Let $p>2$ be a prime. For $n \in \mathbb{Z}^+$ we can define \begin{equation} F(n) = (-1)^n \prod_{1<i<n, p\nmid i} i. \end{equation} Since $\mathbb{Z}$ is dense in $\mathbb{Z}_p$, we can extend $F$ uniquely to a continuous function on $\mathbb{Z}_p$, which is the $p$-adic gamma function $\Gamma_p$.

Is there any good way to explicitly compute $\Gamma_p(a/b)$ where $a,b \in \mathbb{Z}$ such that $a/b \in \mathbb{Z}_p^{\times}$? Specifically, I'm looking at $\Gamma_5(4/11)$.

A bad idea is to evaluate $F(a_n)$, where $a_n$ are the partial sums of the $5$-adic expansion of $4/11$. I don't think this would work since $F(a_n)$ becomes too difficult to compute the further you go in the expansion.

I'm aware that some values can be computed via the Gross-Koblitz formula, but I don't know how or if I can apply it to my situation.

Any help is appreciated!

Edit: As Henri Cohen points out, you can get Sage to do this:

[Input] Sage: R=Zp(5)
        Sage: x = R(4/11)
        Sage: x.gamma('pari')
[Output]      1 + 3*5 + 4*5^3 + 4*5^4 + 4*5^6 + 3*5^7 + 2*5^8 + 4*5^9 
              + 4*5^10 + 5^11 + 4*5^13 + 5^15 + 5^16 + 3*5^17 + 2*5^18 
              + 5^19 + O(5^20)
  • $\begingroup$ How accurately do you want to know $\Gamma_5(4/11)$? $\endgroup$ – KConrad Feb 6 at 5:09
  • $\begingroup$ My goal is to figure out the multiplicative order of $\Gamma_5(4/11)$ modulo $5$, so I suppose I only need to know it within the first $p$-adic digit??? $\endgroup$ – matt stokes Feb 6 at 5:18
  • 2
    $\begingroup$ Yes: $|\Gamma_5(x)-\Gamma_5(y)|_5 \leq |x-y|_5$ for all $x$ and $y$ in $\mathbf Z_5$. Thus $\Gamma_5(4/11) \equiv \Gamma_5(4) \bmod 5$. $\endgroup$ – KConrad Feb 6 at 5:53
  • 1
    $\begingroup$ @KConrad Ah, I feel silly. I should have made it clear what I was trying to do in my question. Thank you for your help! (p.s. your expository papers helped me pass my comprehensive exam). $\endgroup$ – matt stokes Feb 6 at 6:03

The $p$-adic gamma function is implemented in Pari/GP (hence available in Sage). The algorithm used is due to Fernando Rodriguez-Villegas (but probably predates him) and is explained in detail both in his book on experimental mathematics published at Oxford, and in my Springer GTM 240.

  • $\begingroup$ Thanks for pointing this out! I wanted to use Sage to calculate this but couldn't find the command. Wasn't looking hard enough. After reading your post I found what I was looking for. Sage is telling me $\Gamma_5(4/11) = 1 + 3\cdot5 + 4\cdot5^3 + 4\cdot5^4 + 4\cdot5^6 + 3\cdot5^7 + 2\cdot5^8 + 4\cdot5^9 + 4\cdot5^{10} + 5^{11} + 4\cdot5^{13} + 5^{15} + 5^{16} + 3\cdot5^{17} + 2\cdot5^{18} + 5^{19} + O(5^{20})$. $\endgroup$ – matt stokes Feb 6 at 15:04
  • $\begingroup$ Your Sage script given above is one of the reasons I don't like Sage. In Pari/GP you simply type gamma(4/11+O(5^20)). Thanks for accepting my answer! $\endgroup$ – Henri Cohen Feb 6 at 21:53

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