relation?

Thanks a lot!

Edit: Some thinking led me to the following conclusion: Using continuity of $x\mapsto {x\choose p^n}$ as a function $\mathbb{Z}_p\rightarrow \mathbb{Q}_p$ and Lucas' Theorem it follows that $\alpha=\alpha_2$. Does this seem correct?

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It is not very clear to me what you mean by "reduction modulo $p$". Classes mod $p$ live in the finite field ${\Bbb F}_p$ and the sums $\alpha_1$ and $\alpha_2$ wouldn't make much sense there. Are you saying that you are taking some fixed representants of the classes mod $p$ such as $\{0,1,\ldots,p-1\}$ ? – Andrea Mori Mar 2 '11 at 16:39
Oh, yes, the unique nonnegative representative $<p$. – user7698 Mar 2 '11 at 17:55
You can determine whether $p$ divides $\binom{m+n}{n}$ by seeing whether there are any carries when adding $m$ and $n$ in base $p$. $\binom{\alpha}{p^n}$ is particularly simple, since $p^n$ in base $p$ is very easy. – Arturo Magidin Mar 2 '11 at 20:30

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