Largest power of $p$ which divides $F_p=\binom{p^{n+1}}{p^n}-\binom{p^{n}}{p^{n-1}}$

Question

I would like to know your comments in order to obtain the largest power of the prime numberr $p$ which divides $$ F_p= \binom{p^{n+1}}{p^n}-\binom{p^{n}}{p^{n-1}}. $$ I proved the largest power that divided $F_2$ is $3n$. For the case $p=2$, I could use Vandermonde's Identity. So $$\binom{2^{n+1}}{2^n}-\binom{2^{n}}{2^{n-1}}\\ =\sum_{j=0}^{2^{n+1}}\binom{2^{n}}{j}\binom{2^{n}}{2^{n}-j}-\binom{2^{n-1}}{j}\binom{2^{n-1}}{2^{n-1}-j} $$

Answer 1

It is known that $$F_2\equiv 2^{3r}\pmod{2^{3r+3}}\quad (r\ge 2),\quad F_3\equiv 3^{3r+1}\pmod{3^{3r+3}} \quad( r\ge 1),$$ $$F_p\equiv \Delta_p p^{3r+2}\pmod{p^{3r+3}} \quad (p\ge 5, r\ge 2),$$ where $\Delta_p$ depends only on $p$ (see Jacobsthal, Zahlentheoretische Eigenschaften der Binomialkoeffizienten, Norske Vid. Selsk., Skr. 1942, No. 4, 28 S. (1945).). More generally $$\binom{np}{kp}\equiv\binom{n}{k}\pmod{p^{3+\mathrm{ord}_pn+\mathrm{ord}_pk+\mathrm{ord}_p(n-k)+\mathrm{ord}_p\binom{n}{k}}}.$$ It is also known, see Granville, Arithmetic properties of binomial coefficients. I. Binomial coefficients modulo prime powers that the last exponent can only be increased if $p$ devides $B_{p-3}$, the $(p-3)$rd Bernoulli number.

See also discussion at Binomial supercongruences: is there any reason for them?

Answer 2

In point of fact, it was W. Ljunggren who proved in his 1942 paper "En egenskap ved de midtre binomialkoeffisienter" (Norsk. Mat. Tidsskrift 24 (1942), 18-22) that

$$p^{3n+\varepsilon} \mid F_{p}$$

where $\varepsilon=0$ if $p=2$, $\varepsilon = 1$ if $p=3$, and $\varepsilon =2$ if $p \geq 5$.

You can find the following sketch of Ljunggren's proof in pages 563-564 of the second volume of The Collected Papers of Wilhelm Ljunggren:

In the identity

$$F(y) = \prod_{\substack{\lambda =1\\ (\lambda,p)=1}}^{pm}(y-\lambda) = y^{t} -A_{1}y^{t-1} + A_{2}y^{t-2} - \ldots + A_{t-2}y^{2}-A_{t-1}y+A_{t} \quad \quad(*)$$

where $m=p^{n-1}$, $t=\varphi(p^{n})$ and $n>1$ for $p=2$, $A_{t-1}$ is divisible by $p^{\varepsilon} m^{2}$. This is a consequence of a theorem of Leudesdorf. In ($\ast$) we first put $y=pm$. Dividing by $pm$ and making use of the fact that $A_{t-1}$ is divisible by $p^{\varepsilon} \cdot m^{2}$ we easily find that $A_{t-2}$ is divisible by $mp^{\varepsilon-1}$. We then put $y=p^{2}m$ and thus get that $F(p^{2}m)-A_{t}$ is divisible by $m^{3}p^{\varepsilon+2}$. We then can give the expression [that defines $F_{p}$] in the form $$\binom{p^{2}m}{pm} - \binom{pm}{m} = p\binom{pm-1}{m-1} \frac{F(p^{2}m)-A_{t}}{A_{t}}.$$ This gives the proof of our... theorem because $(A_{t},p)=1$.

Answer 3

Although the question was completely answered by others, I want to provide some more input.

1. An illuminating $p$-adic point of view is given in Section 7.1.6, "The Kazandzidis Congruences", of the (excellent) book "A Course in $p$-adic Analysis" by Alain M. Robert. The main result of the section is the following result, mentioned by Alexei Ustinov, and attributed there to Kazandzidis:

$$\forall p > 2: \binom{pn}{pk} \equiv \binom{n}{k} \pmod {p^{2+\varepsilon}nk(n-k)\binom{n}{k}\mathbb{Z}_p}, \quad \varepsilon = 1_{p>3}.$$

The section is part of the chapter on the Morita $p$-adic Gamma function $\Gamma_p: \mathbb{Z}_p \to \mathbb{Z}_p$, which is a continuous function given on integers $n>2$ by $$\Gamma_p(n)=(-1)^n \prod_{1\le j <n, p\nmid j} j,$$ at least when $p$ is an odd prime. Legendre's formula shows that the $p$-adic valuation of both terms in $F_p$ is exactly $1$, and so it remains to understand the $p$-adic valuation of $$\binom{p^{n+1}}{p^n} / \binom{p^n}{p^{n-1}} -1,$$ which is a difference of two elements of $\mathbb{Z}_p^{*}$. As mentioned in the section, L. van Hamme had observed that $$\binom{pa}{pb} / \binom{a}{b} =\frac{\Gamma_p(pa)}{\Gamma_p(pb)\Gamma_p(p(a-b))},$$ and so it remains to compute $$| \frac{\Gamma_p(pa)}{\Gamma_p(pb)\Gamma_p(p(a-b))} - 1|_p,$$ which explains the link with $p$-adic analysis. Properties of the logarithm in $\mathbb{Z}$ show that in fact the above valuation is exactly $$|\log \Gamma_p(pa) - \log \Gamma_p(pb) - \log \Gamma_p(p(a-b))|_p.$$ Now it just a matter of finding out what is the Taylor expansion of $f(x):=\log \Gamma(px)$. It can be shown that $f$ is an odd function, as so $$f(x) = \sum_{i \ge 1} a_i x^{2i-1},$$ and then $$f(a)-f(b)-f(a-b)= ab(a-b)\sum_{i \ge 2} a_i \cdot \frac{a^{2i-1}-b^{2i-1}-(a-b)^{2i-1}}{ab(a-b)}.$$ (Note the vanishing of the linear term $i=1$!) The usual properties of the $p$-adic valuation yield that $$|f(a)-f(b)-f(a-b)|_p \le |ab(a-b)|_p \cdot \max_{i \ge 2} |a_i|_p \cdot |\frac{a^{2i-1}-b^{2i-1}-(a-b)^{2i-1}}{ab(a-b)}|_p.$$ Bounding $\max_{i \ge 2} |a_i|_p \cdot |\frac{a^{2i-1}-b^{2i-1}-(a-b)^{2i-1}}{ab(a-b)}|_p$ is somewhat technical, and it is related to the Bernoulli numbers, as mentioned in Alexei's answer. It can be shown that the term $i=2$ satisfies $|a_2\cdot 3|_p \le p^{-3+1_{p=3}}$ and that the rest of the terms have valuation smaller or equal to that. This yields the Kazandzisids Congruences by the previous arguments. by plugging $a=p^n,b=p^{n-1}$, we obtain a bound on the $p$-adic valuation of your $F_p$.

2. An easy-to-follow and elementary approach to the Kazandzidis Congruences, which gives a (slightly) weaker result, is given in Lemma A of the paper "Some Congruences For Generalized Euler Numbers", by I. Gessel (1983):

Let $p$ be a prime. Let $\epsilon=1$ if $p$ is 2 or 3, and $\epsilon=0$ if $p$ is greater than $3$. Then $\binom{p^ka}{p^k b} \equiv \binom{p^{k-1}a}{p^{k-1}b} \pmod {p^{3k-\epsilon}}$

It is quite possible that following Gessel's proof with the specific choice $b=1,a=p$ might allow you to recover the valuation given by Kazandzidis' result.