Explicit formulas for the $p$-adic valuations of Fibonomial coefficients II

1.   Introduction

The Fibonacci sequence $(F_n)_{n\geq 1}$ is given by the recurrence relation $F_n = F_{n-1} + F_{n-2}$ for $n\geq 3$ with the initial values $F_1 = F_2 = 1$. For each $m\geq 1$ and $1\leq k \leq m$, the Fibonomial coefficients ${m \choose k}_F$ is defined by

Similar to the binomial coefficients, we define ${m \choose k}_F = 1$ if $k = 0$ and ${m \choose k}_F = 0$ if $k > m$, and it is well-known that ${m \choose k}_F$ is always an integer for every $m \geq 1$ and $k\geq 0$.

Recently, there has been an increasing interest in the study of Fibonomial coefficients. Marques and Trojovský ^[25,26] start the investigation on the divisibility of Fibonomial coefficients by determining the integers $n\geq 1$ such that ${pn \choose n}_F$ is divisible by $p$ for $p = 2, 3$. Marques, Sellers, and Trojovský ^[24] show that $p$ divides ${p^{a+1}\choose p^a}_F$ for $p\equiv \pm 2\pmod 5$ and $a\geq 1$. Marques and Trojovsk' ^[27] and Trojovský ^[42] extend their results further and obtained the $p$-adic valuation of ${p^{a+1}\choose p^a}_F$ in ^[42]. Then Ballot [2,Theorem 2] generalizes the Kummer-like theorem of Knuth and Wilf ^[22] and uses it to give a generalization of Marques and Trojovský's results. In particular, Ballot [2,Theorems 3.6,5.2,and 5.3] finds all integers $n$ such that $p\mid {pn \choose n}_U$ for any nondegenerate fundamental Lucas sequence $U$ and $p = 2, 3$ and for $p = 5, 7$ in the case $U = F$. Phunphayap and Pongsriiam ^[31] provide the most general formula for the $p$-adic valuation of Fibonomial coefficients in the most general form ${m\choose n}_F$. For other recent results on the divisibility properties of the Fibonacci numbers, the Fibonomial coefficients, and other combinatorial numbers, see for example ^{[3,4,5,11,12,13,16,17,28,30,32,33,34,37,38,41,43]}. For some identities involving Fibonomial coefficients and generalizations, we refer the reader to the work of Kilic and his coauthors ^{[7,8,18,19,20,21]}. For the $p$-adic valuations of Eulerian, Bernoulli, and Stirling numbers, see ^{[6,9,14,23,40]}. Hence the relation $p \mid {p^a n \choose n}_F$ has been studied only in the case $p = 2, 3, 5, 7$ and $a = 1$.

In this article, we extend the investigation on ${p^a n \choose n}_F$ to the case of any prime $p$ and any positive integer $a$. Replacing $n$ by $p^a$ and $p^a$ by $p$, this becomes Marques and Trojovský's results ^[27,42]. Substituting $a = 1$, $p\in\{2, 3, 5, 7\}$, and letting $n$ be arbitrary, this reduces to Ballot's theorems ^[2]. So our results are indeed an extension of those previously mentioned. To obtain such the general result for all $p$ and $a$, the calculation is inevitably long but we try to make it as simple as possible. As a reward, we can easily show in Corollaries 3.3 and 3.4 that ${4n \choose n}_F$ is odd if and only if $n$ is a nonnegative power of 2, and ${8n \choose n}_F$ is odd if and only if $n = (1 + 3 \cdot 2^{k})/7$ for some $k \equiv 1 \pmod{3}$.

We organize this article as follows. In Section 2, we give some preliminaries and results which are needed in the proof of the main theorems. In Section 3, we calculate the $p$-adic valuation of ${p^a n \choose n}_F$ for all $a$, $p$, and $n$, and use it to give a characterization of the positive integers $n$ such that ${p^a n \choose n}_F$ is divisible by $p$ where $p$ is any prime which is congruent to $\pm 2 \pmod{5}$. Remark that there also is an interesting pattern in the $p$-adic representation of the integers $n$ such that ${pn \choose n}_F$ is divisible by $p$. The proof is being prepared but it is a bit too long to include in this paper. We are trying to make it simpler and shorter and will publish it in the future. For more information and some recent articles related to the Fibonacci numbers, we refer the readers to ^{[15,35,36,39]} and references therein.

2.   Preliminaries and lemmas

Throughout this article, unless stated otherwise, $x$ is a real number, $p$ is a prime, $a, b, k, m, n, q$ are integers, $m, n \geq 1$, and $q \geq 2$. The $p$-adic valuation (or $p$-adic order) of $n$, denoted by $\nu_p (n)$, is the exponent of $p$ in the prime factorization of $n$. In addition, the order (or the rank) of appearance of $n$ in the Fibonacci sequence, denoted by $z(n)$, is the smallest positive integer $m$ such that $n\mid F_m$, $\lfloor x\rfloor$ is the largest integer less than or equal to $x$, $\{ x\}$ is the fractional part of $x$ given by $\{ x\} = x-\lfloor x\rfloor$, $\lceil x\rceil$ is the smallest integer larger than or equal to $x$, and $a\bmod{m}$ is the least nonnegative residue of $a$ modulo $m$. Furthermore, for a mathematical statement $P$, the Iverson notation $[P]$ is defined by

We define $s_q (n)$ to be the sum of digits of $n$ when $n$ is written in base $q$, that is, if $n = (a_k a_{k-1} \ldots a_0)_q = a_k q^k + a_{k-1} q^{k-1} + \cdots + a_0$ where $0\leq a_i < q$ for every $i$, then $s_q (n) = a_k + a_{k-1} + \cdots + a_0$. Next, we recall some well-known and useful results for the reader's convenience.

Lemma 1. Let $p\neq 5$ be a prime. Then the following statements hold.

${\rm(i)}$ $n\mid F_m$ if and only if $z(n)\mid m$

$\rm(ii)$ $z(p)\mid p+1$ if and only if $p\equiv \pm 2 \pmod{5}$ and $z(p)\mid p-1$, otherwise.

$\rm(iii)$ $\gcd (z(p), p) = 1$.

Proof. These are well-known. See, for example, in [31,Lemma 1] for more details.

Lemma 2. (Legendre's formula) Let $n$ be a positive integer and let $p$ be a prime. Then

We will deal with a lot of calculations involving the floor function. So we recall the following results, which will be used throughout this article, sometimes without reference.

Lemma 3. For $k\in\mathbb{Z}$ and $x\in\mathbb{R}$, the following holds

$\rm(i)$ $\lfloor k+x\rfloor = k + \lfloor x\rfloor$,

$\rm(ii)$ $\{k+x\} = \{x\}$,

$\rm(iii)$ $\lfloor x\rfloor + \lfloor -x \rfloor = \begin{cases} -1, & \text{if $x\not\in\mathbb{Z}$; }\\ 0, & \text{if $x\in\mathbb{Z}$, } \end{cases}$

$\rm(iv)$ $0 \leq \{ x \} < 1$ and $\{ x \} = 0$ if and only if $x \in \mathbb{Z}$.

$\rm(v)$ $\lfloor x+y\rfloor = \begin{cases} \lfloor x\rfloor +\lfloor y\rfloor, & \mathit{\text{if $\{x\}+\{y\} < 1$; }}\\ \lfloor x\rfloor +\lfloor y\rfloor+1, & \mathit{\text{if $\{x\}+\{y\}\geq 1$, }} \end{cases}$

$\rm(vi)$ $\left\lfloor\frac{\lfloor x\rfloor}{k}\right\rfloor = \left\lfloor\frac{x}{k}\right\rfloor$ for $k\geq 1$.

Proof. These are well-known and can be proved easily. For more details, see in [10,Chapter 3]. We also refer the reader to ^[1,29] for a nice application of these properties.

The next three theorems given by Phunphayap and Pongsriiam ^[31] are important tools for obtaining the main results of this article.

Theorem 4. [31,Theorem 7] Let $p$ be a prime, $a\geq 0$, $\ell \geq 0$, and $m\geq 1$. Assume that $p\equiv \pm 1\pmod{m}$ and $\delta = [\ell \not\equiv 0\pmod{m}]$ is the Iverson notation. Then

Theorem 5. [31,Theorem 11 and Corollary 12] Let $0\leq k\leq m$ be integers. Then the following statements hold.

$\rm(i)$ Let $A_2 = \nu_2\left(\left\lfloor \frac{m}{6} \right\rfloor!\right) - \nu_2\left(\left\lfloor \frac{k}{6} \right\rfloor!\right) -\nu_2\left(\left\lfloor \frac{m-k}{6} \right\rfloor!\right)$. If $r = m \bmod{6}$ and $s = k \bmod{6}$, then

$\rm(ii)$ $\nu_5\left({m \choose k}_F\right) = \nu_5\left({m \choose k}\right)$.

$\rm(iii)$ Suppose that $p$ is a prime, $p\neq 2$, and $p\neq 5$. If $m' = \left\lfloor\frac{m}{z(p)}\right\rfloor$, $k' = \left\lfloor\frac{k}{z(p)}\right\rfloor$, $r = m\bmod z(p)$, and $s = k\bmod z(p)$, then

Theorem 6. [31,Theorem 13] Let $a$, $b$, $\ell_1$, and $\ell_2$ be positive integers and $b\geq a$. For each $p \neq 5$, assume that $\ell_1 p^b > \ell_2 p^a$ and let $m_p = \left\lfloor\frac{\ell_1 p^{b-a}}{z(p)}\right\rfloor$ and $k_p = \left\lfloor\frac{\ell_2}{z(p)}\right\rfloor$. Then the following statements hold.

$\rm(i)$ If $a\equiv b \pmod{2}$, then $\nu_2\left({\ell_1 2^b \choose \ell_2 2^a}_F \right)$ is equal to

and if $a\not\equiv b \pmod{2}$, then $\nu_2\left({\ell_1 2^b \choose \ell_2 2^a}_F \right)$ is equal to

$\rm(ii)$ Let $p\neq 5$ be an odd prime and let $r = \ell_1 p^b \mod{z(p)}$ and $s = \ell_2 p^a \mod{z(p)}$. If $p\equiv \pm 1\pmod{5}$, then

and if $p\equiv \pm 2\pmod{5}$, then $\nu_p\left({\ell_1 p^b \choose \ell_2 p^a}_F \right)$ is equal to

In fact, Phunphayap and Pongsriiam ^[31] obtain other results analogous to Theorems 2.5 and 2.6 too but we do not need them in this article.

3.   Main results

We begin with the calculation of the 2-adic valuation of ${2^a n \choose n}_F$ and then use it to determine the integers $n$ such that ${2n \choose n}_F, {4n \choose n}_F, {8n \choose n}_F$ are even. Then we calculate the $p$-adic valuation of ${p^a n \choose n}_F$ for all odd primes $p$. For binomial coefficients, we know that $\nu_2 \left({2n \choose n}\right) = s_2 (n)$. For Fibonomial coefficients, we have the following result.

Theorem 7. Let $a$ and $n$ be positive integers, $\varepsilon = [n \not \equiv 0 \pmod{3}]$, and $A = \left\lfloor \frac{(2^a - 1)n}{3\cdot 2^{\nu_2 (n)}} \right\rfloor$. Then the following statements hold.

$\rm(i)$ If $a$ is even, then

where $\delta = [n \bmod{6} = 3, 5]$. In other words, $\delta = 1$ if $n \equiv 3, 5 \pmod{6}$ and $\delta = 0$ otherwise.

$\rm(ii)$ If $a$ is odd, then

where $\delta = \frac{(n\bmod{6}) - 1}{2} [2\nmid n] + \left\lceil \frac{\nu_2 (n) + 3 - n \bmod{3}}{2} \right\rceil [n \bmod 6 = 2, 4]$. In other words, $\delta = \frac{(n \bmod{6}) - 1}{2}$ if $n$ is odd, $\delta = 0$ if $n \equiv 0 \pmod{6}$, $\delta = \left\lceil \frac{\nu_2 (n)}{2} \right\rceil + 1$ if $n \equiv 4 \pmod{6}$, and $\delta = \left\lceil \frac{\nu_2 (n) + 1}{2} \right\rceil$ if $n \equiv 2 \pmod{6}$.

Proof. The second equalities in (3.1) and (3.2) follow from Legendre's formula. So it remains to prove the first equalities in (3.1) and (3.2). To prove (ⅰ), we suppose that $a$ is even and divide the consideration into two cases.

Case 1. $2 \nmid n$. Let $r = 2^a n \bmod{6}$ and $s = n \bmod{6}$. Then $s \in \{1, 3, 5\}$, $r \equiv 2^a n \equiv 4n \equiv 4s \pmod{6}$, and therefore $(r, s) = (4, 1), (0, 3), (2, 5)$. In addition, $A = \left\lfloor \frac{(2^a - 1)n}{3} \right\rfloor = \frac{(2^a - 1)n}{3}$ and $\delta = [s = 3, 5]$. By Theorem 5(ⅰ), the left–hand side of (3.1) is $A_2$ if $s = 1$ and $A_2 + 2$ if $s = 3, 5$, where $A_2 = \nu_{2}\left(\left\lfloor \frac{2^a n}{6} \right\rfloor! \right) - \nu_2 \left(\left\lfloor \frac{n}{6} \right\rfloor! \right) - \nu_2 \left(\left\lfloor \frac{(2^a - 1)n}{6} \right\rfloor! \right)$. We obtain by Theorem 4 that

By Legendre's formula and Lemma 3, we have

From the above observation, we obtain

It is now easy to check that $A_2$ (if $s = 1$), $A_2 + 2$ (if $s = 3, 5$) are the same as $\delta + A - \frac{a}{2}\varepsilon - \nu_2 (A!)$ in (3.1). So (3.1) is verified.

Case 2. $2 \mid n$. We write $n = 2^b \ell$ where $2 \nmid \ell$ and let $m = \left\lfloor \frac{2^a \ell}{3} \right\rfloor$, $k = \left\lfloor \frac{\ell}{3} \right\rfloor$, $r = 2^a \ell \bmod{3}$, and $s = \ell \bmod{3}$. Since $a$ is even, $r = s$. Then we apply Theorem 6(ⅰ) to obtain

We see that $\ell \not\equiv 0 \pmod{3}$ if and only if $n \not\equiv 0\pmod{3}$. In addition, $A = \frac{(2^a - 1)\ell}{3}$ and $\delta = 0$. By Theorem 4, we have

In addition,

So $\nu_2 ((m-k)!) = \nu_2 (A!)$. Substituting these in (3.3), we obtain (3.1). This completes the proof of (ⅰ).

To prove (ⅱ), we suppose that $a$ is odd and divide the proof into two cases.

Case 1. $2\nmid n$. This case is similar to Case 1 of the previous part. So we let $r = 2^a n \bmod{6}$ and $s = n \bmod{6}$. Then $s \in \{1, 3, 5\}$, $r \equiv 2^a n \equiv 2n \equiv 2s \pmod{6}$, $(r, s) = (2, 1), (0, 3), (4, 5)$, $\delta = \frac{s-1}{2}$, and the left–hand side of (3.2) is $A_2$ if $s = 1$, $A_2 + 2$ if $s = 3$, and $A_2 + 3$ if $s = 5$, where $A_2 = \nu_2 \left(\left\lfloor \frac{2^a n}{6} \right\rfloor! \right) - \nu_2 \left(\left\lfloor \frac{n}{6} \right\rfloor! \right) - \nu_2 \left(\left\lfloor \frac{(2^a - 1)n}{6} \right\rfloor! \right)$. In addition, we have

Therefore

Furthermore,

which implies that $A = \frac{(2^a - 1)n}{3} - \frac{r}{6}$. Then

It is now easy to check that $A_2$ (if $s = 1$), $A_2 + 2$ (if $s = 3$), and $A_2 +3$ (if $s = 5$), are the same as $\delta + A - \frac{a-1}{2}\varepsilon - \nu_2 (A!)$ in (3.2). So (3.2) is verified.

Case 2. $2\mid n$. This case is similar to Case 2 of the previous part. So we write $n = 2^{b} \ell$ where $2 \nmid \ell$ and let $m = \left\lfloor \frac{2^a \ell}{3} \right\rfloor$, $k = \left\lfloor \frac{\ell}{3} \right\rfloor$, $r = 2^a \ell \bmod{3}$, and $s = \ell \bmod{3}$. We obtain by Theorem 6 that $\nu_2 \left({2^a n \choose n}_F \right)$ is equal to

By Theorem 4, we have

Since $(2^a - 1)\ell \equiv \ell \pmod{3}$, $\left\{ \frac{(2^a - 1)\ell }{3} \right\} = \left\{ \frac{\ell}{3} \right\}$. This implies that $\nu_2 (m!) = A - \frac{a-1}{2}\varepsilon + \nu_2(k!)$. In addition, $(r, s) = (0, 0), (2, 1), (1, 2)$, and

From the above observation, we obtain

Substituting this in (3.4), we see that

Recall that $n = 2^b \ell \equiv (-1)^b \ell \pmod{3}$. So (3.5) implies that

which is the same as (3.2). This completes the proof.

We can obtain the main result of Maques and Trojovský ^[25] as a corollary.

Corollary 8. (Marques and Trojovský ^[25]) ${2n \choose n}_F$ is even for all $n \geq 2$.

Proof. Let $n \geq 2$ and apply Theorem 7 with $a = 1$ to obtain $\nu_2 \left({2n \choose n}_F \right) = \delta + s_2 (A)$. If $n \not \equiv 0, 1 \pmod{6}$, then $\delta > 0$. If $n \equiv 0 \pmod{6}$, then $n \geq 3\cdot 2^{\nu_2 (n)}$, and so $A \geq 1$ and $s_2 (A) > 0$. If $n \equiv 1 \pmod{6}$, then $A = \left\lfloor \frac{n}{3} \right\rfloor > 1$ and so $s_2 (A) > 0$. In any case, $\nu_2 \left({2n \choose n}_F \right) > 0$. So ${2n \choose n}_F$ is even.

Corollary 9. Let $n \geq 2$. Then ${4n \choose n}_F$ is even if and only if $n$ is not a power of 2. In other words, for each $n \in \mathbb{N}$, ${4n \choose n}_F$ is odd if and only if $n = 2^k$ for some $k\geq 0$.

Proof. Let $\delta$, $\varepsilon$, and $A$ be as in Theorem 7. If $n = 2^k$ for some $k \geq 1$, then we apply Theorem 7 with $a = 2$, $\delta = 0$, $\varepsilon = 1$, $A = 1$ leading to $\nu_2 \left({4n \choose n}_F \right) = 0$, which implies that ${4n \choose n}_F$ is odd.

Suppose $n$ is not a power of 2. By Theorem 7, $\nu_2 \left({4n \choose n}_F \right) = \delta + s_2 (A) - \varepsilon \geq s_2 (A) - 1$. Since $n$ is not a power of 2, the sum $s_2 (n) \geq 2$. It is easy to see that $s_2 (m) = s_2 (2^c m)$ for any $c, m \in \mathbb{N}$. Therefore $s_2 (A) = s_2 \left(\frac{n}{2^{\nu_2 (n)}} \right) = s_2 \left(2^{\nu_2 (n)} \cdot \frac{n}{2^{\nu_2 (n)}} \right) = s_2 (n) \geq 2$, which implies $\nu_2 \left({4n \choose n}_F \right) \geq 1$, as required.

Observe that $2, 2^2, 2^3$ are congruent to $2, 4, 1 \pmod{7}$, respectively. This implies that if $k \geq 1$ and $k \equiv 1 \pmod{3}$, then $(1 + 3 \cdot 2^k)/7$ is an integer. We can determine the integers $n$ such that ${8n \choose n}_F$ is odd as follows.

Corollary 10. ${8n \choose n}_F$ is odd if and only if $n = \frac{1 + 3 \cdot 2^k}{7}$ for some $k \equiv 1 \pmod{3}$.

Proof. Let $a, \delta, A, \varepsilon$ be as in Theorem 7. We first suppose $n = (1 + 3 \cdot 2^k)/7$ where $k \geq 1$ and $k \equiv 1 \pmod{3}$. Then $n \equiv 7n \equiv 1 + 3 \cdot 2^k \equiv 1 \pmod{6}$. Then $a = 3$, $\varepsilon = 1$, $\delta = 0$, $A = 2^k$, and so $\nu_{2} \left({8n \choose n}_F \right) = 0$. Therefore ${8n \choose n}_F$ is odd. Next, assume that ${8n \choose n}_F$ is odd. Observe that $A \geq 2$ and $s_2 (A) > 0$. If $n \equiv 0 \pmod{3}$, then $\varepsilon = 0$ and $\nu_2 \left({8n \choose n}_F \right) = \delta + s_2(A) > 0$, which is not the case. Therefore $n \equiv 1, 2 \pmod{3}$, and so $\varepsilon = 1$. If $n\equiv 0 \pmod{2}$, then $\delta = \left\lceil \frac{\nu_2 (n) + 3 - n \bmod{3}}{2} \right\rceil \geq 1$, and so $\left({8n \choose n} \right)_F \geq s_2 (A) > 0$, which is a contradiction. So $n \equiv 1 \pmod{2}$. This implies $n \equiv 1, 5 \pmod{6}$. But if $n \equiv 5 \pmod{6}$, then $\delta \geq 2$ and $\nu_2 \left({8n \choose n}_F \right) > 0$, a contradiction. Hence $n \equiv 1 \pmod{6}$. Then $\delta = 0$. Since $s_2 (A) - 1 = \nu_2 \left({8n \choose n}_F \right) = 0$, we see that $A = 2^k$ for some $k \geq 1$. Then $\frac{7n - 1}{3} = \left\lfloor \frac{7n}{3} \right\rfloor = A = 2^k$, which implies $n = \frac{1 + 3 \cdot 2^k}{7}$, as required.

Theorem 11. For each $a, n \in \mathbb{N}$, $\nu_5 \left({5^a n \choose n}_F \right) = \nu_5 \left({5^a n \choose n} \right) = \frac{s_5 ((5^a - 1)n)}{4}$. In particular, ${5^an \choose n}_F$ is divisible by 5 for every $a, n \in \mathbb{N}$.

Proof. The first equality follows immediately from Theorem 5(ⅱ). By Legendre's formula, $\nu_5 \left({n \choose k} \right) = \frac{s_5 (k) + s_5 (n - k) - s_5 (n)}{4}$ for all $n \geq k \geq 1$. So $\nu_5 \left({5^a n \choose n}_F \right)$ is

Theorem 12. Let $p \neq 2, 5$, $a, n \in \mathbb{N}$, $r = p^a n \bmod{z(p)}$, $s = n \bmod{z(p)}$, and $A = \left\lfloor \frac{n(p^a - 1)}{p^{\nu_p (n)}z(p)} \right\rfloor$. Then the following statements hold.

$\rm(i)$ If $p \equiv \pm 1 \pmod{5}$, then $\nu_p \left({p^a n \choose n}_F \right)$ is equal to

$\rm(ii)$ If $p \equiv \pm 2 \pmod{5}$ and $a$ is even, then $\nu_p \left({p^a n \choose n}_F \right)$ is equal to

$\rm(iii)$ If $p \equiv \pm 2 \pmod{5}$ and $a$ is odd, then $\nu_p \left({p^a n \choose n}_F \right)$ is equal to

where $\delta = \left(\left\lfloor \frac{\nu_p (n)}{2} \right\rfloor + [2 \nmid \nu_p(n)][r > s] + [r < s]\nu_p(F_{z(p)}) \right)[r \neq s]$, or equivalently, $\delta = 0$ if $r = s$, $\delta = \left\lfloor \frac{\nu_p (n)}{2} \right\rfloor + \nu_p (F_{z(p)})$ if $r < s$, and $\delta = \left\lceil \frac{\nu_p (n)}{2} \right\rceil$ if $r > s$.

Proof. We first prove (ⅰ) and (ⅱ). So we suppose that the hypothesis of (ⅰ) or (ⅱ) is true. By writing $\nu_p (A!) = \frac{A - s_p (A)}{p - 1}$, we obtain the equalities in (3.6) and (3.7). By Lemma 1(ⅱ), $p^a \equiv 1\pmod{z(p)}$. Then $r = s$.

Case 1. $p \nmid n$. Let $m = \left\lfloor\frac{p^a n}{z(p)}\right\rfloor$ and $k = \left\lfloor\frac{n}{z(p)}\right\rfloor$. Then we obtain by Theorem 5(ⅲ) that

By Lemma 1(ⅱ) and Theorem 4, we see that if $p \equiv \pm 1 \pmod{5}$, then $p \equiv 1 \pmod{z(p)}$ and

and if $p \equiv \pm 2 \pmod{5}$ and $a$ is even, then $p \equiv -1 \pmod{z(p)}$ and

Since $z(p) \mid p^a - 1$ and $p\nmid n$, $A = \frac{n(p^a -1)}{z(p)}$. Therefore

Substituting (3.10), (3.11), and (3.12) in (3.9), we obtain (3.6) and (3.7).

Case 2. $p \mid n$. Let $n = p^b \ell$ where $p \nmid \ell$, $m = \left\lfloor\frac{\ell p^a}{z(p)}\right\rfloor$, and $k = \left\lfloor\frac{\ell}{z(p)}\right\rfloor$. Since $r = s$, we obtain by Theorem 6 that $\nu_p \left({p^a n \choose n}_F \right)$ is equal to

Since $\gcd (p, z(p)) = 1$, we see that $\ell \equiv 0 \pmod{z(p)} \Leftrightarrow n \equiv 0 \pmod{z(p)} \Leftrightarrow s = 0$. Similar to Case 1, we have $\nu_p(m!) = \frac{\ell(p^a - 1)}{z(p)(p-1)} - a\left\{ \frac{\ell}{z(p)} \right\} + \nu_p (k!)$ if $p \equiv \pm 1 \pmod{5}$, $\nu_p (m!) = \frac{\ell(p^a - 1)}{z(p)(p-1)} - \frac{a}{2}[s \neq 0] + \nu_p\left(k!\right)$ if $p \equiv \pm 2 \pmod{5}$ and $a$ is even, $\ell p^a \equiv \ell \pmod{z(p)}$, $A = \frac{\ell (p^a - 1)}{z(p)}$, and $m-k = A$. So (3.13) leads to (3.6) and (3.7). This proves (ⅰ) and (ⅱ).

To prove (ⅲ), suppose that $p \equiv \pm 2 \pmod{5}$ and $a$ is odd. By Lemma 1(ⅱ), $p \equiv -1 \pmod{z(p)}$. In addition, $\frac{p^a - 1}{p - 1} = p^{a-1} + p^{a-2} + \ldots + 1 \equiv 1 \pmod{z(p)}$. We divide the consideration into two cases.

Case 1. $p \nmid n$. This case is similar to Case 1 of the previous part. So we apply Theorems 4 and 5(ⅲ). Let $m = \left\lfloor\frac{p^a n}{z(p)}\right\rfloor$ and $k = \left\lfloor\frac{n}{z(p)}\right\rfloor$. Then

Since $\frac{p^a - 1}{p - 1} \equiv 1 \pmod{z(p)}$, $\frac{n(p^a - 1)}{p-1} \equiv n \pmod{z(p)}$. This implies that $\left\{ \frac{n(p^a - 1)}{z(p)(p-1)} \right\} = \left\{ \frac{n}{z(p)} \right\}$. Therefore

From the above observation, if $r\geq s$, then $A = m-k$ and

which leads to (3.8). If $r < s$, then $A = m - k - 1$, $\left\lfloor \frac{p^a n - n + z(p)}{z(p)} \right\rfloor = A + 1$, and $\nu_p \left({p^a n \choose n}_F \right)$ is equal to

which is the same as (3.8).

Case 2. $p \mid n$. Let $n = p^b \ell$ where $p \nmid \ell$, $m = \left\lfloor \frac{\ell p^a}{z(p)} \right\rfloor$, and $k = \left\lfloor \frac{\ell}{z(p)} \right\rfloor$. Similar to Case 1, $s = 0 \Leftrightarrow \ell \equiv 0 \pmod{z(p)}$. In addition, $\frac{\ell(p^a - 1)}{p - 1} \equiv \ell \pmod{z(p)}$, and so we obtain by Theorem 4 that $\nu_p (m!) = \left\lfloor \frac{A}{p-1} \right\rfloor - \frac{a-1}{2}[s \neq 0] + \nu_p (k!)$. The calculation of $\nu_p \left({p^a n \choose n}_F \right) = \nu_p \left({\ell p^{a+b} \choose \ell p^b}_F \right)$ is done by the applications of Theorem 6 and is divided into several cases. Suppose $r = s$. Then $p^{a+b} \ell \equiv p^a n \equiv r \equiv s \equiv n \equiv p^b \ell \pmod{z(p)}$. Since $(p, z(p)) = 1$, this implies $\ell p^a \equiv \ell \pmod{z(p)}$. Therefore $A = \left\lfloor \frac{\ell p^a - \ell}{z(p)} \right\rfloor = \frac{\ell p^a - \ell}{z(p)} = m - k$ and

which is (3.8). Obviously, if $\ell \equiv 0 \pmod{z(p)}$, then $r = s$, which is already done. So from this point on, we assume that $r \neq s$ and $\ell \not\equiv 0 \pmod{z(p)}$. Recall that $p \equiv -1 \pmod{z(p)}$ and $a$ is odd. So if $b$ is odd, then

Similarly, if $b$ is even, then $r = \ell p^a \bmod{z(p)}$, $s = \ell \bmod{z(p)}$, and $A = m - k + \left\lfloor \frac{r - s}{z(p)} \right\rfloor$. Let $R = \left\lfloor \frac{A}{p-1} \right\rfloor - \frac{a-1}{2}[s \neq 0] - \nu_p (A!) + \delta$ be the quantity in (3.8). From the above observation and the application of Theorem 6, we obtain $\nu_p \left({p^a n \choose n}_F \right)$ as follows. If $r > s$ and $b$ is even, then $A = m - k$ and

If $r > s$ and $b$ is odd, then $A = m - k - 1$ and

If $r < s$ and $b$ is even, then $A = m - k - 1$ and

If $r < s$ and $b$ is odd, then $A = m - k$ and

This completes the proof.

In the next two corollaries, we give some characterizations of the integers $n$ such that ${p^a n \choose n}_F$ is divisible by $p$.

Corollary 13. Let $p$ be a prime and let $a$ and $n$ be positive integers. If $n \equiv 0 \pmod{z(p)}$, then $p \mid {p^a n \choose n}_F$.

Proof. We first consider the case $p \neq 2, 5$. Assume that $n \equiv 0 \pmod{z(p)}$ and $r$, $s$, $A$, and $\delta$ are as in Theorem 12. Then $\frac{n}{p^{\nu_p(n)}z(p)}, \frac{A}{p-1} \in \mathbb{Z}$, $r = s = 0$, and $\delta = 0$. Every case in Theorem 12 leads to $\nu_p \left({p^a n \choose n}_F \right) = \frac{s_p (A)}{p-1} > 0$, which implies $p \mid {p^a n \choose n}_F$. If $p = 5$, then the result follows immediately from Theorem 11. If $p = 2$, then every case of Theorem 7 leads to $\nu_2 \left({2^a n \choose n}_F \right) \geq s_2 (A) > 0$, which implies the desired result.

Corollary 14. Let $p \neq 2, 5$ be a prime and let $a$, $n$, $r$, $s$, and $A$ be as in Theorem 12. Assume that $p \equiv \pm 2 \pmod{5}$ and $n \not \equiv 0 \pmod{z(p)}$. Then the following statements hold.

$\rm(i)$ Assume that $a$ is even. Then $p \mid {p^a n \choose n}_F$ if and only if $s_p (A) > \frac{a}{2} (p - 1)$.

$\rm(ii)$ Assume that $a$ is odd and $p \nmid n$. If $r < s$, then $p \mid {p^a n \choose n}_F$. If $r \geq s$, then $p \mid {p^a n \choose n}_F$ if and only if $s_p (A) \geq \frac{a + 1}{2} (p - 1)$.

$\rm(iii)$ Assume that $a$ is odd and $p \mid n$. If $r\neq s$, then $p \mid {p^a n \choose n}_F$. If $r = s$, then $p \mid {p^a n \choose n}_F$ if and only if $s_p (A) \geq \frac{a + 1}{2} (p - 1)$.

Proof. We use Lemmas 2 and 3 repeatedly without reference. For (ⅰ), we obtain by (3.7) that

This proves (ⅰ). To prove (ⅱ) and (ⅲ), we let $\delta$ be as in Theorem 12 and divide the consideration into two cases.

Case 1. $p \nmid n$. If $r < s$, then we obtain by Theorem 5(ⅲ) that $\nu_p \left({p^a n \choose n}_F \right) \geq \nu_p (F_{z(p)}) \geq 1$. Suppose $r \geq s$. Then $\delta = 0$ and (3.8) is

If $s_p(A) \geq \frac{a+1}{2}(p - 1)$, then (3.8) implies that

Similarly, if $s_p (A) < \frac{a+1}{2}(p-1)$, then $\nu_p \left({p^a n \choose n}_F \right) < 1 - \left\{ \frac{A}{p-1} \right\} \leq 1$. This proves (ⅱ).

Case 2. $p \mid n$. We write $n = p^b \ell$ where $p \nmid \ell$. Then $b \geq 1$. Recall that $\nu_p (F_{z(p)}) \geq 1$. If $r \neq s$, then Theorem 6 implies that $\nu_p \left({p^a n \choose n} \right) \geq \frac{b}{2}$ if $b$ is even and it is $\geq \frac{b + 1}{2}$ if $b$ is odd. In any case, $\nu_p \left({p^a n \choose n}_F \right) \geq 1$. So $p \mid {p^a n \choose n}_F$. If $r = s$, then $\delta = 0$ and we obtain as in Case 1 that $p \mid {p^a n \choose n}_F$ if and only if $s_p (A) \geq \frac{a + 1}{2} (p - 1)$. This proves (ⅲ).

Corollary 15. Let $p \neq 2, 5$ be a prime and let $A = \frac{n(p-1)}{p^{\nu_p (n)}z(p)}$. Assume that $p \equiv \pm 1 \pmod{5}$. Then $p \mid {pn \choose n}_F$ if and only if $s_p (A) \geq p-1$.

Proof. We remark that by Lemma 1(ⅱ), $A$ is an integer. Let $x = \frac{n}{p^{\nu_p (n)}z(p)}$. We apply Theorem 12(ⅰ) with $a = 1$. If $s_p (A) \geq p-1$, then (3.6) implies that $\nu_p \left({pn \choose n}_F \right) \geq 1 - \{x\} > 0$. If $s_p (A) < p - 1$, then $\nu_p \left({pn \choose n}_F \right) < 1 - \{x\} \leq 1$. This completes the proof.

4.   Conclusions

We give exact formulas for the $p$-adic valuations of Fibonomial coefficients of the form ${p^an\choose n}_F$ for all primes $p$ and $a, n\in \mathbb N$. Then we use it to characterize the integers $n$ such that ${p^a n \choose n}_F$ is divisible by $p$.

Acknowledgments

Phakhinkon Phunphayap receives a scholarship from Science Achievement Scholarship of Thailand(SAST). This research was jointly supported by the Thailand Research Fund and the Faculty of Science Silpakorn University, grant number RSA5980040.

Conflict of interest

The authors declare that there is no conflict of interests regarding the publication of this article.