Divisibility of Fibonomial coefficients in terms of their digital representations and applications

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 coefficient ${m \choose k}_F$ is defined by

As usual, if $m = k$, then the empty product $F_1 F_2 \cdots F_{m - k}$ is defined to be 1, and similar to the binomial coefficients, we let ${m \choose k}_F = 1$ if $k = 0$ and ${m \choose k}_F = 0$ if $k > m$. Then it is well known that ${m \choose k}_F$ is always an integer for every $m \geq 1$ and $k\geq 0$.

There has been some interest in the study of certain generalizations of binomial coefficients such as the Fibonomial or Lucasnomial coefficients. For instance, Marques and Trojovský ^[9] determined the integers $n \geq 1$ such that ${3n \choose n}_F$ is divisible by $3$. Then Ballot ^[1] largely extended Marques and Trojovský's results by characterizing all integers $n \geq 1$ such that ${pn \choose n}_U$ is divisible by $p$ for any nondegenerate fundamental Lucas sequence $U$ and $p = 2, 3$ and for $p = 5, 7$ in the case $U = F$. Ballot ^[1] also proved that $E_3 (x) = O (\log x)$ and $E_7 (x) = O \left((\log x)^3 \right)$. Here and throughout this article, $E_p (x)$ denotes the number of positive integers $n$ less than or equal to $x$ such that ${pn \choose n}_F$ is not divisible by $p$. In other words,

In particular, we ^[13,14] have recently provided an explicit formula for the $p$-adic valuation of certain Fibonomial coefficients, and have used it in the investigation of the integers $n \geq 1$ such that ${p^a n \choose n}_F$ is divisible by $p$ for any prime $p \equiv \pm 2 \pmod{5}$ and any integer $a \geq 1$, and also for primes $p \equiv \pm 1 \pmod{5}$ and $a = 1$ in terms of the sum of digit function.

In this article, we give characterizations for the integers $n \geq 1$ such that ${pn \choose n}_F$ is divisible by $p$ for any prime $p \neq 2, 5$ in terms of the digital representation of $n$. Then we use it in the calculation for asymptotic formulas of $E_p(x)$ for all primes $p$. This extends many results in the literature which focus only on small primes $p \leq 7$.

We organize this article as follows. In Section 2, we recall some definitions and useful results. In Section 3, we prove our main theorems and give some examples. For more information on Fibonacci numbers, Fibonomial coefficients, and generalizations, we refer the reader to some recent articles by Ballot ^[2,3,4], Chu and Kiliç ^[5], Kiliç and Akkus ^[7], Kiliç and Prodinger ^[8], Onphaeng and Pongsriiam ^[10,11,12], and Pongsriiam ^[15,16].

2.   Preliminaries and lemmas

Throughout this article, unless stated otherwise, $x$ is a positive real number, $p$ is a prime, $a, b, k, m, n, q, r$ are integers, $m, n \geq 1$, $q \geq 2$, $\lfloor x\rfloor$ is the largest integer less than or equal to $x$, $\{ x\}$ is the fractional part of $x$ given by $\{ x\} = x-\lfloor x\rfloor$, $a\bmod{m}$ is the least nonnegative residue of $a$ modulo $m$, and $\log x$ is the natural logarithm of $x$. The $p$-adic valuation 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 $n\mid F_m$. Furthermore, we define $s_q (n)$ to be the sum of digits of $n$ when $n$ is written in base $q$, that is, if

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. 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 $p\equiv \pm 2 \pmod{5}$.

${\rm(iii)}$ $z(p)\mid p-1$ if and only if $p \equiv \pm 1 \pmod{5}$.

${\rm(iv)}$ If $p \neq 5$, then $\gcd (z(p), p) = 1$.

Proof. These are well known. See, for example, in [13,Lemma 1] for more details.

We will deal with a lot of calculations involving the floor function. So it is useful to recall the following results, which will be applied throughout this article without further reference.

Lemma 2. For $k\in\mathbb{Z}$ and $x\in\mathbb{R}$, the following statements hold.

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

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

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

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

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

(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 [6,Chapter 3]. We also refer the reader to ^[11] for a nice application of these properties.

The next three lemmas are important tools for obtaining the characterizations of the integers $n$ such that ${pn \choose n}_F$ is divisible by $p$.

Lemma 3. [14,Corollary 13] Suppose that $p \neq 2, 5$ and $a$, $n$ are positive integers. If $n \equiv 0 \pmod{z(p)}$, then $p \mid {p^a n \choose n}_F$.

Lemma 4. [14,Corollary 14] Let $p \neq 2, 5$, $p \equiv \pm 2 \pmod{5}$, $a, n \in \mathbb{N}$, $r = p^a n \bmod{z(p)}$, $s = n \bmod{z(p)}$, $A = \left\lfloor \frac{n(p^a -1)}{p^{\nu_p (n)}z(p)} \right\rfloor$, and $n \not \equiv 0 \pmod{z(p)}$. Then the following statements hold.

${\rm(i)}$ 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 $s_p (A) \geq \frac{a + 1}{2} (p - 1)$.

${\rm(ii)}$ 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 $s_p (A) \geq \frac{a + 1}{2} (p - 1)$.

Lemma 5. [14,Corollary 15] Let $p \neq 2, 5$, $p \equiv \pm 1 \pmod{5}$, and $A = \frac{n(p-1)}{p^{\nu_p (n)}z(p)}$. Then $p \mid {pn \choose n}_F$ if and only if $s_p (A) \geq p-1$.

Lemma 6. Let $k \geq 0$, $q \geq 2$, $1\leq a \leq q-1$, and $0 \leq b\leq q-1$. Then

Proof. When $b = 0$, the result is obvious. So we assume that $b\geq 1$. If $a = 1$, then we write $a(q-1)q^k + bq^k = q^{k+1} + (b-1)q^k$. If $a\geq 2$ and $b\leq a-1$, then we write $a(q-1)q^k + bq^k = (a-1)q^{k+1} + (q-a+b)q^k$. If $a\geq 2$ and $b\geq a$, then we write $a(q-1)q^k + bq^k = aq^{k+1} + (b-a)q^k$. In each case, $s_q(a(q-1)q^k + bq^k)$ is equal to, respectively, $1+b-1 = b$, $a-1+q-a+b = q+b-1$, and $a+b-a = b$. In any case, it is at least $b$.

Lemma 7. Let $p \geq 3$, $p\equiv \pm 2 \pmod{5}$, $0\leq a \leq \frac{p-1}{2}$, and $1 \leq k \leq \frac{z(p)}{2}$. Then $az(p) + k\equiv 0 \pmod{p}$ if and only if $a = \frac{p-1}{2}$, $z(p)$ is even, and $k = \frac{z(p)}{2}$. In particular, if $a < \frac{p - 1}{2}$, then $az(p) + k \not\equiv 0 \pmod{p}$.

Proof. From the assumption, we have

Suppose that $az(p) + k \equiv 0 \pmod{p}$. Then $az(p) + k = np$ for some $1\leq n \leq \frac{z(p)}{2}$. Since $p\equiv \pm 2 \pmod{5}$, we obtain by Lemma 1 that $p\equiv -1 \pmod{z(p)}$. Then $k\equiv az(p) + k \equiv np \equiv -n \pmod{z(p)}$. So there exists $m \in \mathbb{N}$ such that $k = mz(p) - n$. Therefore

This implies that $k = \frac{z(p)}{2}$, $z(p)$ is even, $m = 1$, and $n = \frac{z(p)}{2}$. Since $az(p) + k = np$, we also obtain $a = \frac{p-1}{2}$. The converse can be verified easily. This completes the proof.

We introduce the following notation for convenience.

Definition 8. Let $q$ and $i$ be integers such that $q \geq 2$ and $0\leq i \leq q-1$. We define

In other words, $H(q, i)$ is the set of nonnegative integers $n$ such that the $q$-adic representation of $n$ is increasing (from the left to the right), and the last digit (the rightmost digit) is equal to $i$.

For example, if $q = 10$ and $i = 3$, then $111122233$ and $11111333$ are in $H(10, 3)$ but $213$ and $1234$ are not in $H(10, 3)$.

Definition 9. For positive integers $k$ and $q$, we define

The next lemma is usually called stars and bars problem. Recall that if a set $A$ has exactly $n$ distinct elements, then the number of all possible ways in choosing $m$ elements from $A$ with repetitions allowed is ${n + m - 1 \choose m}$. We have the following lemma.

Lemma 10. Let $k \geq 1$, $q \geq 2$, and $1 \leq t \leq q - 1$ be integers. Then

Proof. This is stars and bars problem. The set $A$ is $\{ 1, 2, 3, \ldots, t \}$. We would like to choose $k$ elements from $A$ with repetitions allowed. So the number of ways, as recalled above, is ${t + k - 1 \choose k}$, which proves this lemma.

Lemma 11. Let $q \geq 2$ and $1 \leq t \leq q-1$ be integers. Then

Consequently,

where the implied constant depends at most on $t$.

Proof. The conditions $0 \leq m < q^r$ and $m \in \bigcup_{i = 0}^{t} H(q, i)$ mean that $m = (a_r a_{r-1} \cdots a_1)_q$ and $0 \leq a_r \leq a_{r - 1}\leq \cdots \leq a_1 \leq t$. So this is also stars and bars problem. The set $A$ is $\{0, 1, 2, \ldots, t\}$. We would like to choose $r$ elements from $A$ with repetitions allowed. Therefore the number of ways is ${t + 1 + r - 1 \choose r} = {r + t \choose r}$, which proves the first part. Next,

where $P(r)$ is a polynomial in $r$ of degree $t - 1$ with the coefficients depending only on $t$. Therefore $P(r) = O\left(r^{t - 1} \right)$ and the implied constant depends at most on $t$. This completes the proof.

3.   Main results

In this section, we begin with a property of the sum of digit function. Then we use it in the study of the divisibility $p \mid {pn \choose n}_F$ in terms of the digital representation of $n$. After that, we determine an asymptotic formula for $E_p (x)$.

Theorem 12. Let $m \geq 0$, $q\geq 2$, and $1 \leq k \leq z(q) - 1$. Then

Proof. Let $H = \bigcup_{i = 0}^{t(q, k)} H(q, i)$ and $t = t(q, k)$. Since $k < z(q)$, we see that $t < q-1$. If $m = 0$, then $m \in H(q, 0) \subseteq H$ and $s_q((q-1)m + t) = s_q (t) = t < q - 1$, so we are done. From this point on, we assume that $m \geq 1$. To prove this theorem, we first show that

We prove (3.1) by induction on $r$ where $r$ is the number of digits in the $q$-adic expansion of $m$. For $r = 1$, we let $m = a$, $1 \leq a\leq q-1$, $a \not\in H$, and write

Observe that $i\in H(q, i) \subseteq H$ for each $0 \leq i\leq t$. Since $a \not\in H$, we see that $a > t$ which implies $0 \leq q - a + t \leq q - 1$. Therefore $s_q \left((q-1)m + t\right) = a - 1 + q - a + t \geq q-1$. Next, let $r\geq 1$ and suppose that (3.1) holds for any $m\in\mathbb{N}$ such that the number of digits of $m$ in its $q$-adic expansion is less than or equal to $r$. Assume that $m = (a_{r+1}a_{r}\cdots a_1)_q$, $a_{r+1}\neq 0$, $0\leq a_i < q$ for all $i$, and $m\not\in H$. Let $m_1 = (a_r a_{r-1}\cdots a_1)_q$.

Case 1. $m_1 \in H$. If $r = 1$, let $m_2 = 0$ and if $r \geq 2$, we let $m_2 = (a_{r-1}a_{r-2}\cdots a_1)_q$. Then we write $(q-1)m + t$ as $(q - 1)(a_{r+1} q^{r} + a_r q^{r - 1} + m_2) + t$, which is equal to

Since $m_1 \in H$, $a_{r}\leq a_i \leq t$ for all $1\leq i\leq r$. So we have

Therefore

Thus

Since $m \not\in H$ and $m_1 \in H$, $a_{r+1} > a_r$. Thus $0 \leq q - a_{r+1} + a_r < q$. From this and from (3.2) and (3.3), we obtain that $s_q\left((q-1)m + t \right)$ is equal to

Case 2. $m_1 \not\in H$. Since $(q-1)m_1 + t < (q-1)q^r + q - 1 < q^{r+1}$, we write $(q-1)m_1 + t = (b_{r+1}b_r \cdots b_1)_q$ where $b_{r+1}$ may be zero. Since the number of digits in the $q$-adic representation of $m_1$ is less than or equal to $r$, we can apply the induction hypothesis on $m_1$ to obtain

Next we write

By the above equation, Lemma 6, and (3.4), we obtain $s_q \left((q-1)m + t \right) =$

This proves (3.1). To prove the converse, assume that $m \in H$ and let $a = m \bmod{q}$ be the least nonnegative residue of $m$ modulo $q$. Then $a$ is the last digit of $m$ in its $q$-adic expansion. Since $m\geq 1$ and $m \in H$, we see that $1\leq a \leq t$ and $m \in H(q, a)$. So the possible digits in the $q$-adic representation of $m$ with nonzero leading digit are $1, 2, 3, \ldots, a$. Therefore we can write $m$ as

where $n_1, n_2, \ldots, n_a$ are nonnegative integers and the empty sum is defined to be zero. So, for instance, if $a - 1$ does not appear as a digit in the $q$-adic representation of $m$, then we let $n_{a-1} = 0$ and the second sum in (3.5) is 0. For $0 \leq i \leq a-1$, let $d_i = \left(\sum_{i+1 \leq j \leq a } n_j \right) +1$. By (3.5), $m$ is equal to

Then $(q-1)m + t = \sum_{i = 0}^{a-1} q^{d_i} - a + t$. Since $d_i \geq 1$ for all $i$ and $0 \leq t -a < q-1$, we see that $s_q\left((q-1)m + t \right)$ is equal to

This completes the proof.

Recall that we ^[14] previously gave a characterization for the divisibility $p \mid {pn \choose n}_F$ in terms of the sum of digits function. We are now ready to characterize it in terms of a digital representation. We first prove it for the prime $p \equiv \pm 2 \pmod{5}$ in the next theorem.

Theorem 13. Let $p$ be an odd prime, $p \equiv \pm 2 \pmod{5}$, and $n$ a positive integer. Then $p\mid {pn \choose n}_F$ if and only if $n$ is not of the form

Proof. We first assume that $p\nmid {pn \choose n}_F$. To show that $n$ can be written as in (3.6), let $k = n \bmod{z(p)}$, $t = t(p, k)$, and $H = \bigcup_{i = 0}^{t} H(p, i)$.

Case 1. $p \mid n$. We write $n = p^{a}\ell$ where $a, \ell \in \mathbb{N}$ and $p \nmid \ell$. By Lemma 3, we obtain $n \not \equiv 0 \pmod{z(p)}$. Then by Lemma 4(ⅱ), we have

By Lemma 1, we know that $p \equiv -1 \pmod{z(p)}$, and so $n \equiv pn \equiv -n \pmod{z(p)}$. Therefore $z(p) \mid 2n$ and $z(p) \nmid n$. This implies

Then $k = \frac{z(p)}{2}$, $t = \frac{p-1}{2}$, $\ell = z(p) m_1 + k$ for some $m_1 \geq 0$, and $\left\lfloor\frac{\ell (p-1)}{z(p)}\right\rfloor = (p-1)m_1 + t$. Since $s_p \left((p-1)m_1 + t \right) < p-1$, we obtain by Theorem 12 that $m_1 \in H$. In addition, we obtain that

Since $m_1 \in H$, so is $m$. Hence $n$ is of the form (3.6).

Case 2. $p \nmid n$. This case is similar to Case 1. By Lemmas 3 and 2.4(ⅰ), we obtain

Since $p \equiv -1 \pmod{z(p)}$, $pn \equiv -n \pmod{z(p)}$. Therefore $n \bmod{z(p)} \leq (-n) \bmod{z(p)} = z(p) - \left(n \bmod{z(p)} \right)$. Then $n \bmod{z(p)} \leq \frac{z(p)}{2}$. Then

Since $s_p \left(\left\lfloor \frac{n(p-1)}{z(p)} \right\rfloor \right) < p - 1$, we obtain by Theorem 12 that $m \in H$. Therefore $n$ is of the form (3.6). This proves the converse of this theorem.

For the other direction, assume that $n$ is of the form (3.6). We still let $t = t(p, k)$ and $H = \bigcup_{i = 0}^{t} H(p, i)$, and separate the consideration into two cases.

Case 3. $k < \frac{z(p)}{2}$. Then $0\leq t\leq \frac{p-3}{2}$. Let $m = (a_r a_{r-1} \cdots a_1)_p$ be the $p$-adic expansion of $m$. Since $m \in H$ and $0\leq t \leq \frac{p-3}{2}$, we see that $0\leq a_1\leq \frac{p-3}{2}$. So we obtain by Lemma 7 that

Applying the fact that $p \equiv -1 \pmod{z(p)}$, $n \bmod{z(p)} = k$, and $1 \leq k \leq \frac{z(p)}{2}$, we obtain

Since $m \in H$ and $\left\lfloor \frac{n(p-1)}{z(p)} \right\rfloor = (p-1)m + t$, we obtain by Theorem 12 that

By (3.7), (3.8), (3.9), and Lemma 4(i), we obtain $p\nmid {pn \choose n}_F$.

Case 4. $k = \frac{z(p)}{2}$. Similar to Case 1, we have

If $p\nmid n$, then we obtain by Lemma 4(i) that $p\nmid {pn \choose n}_F$. So suppose that $p\mid n$ and let $a = m\bmod{p}$. Since $m \in H$, we see that $a \leq t = \frac{p-1}{2}$. In addition, $az(p) + k \equiv mz(p) + k \equiv n \equiv 0 \pmod{p}$, so we obtain by Lemma 7 that $a = \frac{p-1}{2}$. Since $m \in H(p, a)$, there are $r \geq 0$ and $m_{2} \in \bigcup_{i = 0}^{\frac{p-3}{2}} H(p, i)$ such that

So we have

Since $m_{2} \in \bigcup_{i = 0}^{\frac{p-3}{2}} H(p, i)\subseteq H$, we obtain by Theorem 12 that

In addition, if $m_{2} \bmod{p} = a_2$, then $0 \leq a_2 \leq \frac{p-3}{2}$ and we obtain by Lemma 7 that

Since $n = (z(p)m_{2} + k)p^{r+1}$ and $z(p)m_{2} + k \not \equiv 0 \pmod{p}$, we obtain $r+1 = \nu_p (n)$. In addition,

Therefore $p\nmid {pn \choose n}_F$, by Lemma 4(ⅱ). This completes the proof.

By Theorem 13, we immediately obtain the following corollary.

Corollary 14. If $p > 2$, $p \equiv \pm 2 \pmod{5}$, and $n\bmod{z(p)} > \frac{z(p)}{2}$, then $p\mid {pn \choose n}_F$.

If $n \bmod{z(p)} < \frac{z(p)}{2}$, then we may still have $p \mid {pn \choose n}_F$ as shown in the next corollary.

Corollary 15. Let $p$ be an odd prime, $p \equiv \pm 2 \pmod{5}$, $p\mid n$, and $n\bmod{z(p)} \neq \frac{z(p)}{2}$. Then $p\mid {pn \choose n}_F$.

Proof. Suppose for a contradiction that $p\nmid {pn \choose n}_F$. Then we obtain by Theorem 13 that $n = z(p)m + k$, $1\leq k \leq \frac{z(p)}{2}$, and $m \in H$ where $H = \bigcup_{i = 0}^{t(p, k)} H(p, i)$. Since $n\bmod{z(p)} \neq \frac{z(p)}{2}$, $k < \frac{z(p)}{2}$. This implies that $t(p, k)\leq \frac{p-3}{2}$. Let $m = (a_r a_{r-1}\cdots a_1)_p$ be the $p$-adic representation of $m$. Since $m \in H$ and $t(p, k)\leq \frac{p-3}{2}$, we see that $0\leq a_1 \leq \frac{p-3}{2}$. By Lemma 7, we obtain $a_1 z(p) + k \not\equiv 0 \pmod{p}$. Therefore

which contradicts the assumption that $p\mid n$. Hence the proof is complete.

Next, we give a characterization for the divisibility $p \mid {pn \choose n}_F$ when $p \equiv \pm 1 \pmod{5}$.

Theorem 16. Let $p$ be an odd prime such that $p \equiv \pm 1 \pmod{5}$ and let $n$ be a positive integer. Then $p\mid {pn \choose n}_F$ if and only if $n$ is not of the form

Proof. Let $A = \frac{n(p-1)}{p^{\nu_p (n)}z(p)}$. Similar to the proof of Theorem 13, we first assume that $p \nmid {pn \choose n}_F$ and let $k = n \bmod{z(p)}$. Then $n = z(p)m + k$ for some $m \geq 0$, and by Lemma 3, $k \neq 0$. So $1 \leq k \leq z(p) - 1$. If remains to show that $m \in \bigcup_{i = 0}^{t(p, k)} H(p, i)$. Since $p \equiv \pm 1 \pmod{5}$, we obtain by Lemma 1 that $z(p) \mid p - 1$. This implies $t(p, k) = \frac{k(p-1)}{z(p)}$. By Lemma 5, we have

By Lemma 12, $m \in \bigcup_{i = 0}^{t(p, k)} H(p, i)$, as required. Next, if $n$ is of the form (3.10), then we apply Theorem 12 to obtain

and then use Lemma 5 to conclude that $p \nmid {pn \choose n}_F$. This completes the proof.

Next we apply Theorems 13 and 16 to determine an asymptotic formula for $E_p (x)$.

Theorem 17. Let $p$ be an odd prime, $p \equiv \pm 2 \pmod{5}$, and $t = t\left(p, \left\lfloor \frac{z(p)}{2} \right\rfloor \right)$. Then uniformly for $x \geq 2$,

and consequently,

where the implied constants depend at most on $p$.

Proof. In this proof, the implied constants in each estimate depend at most on $p$. By Theorem 13, we obtain

For each $1 \leq k \leq \frac{z(p)}{2}$, let $r_k$ be the number of digits in the $p$-adic expansion of $\left\lfloor \frac{x - k}{z(p)} \right\rfloor$ and let $r = r_{\left\lfloor \frac{z(p)}{2} \right\rfloor}$. Then

By (3.11) and Lemma 11,

In addition,

Therefore

Recall that $t = t\left(p, \left\lfloor \frac{z(p)}{2} \right\rfloor \right)$. Since $p\equiv \pm 2 \pmod{5}$, we obtain by Lemma 1 that $z(p) = p + 1$ or $z(p) \leq \frac{p+1}{2}$. If $z(p) = p + 1$, then $z(p)$ is even and for $1 \leq k \leq \frac{z(p)}{2} - 1$,

If $z(p) \leq \frac{p + 1}{2}$, then for $1 \leq k \leq \frac{z(p)}{2} - 1$,

where $s = \left\lfloor \frac{z(p)}{2}\right\rfloor (p - 1) \bmod{z(p)}$. In any case, $t(p, k) \leq t - 1$ for $1 \leq k \leq \frac{z(p)}{2} - 1$. In addition,

Therefore $r_{k}^{t(p, k)} \ll r^{t - 1} \ll (\log x)^{t-1}$ for any $1 \leq k \leq \frac{z(p)}{2} - 1$. Therefore

Thus (3.12) implies that

The rest is now obvious. So the proof is completes.

Theorem 18. Let $p$ be an odd prime, $p \equiv \pm 1 \pmod{5}$, and $t = t\left(p, z(p)-1 \right)$. Then uniformly for $x \geq 2$,

and consequently,

where the implied constants depend at most on $p$.

Proof. The proof is similar to that of Theorem 17, so we omit some details, and the implied constants in the following estimates depend at most on $p$. We obtain by Theorem 16 that

For each $1 \leq k \leq z(p) - 1$, let $r_k$ be the number of digits in the $p$-adic expansion of $\left\lfloor \frac{x - k}{z(p)} \right\rfloor$ and let $r = r_{z(p) - 1}$. Then $r_k = \left\lfloor \log_p \left\lfloor \frac{x - k}{z(p)} \right\rfloor \right\rfloor + 1$ for all $1 \leq k \leq z(p) - 1$. Similar to the proof of Theorem 17, we apply Lemma 11 to obtain

Recall that

Since $p\equiv \pm 1 \pmod{5}$, we obtain by Lemma 1 that $z(p) = p - 1$ or $z(p) \leq \frac{p-1}{2}$. If $z(p) = p - 1$, then for $1 \leq k \leq z(p) - 2$, we have

If $z(p) \leq \frac{p - 1}{2}$, then we obtain by Lemma 2(v) that for $1 \leq k \leq z(p) - 2$,

In any case, $t(p, k) \leq t - 1$ for all $1 \leq k \leq z(p) - 2$. In addition,

Therefore $r_{k}^{t(p, k)} \ll r^{t-1} \ll (\log x)^{t - 1}$ for any $1 \leq k \leq z(p) - 2$. From (3.14) and this observation, we obtain the desired results.

We give an example to show an application of our results. Also see ^[1,9,14] for the characterization of the divisibility $p \mid {pn \choose n}_F$ when $p = 2, 3, 5, 7$.

Example 19. Let $n \in \mathbb{N}$. Then $11 \nmid {11n \choose n}_F$ if and only if

In addition, $13 \nmid {13n \choose n}_F$ if and only if

Furthermore,

Proof. We have $z(11) = 10$, $z(13) = 7$, $t(11, k) = k$ for $1 \leq k \leq 9$, $t(13, k) = 2k - 1$ for $1 \leq k \leq 3$. Applying Theorems 13, 16, 17, and 18, we immediately obtain the desired results.

4.   Conclusions and a future project

We give characterizations for the integers $n \geq 1$ such that ${pn \choose n}_F$ is divisible by $p$ for any prime $p \neq 2, 5$ in terms of the digital representation of $n$. We also obtain asymptotic formulas of $E_p(x)$ for all primes $p$, extending many results in the literature which focus only on small primes $p \leq 7$. For a future project, we may be able to extend the results to ${pn \choose n}_U$ for any nondegenerate fundamental Lucas sequences $U$ and any prime $p$.

Acknowledgments

We thank the referees for their comments which improve the quality of this paper. This project is funded by the National Research Council of Thailand (NRCT), grant number NRCT5-RSA63021-02.

Conflict of interest

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