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, 2021 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??? $\endgroup$ Feb 6, 2021 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, 2021 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). $\endgroup$ Feb 6, 2021 at 6:03

1 Answer 1


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 $\Gamma_5(4/11) = 1 + 3\cdot5 + 4\cdot5^3 + 4\cdot5^4 + 4\cdot5^6 + 3\cdot5^7 + 2\cdot5^8 + 4\cdot5^9 + 4\cdot5^{10} + 5^{11} + 4\cdot5^{13} + 5^{15} + 5^{16} + 3\cdot5^{17} + 2\cdot5^{18} + 5^{19} + O(5^{20})$. $\endgroup$ Feb 6, 2021 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! $\endgroup$ Feb 6, 2021 at 21:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.