Explicit Formulas for the p-adic Valuations of Fibonomial Coefficients

Phakhinkon Phunphayap and Prapanpong Pongsriiam^{1} Department of Mathematics Faculty of Science Silpakorn University Nakhon Pathom 73000 Thailand

                     phakhinkon@gmail.com
        pongsriiam p@silpakorn.edu, prapanpong@gmail.com

Abstract We obtain explicit formulas for the p-adic valuations of Fibonomial coefficients

which extend some results in the literature.

1 Introduction

The Fibonacci sequence (F_{n})_{n≥1} is given by the recurrence relation F_{n} = F_{n−1} + F_{n−2} for n ≥ 3 with the initial values F_{1} = F_{2} = 1. For each m ≥ 1 and 1 ≤ k ≤ m, the Fibonomial coefficients \binom{m}{k}_{F} are defined by

\binom{m}{k}_{F} = \frac{F_{1}F_{2}F_{3} ⋯ F_{m}}{(F_{1}F_{2}F_{3} ⋯ F_{k})(F_{1}F_{2}F_{3} ⋯ F_{m−k})} = \frac{F_{m−k+1}F_{m−k+2} ⋯ F_{m}}{F_{1}F_{2}F_{3} ⋯ F_{k}} ,

where F_{n} is the nth Fibonacci number. If k = 0, we define \binom{m}{k}_{F} = 1 and if k > m, we define \binom{m}{k}_{F} = 0. It is well known that \binom{m}{k}_{F} is an integer for all positive integers m and k. So it is natural to consider the divisibility properties and the p-adic valuation of \binom{m}{k}_{F}. As usual, p always denotes a prime and the p-adic valuation (or p-adic order) of a positive integer n, denoted by ν_{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 k such that n | F_{k}. The Fibonacci sequence and the triangle of Fibonomial coefficients are, respectively, \underline{A000045} and \underline{A010048} in OEIS [25]. Also see \underline{A055870} and \underline{A003267} for signed Fibonomial triangle and central Fibonomial coefficients, respectively.

1Prapanpong Pongsriiam receives financial support jointly from The Thailand Research Fund and Faculty of Science Silpakorn University, grant number RSA5980040. Prapanpong Pongsriiam is the corresponding author.

In 1989, Knuth and Wilf [8] gave a short description of the p-adic valuation of \binom{m}{k}_{C} where C is a regularly divisible sequence. However, this does not give explicit formulas for \binom{m}{k}_{F}. Then recently, there has been some interest in explicitly evaluating the p-adic valuation of Fibonomial coefficients of the form \binom{p^{b}}{p^{a}}_{F}. For example, Marques and Trojovský [10, 11] and Marques, Sellers, and Trojovský [12] deal with the case b = a + 1, a ≥ 1. Ballot [2, Theorem 4.2] extends the Kummer-like theorem of Knuth and Wilf [8, Theorem 2], which gives the p-adic valuation of Fibonomials, to all Lucasnomials, and, in particular, uses it to determine explicitly the p-adic valuation of Lucasnomials of the form \binom{p^{b}}{p^{a}}_{U}, for all nondegenerate fundamental Lucas sequences U and all integers b > a ≥ 0, [2, Theorem 7.1].

Note that in the formula given by Marques and Trojovský [11, Theorem 1] for U = F and b = a + 1, only the case of a even is actually explicitly computed. It appears, using the theorem of Ballot [1, Theorem 7.1], that their stated result for a odd is correct only for primes p for which p^{2} does not divide F_{z(p)}, where z(p) is the rank of appearance of p in the Fibonacci sequence. Also see Examples 16 and 18 in this article.

Our purpose is to extend Ballot’s theorem, Theorem 7.1, in the case U = F and b ≥ a > 0 and obtain explicit formulas for \binom{ℓ_{1}p^{b}}{ℓ_{2}p^{a}}_{F}, where ℓ_{1} and ℓ_{2} are arbitrary positive integers such that ℓ_{1}p^{b} > ℓ_{2}p^{a}. This leads us to study the p-adic valuations of integers of the forms

⌊\frac{ℓp^{a}}{m} ⌋! or ⌊\frac{ℓ_{1}p^{b} − ℓ_{2}p^{a}}{m} ⌋!,  (1)

where p ≡ ±1 (mod m). For instance, we obtain in Example 17 the following result: for positive integers a, b, ℓ with b ≥ a, and a prime p distinct from 2 and 5, if p ≡ ±1 (mod 5), then

ν_{p} (\binom{ℓp^{b}}{p^{a}}_{F} ) = \begin{cases}b + ν_{p}(F_{z(p)}) + ν_{p}(ℓ), & if z(p) | ℓ; \\ 0, & otherwise. \end{cases}

Furthermore, if p ≡ ±2 (mod 5), then

ν_{p} (\binom{ℓp^{b}}{p^{a}}_{F} ) = \begin{cases}0, & if ℓ ≡ 1 − 2ε (mod z(p)); \\ b + ν_{p}(F_{z(p)}) + ν_{p}(ℓ), & if ℓ ≡ 0 (mod z(p)); \\ \frac{a}{2}, & if ℓ ≢ 0, 1 − 2ε (mod z(p)) and a is even; \\ \frac{a−1}{2} + ν_{p}(F_{z(p)}), & if ℓ ≢ 0, 1 − 2ε (mod z(p)) and a is odd, \end{cases}

where ε = 1 if a and b have different parity and ε = 0 otherwise. We also obtain the corresponding results for p ∈ {2, 5} in Examples 15 and 19. These extend all the main results in [10, 11, 12] and Ballot’s theorem, Theorem 7.1, in the case U = F.

Recall that for each x ∈ ℝ, ⌊x⌋ is the largest integer less than or equal to x, {x} is the fractional part of x given by {x} = x − ⌊x⌋, and ⌈x⌉ is the smallest integer larger than or equal to x. In addition, we write a mod m to denote the least nonnegative residue of a modulo m. We also use the Iverson notation: if P is a mathematical statement, then

[P] = \begin{cases}1, & if P holds; \\ 0, & otherwise. \end{cases}

For example, [5 ≡ −1 (mod 4)] = 0 and [3 ≡ −1 (mod 4)] = 1.

We organize this article as follows. In Section 2, we give some preliminaries and useful results which are needed in the proof of the main theorems. In Section 3, we give exact formulas for the p-adic valuations of integers (1). In Section 4, we apply the results obtained in Section 3 to Fibonomial coefficients. Our most general theorem is Theorem 13. Finally, in Section 5, we give the p-adic valuations of some specific sub-families of Fibonomial coefficients of type (1), since generally, the more specific the family, the shortest the formula becomes.

For more information related to Fibonacci numbers, we invite the readers to visit the second author’s Researchgate account [23] which contains some freely downloadable versions of his publications [5, 6, 7, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22].

2 Preliminaries and lemmas

Recall that for each odd prime p and a ∈ ℤ, the Legendre symbol (\frac{a}{p}) is defined by

(\frac{a}{p}) = \begin{cases}0, & if p | a; \\ 1, & if a is a quadratic residue of p; \\ −1, & if a is a quadratic nonresidue of p. \end{cases}

Then we have the following result.

Lemma 1. Let p ≠ 5 be a prime and let m and n be positive integers. Then the following statements hold.

(i) If p > 2, then F_{p−( \frac{5}{p})} ≡ 0 (mod p).

(ii) n | F_{m} if and only if z(n) | m.

(iii) z(p) | p + 1 if and only if p ≡ 2 or − 2 (mod 5), and z(p) | p − 1 otherwise.

(iv) gcd(z(p), p) = 1.

Proof. These are well known results. For example, (i) and (ii) can be found in [4, p. 410] and [26], respectively. Then (iii) follows from (i) and (ii). By (iii), z(p) | p ± 1. Since gcd(p, p ± 1) = 1, we obtain gcd(z(p), p) = 1. This proves (iv).

Lengyel’s result and Legendre’s formula given in the following lemmas are important tools in evaluating the p-adic valuation of Fibonomial coefficients. We also refer the reader to [10, 11, 12, 15] for other similar applications of Lengyel’s result.

Lemma 2. (Lengyel [9]) For n ≥ 1, we have

ν_{2}(F_{n}) = \begin{cases}0, & if n ≡ 1, 2 (mod 3); \\ 1, & if n ≡ 3 (mod 6); \\ ν_{2}(n) + 2, & if n ≡ 0 (mod 6), \end{cases}

ν_{5}(F_{n}) = ν_{5}(n), and if p is a prime distinct from 2 and 5, then

ν_{p}(F_{n}) = \begin{cases}ν_{p}(n) + ν_{p}(F_{z(p)}), & if n ≡ 0 (mod z(p)) : \\ 0, & if n ≢ 0 (mod z(p)), \end{cases}

Lemma 3. (Legendre’s formula) Let n be a positive integer and let p be a prime. Then

ν_{p}(n!) = ∑_{k=1}^{∞} ⌊\frac{n}{p^{k}} ⌋ .

In the proof of the main results, we will deal with a lot of calculation involving the floor function. So it is useful to recall the following results.

Lemma 4. For n ∈ ℤ and x ∈ ℝ, the following holds

(i) ⌊n + x⌋ = n + ⌊x⌋,

(ii) {n + x} = {x},

(iii) ⌊x⌋ + ⌊−x⌋ = \begin{cases}−1, & if x ∉ ℤ; \\ 0, & if x ∈ ℤ, \end{cases}

(iv) \{−x\} = \begin{cases}1 − \{x\}, & if x ∉ ℤ; \\ 0, & if x ∈ ℤ, \end{cases}

(v) ⌊x + y⌋ = \begin{cases}⌊x⌋ + ⌊y⌋, & if \{x\} + \{y\} < 1; \\ ⌊x⌋ + ⌊y⌋ + 1, & if \{x\} + \{y\} ≥ 1, \end{cases}

(vi) ⌊\frac{⌊x⌋}{n}⌋ = ⌊\frac{x}{n}⌋ for n ≥ 1.

Proof. These are well-known results and can be proved easily. For more details, see in [1, Exercise 13, p. 72] or in [3, Chapter 3]. We also refer the reader to [14] for a nice application of (v).

The next lemma is used often in counting the number of positive integers n ≤ x lying in a residue class a mod q, see for instance in [24, Proof of Lemma 2.6].

Lemma 5. For x ∈ [1, ∞), a, q ∈ ℤ and q ≥ 1, we have

∑_{\begin{matrix}1≤n≤x \\ n ≡ a (mod q) \end{matrix}} 1 = ⌊\frac{x − a}{q}⌋ − ⌊−\frac{a}{q}⌋ .  (2)

Proof. Replacing a by a + q and applying Lemma 4, we see that the value on the right-hand side of (2) is not changed. Obviously, the left-hand side is also invariant when we replace a by a + q. So it is enough to consider only the case 1 ≤ a ≤ q. Since n ≡ a (mod q), we write n = a + kq where k ≥ 0 and a + kq ≤ x. So k ≤ \frac{x−a}{q}. Therefore

∑_{\begin{matrix}1≤n≤x \\ n ≡ a (mod q) \end{matrix}} 1 = ∑_{0≤k≤ \frac{x−a}{q}} 1 = ⌊\frac{x − a}{q}⌋ + 1 = ⌊\frac{x − a}{q}⌋ − ⌊−\frac{a}{q}⌋ .

It is convenient to use the Iverson notation and to denote the least nonnegative residue of a modulo m by a mod m. Therefore we will do so from this point on.

Lemma 6. Let n and k be integers, m a positive integer, r = n mod m, and s = k mod m. Then

⌊\frac{n − k}{m} ⌋ = ⌊\frac{n}{m}⌋ − ⌊\frac{k}{m}⌋ − [r < s].

Proof. By Lemma 4(i) and the fact that 0 ≤ r < m, we obtain

⌊\frac{n}{m}⌋ = ⌊\frac{n − r}{m} + \frac{r}{m}⌋ = \frac{n − r}{m} + ⌊\frac{r}{m}⌋ = \frac{n − r}{m}.

Similarly, ⌊\frac{k}{m}⌋ = \frac{k−s}{m}. Therefore ⌊\frac{n−k}{m}⌋ is equal to

⌊\frac{n − r}{m} − \frac{k − s}{m} + \frac{r − s}{m}⌋ = \frac{n − r}{m} − \frac{k − s}{m} + ⌊\frac{r − s}{m}⌋ = \begin{cases}⌊\frac{n}{m}⌋ − ⌊\frac{k}{m}⌋ , & if r ≥ s; \\ ⌊\frac{n}{m}⌋ − ⌊\frac{k}{m}⌋ − 1, & if r < s. \end{cases}

3 The p-adic valuation of integers in special forms

In this section, we calculate the p-adic valuation of ⌊\frac{ℓp^{a}}{m} ⌋! and other integers in similar forms.

Theorem 7. Let p be a prime and let a ≥ 0, ℓ ≥ 0, and m ≥ 1 be integers. Assume that p ≡ ±1 (mod m) and let δ = [ℓ ≢ 0 (mod m)]. Then

ν_{p} (⌊\frac{ℓp^{a}}{m} ⌋!) = \begin{cases}\frac{ℓ(p^{a}−1)}{m(p−1)} − a \{\frac{ℓ}{m}\} + ν_{p} (⌊\frac{ℓ}{m}⌋!) , & if p ≡ 1 (mod m); \\ \frac{ℓ(p^{a}−1)}{m(p−1)} − \frac{a}{2}δ + ν_{p} (⌊\frac{ℓ}{m}⌋!) , & if p ≡ −1 (mod m) and a is even; \\ \frac{ℓ(p^{a}−1)}{m(p−1)} − \frac{a−1}{2}δ − \{\frac{ℓ}{m}\} + ν_{p} (⌊\frac{ℓ}{m}⌋!) , & if p ≡ −1 (mod m) and a is odd. \end{cases}

We remark that if m = 1 or 2, then the expressions in each case of this theorem are all equal.

Proof. The result is easily verified when a = 0 or ℓ = 0. So we assume throughout that a ≥ 1 and ℓ ≥ 1. We also use Lemmas 4(i), 4(vi), and 5 repeatedly without reference. By Legendre’s formula, we obtain

ν_{p} (⌊\frac{ℓp^{a}}{m} ⌋!) = ∑_{j=1}^{∞} ⌊\frac{ℓp^{a}}{mp^{j}} ⌋ = ∑_{j=1}^{a} ⌊\frac{ℓp^{a−j}}{m} ⌋ + ∑_{j=a+1}^{∞} ⌊\frac{ℓp^{a−j}}{m} ⌋ = ∑_{j=1}^{a} ⌊\frac{ℓp^{a−j}}{m} ⌋ + ν_{p} (⌊\frac{ℓ}{m}⌋!) .

(3)

From (3), it is immediate that for m = 1, we obtain

ν_{p} ((ℓp^{a})!) = \frac{ℓ(p^{a} − 1)}{p − 1} + ν_{p}(ℓ!).

So we assume throughout that m ≥ 2.

Case 1. p ≡ 1 (mod m). Then, for every k ≥ 0, p^{k} ≡ 1 (mod m) and

⌊\frac{ℓp^{k}}{m} ⌋ = ⌊\frac{ℓ(p^{k} − 1)}{m} + \frac{ℓ}{m}⌋ = \frac{ℓ(p^{k} − 1)}{m} + ⌊\frac{ℓ}{m}⌋ .

Therefore the sum ∑_{j=1}^{a} ⌊\frac{ℓp^{a−j}}{m} ⌋ appearing in (3) is equal to

∑_{j=1}^{a} (\frac{ℓ(p^{a−j} − 1)}{m} + ⌊\frac{ℓ}{m}⌋) = (\frac{ℓ}{m} ∑_{j=1}^{a} p^{a−j}) − a (\frac{ℓ}{m} − ⌊\frac{ℓ}{m}⌋) = \frac{ℓ(p^{a} − 1)}{m(p − 1)} − a \{\frac{ℓ}{m}\} .

Case 2. p ≡ −1 (mod m). Then for k ≥ 0, we have p^{k} ≡ 1 (mod m) if k is even, and p^{k} ≡ −1 (mod m) if k is odd. Therefore

⌊\frac{ℓp^{k}}{m} ⌋ = ⌊\frac{ℓ(p^{k} − 1)}{m} + \frac{ℓ}{m}⌋ = \frac{ℓ(p^{k} − 1)}{m} + ⌊\frac{ℓ}{m}⌋ if k ≥ 0 and k is even,
⌊\frac{ℓp^{k}}{m} ⌋ = ⌊\frac{ℓ(p^{k} + 1)}{m} − \frac{ℓ}{m}⌋ = \frac{ℓ(p^{k} + 1)}{m} + ⌊−\frac{ℓ}{m}⌋ if k ≥ 0 and k is odd.

Therefore the sum ∑_{j=1}^{a} ⌊\frac{ℓp^{a−j}}{m} ⌋ appearing in (3) is equal to

∑_{\begin{matrix}1≤j≤a \\ a−j ≡ 0 (mod 2) \end{matrix}} (\frac{ℓ(p^{a−j} − 1)}{m} + ⌊\frac{ℓ}{m}⌋) + ∑_{\begin{matrix}1≤j≤a \\ a−j ≡ 1 (mod 2) \end{matrix}} (\frac{ℓ(p^{a−j} + 1)}{m} + ⌊−\frac{ℓ}{m}⌋)

= \frac{ℓ}{m} ∑_{1≤j≤a} p^{a−j} − ∑_{\begin{matrix}1≤j≤a \\ j ≡a (mod 2) \end{matrix}} (\frac{ℓ}{m} − ⌊\frac{ℓ}{m}⌋) + ∑_{\begin{matrix}1≤j≤a \\ j ≡ a−1 (mod 2) \end{matrix}} (\frac{ℓ}{m} + ⌊−\frac{ℓ}{m}⌋)

= \frac{ℓ(p^{a} − 1)}{m(p − 1)} + ⌊−\frac{a}{2}⌋ (\frac{ℓ}{m} − ⌊\frac{ℓ}{m}⌋) − ⌊−\frac{a − 1}{2}⌋ (\frac{ℓ}{m} + ⌊−\frac{ℓ}{m}⌋) .  (4)

By Lemma 4(iii), we see that

⌊\frac{ℓ}{m}⌋ + ⌊−\frac{ℓ}{m}⌋ = −[ℓ ≢ 0 (mod m)] = −δ.

Therefore if a is even, then (4) is equal to

\frac{ℓ(p^{a} − 1)}{m(p − 1)} − \frac{a}{2} (\frac{ℓ}{m} − ⌊\frac{ℓ}{m}⌋) + \frac{a}{2} (\frac{ℓ}{m} + ⌊−\frac{ℓ}{m}⌋)

= \frac{ℓ(p^{a} − 1)}{m(p − 1)} + \frac{a}{2} (⌊\frac{ℓ}{m}⌋ + ⌊−\frac{ℓ}{m}⌋) = \frac{ℓ(p^{a} − 1)}{m(p − 1)} − \frac{a}{2}δ,

and if a is odd, then (4) is equal to

\frac{ℓ(p^{a} − 1)}{m(p − 1)} − \frac{a + 1}{2} (\frac{ℓ}{m} − ⌊\frac{ℓ}{m}⌋) + \frac{a − 1}{2} (\frac{ℓ}{m} + ⌊−\frac{ℓ}{m}⌋)

= \frac{ℓ(p^{a} − 1)}{m(p − 1)} + \frac{a − 1}{2} (⌊\frac{ℓ}{m}⌋ + ⌊−\frac{ℓ}{m}⌋) + ⌊\frac{ℓ}{m}⌋ − \frac{ℓ}{m}

= \frac{ℓ(p^{a} − 1)}{m(p − 1)} − (\frac{a − 1}{2}) δ − \{\frac{ℓ}{m}\} .

This completes the proof.

We can combine every case in Theorem 7 into a single form as given in the next corollary. Corollary 8. Assume that p, a, ℓ, m, and δ satisfy the same assumptions as in Theorem 7. Then the p-adic valuation of ⌊\frac{ℓp^{a}}{m} ⌋! is

\frac{ℓ(p^{a} − 1)}{m(p − 1)} − ⌊\frac{a}{2}⌋ δ − \{\frac{ℓ}{m}\} [a ≡ 1 (mod 2)]

+ δ ⌊\frac{a}{2}⌋ (1 − 2 \{\frac{ℓ}{m}\}) [p ≡ 1 (mod m)] + ν_{p} (⌊\frac{ℓ}{m}⌋!) .  (5)

Proof. This is merely a combination of each case from Theorem 7. For example, when p ≡ −1 (mod m), the right-hand side of (5) reduces to

\frac{ℓ(p^{a} − 1)}{m(p − 1)} − ⌊\frac{a}{2}⌋ δ − \{\frac{ℓ}{m}\} [a ≡ 1 (mod 2)] + ν_{p} (⌊\frac{ℓ}{m}⌋!)

= \begin{cases}\frac{ℓ(p^{a}−1)}{m(p−1)} − \frac{a}{2}δ + ν_{p} (⌊\frac{ℓ}{m}⌋!) , & if a is even; \\ \frac{ℓ(p^{a}−1)}{m(p−1)} − (\frac{a−1}{2}) δ − \{\frac{ℓ}{m}\} + ν_{p} (⌊\frac{ℓ}{m}⌋!) , & if a is odd, \end{cases}

which is the same as Theorem 7. The other cases are similar. We leave the details to the reader.

Next we deal with the p-adic valuation of an integer of the form ⌊\frac{ℓ_{1}p^{b}−ℓ_{2}p^{a}}{m} ⌋! where a, b, ℓ_{1}, ℓ_{2}, and m are positive integers. It is natural to assume ℓ_{1}p^{b} − ℓ_{2}p^{a} > 0. In addition, if a = b, then the above expression is reduced to ⌊\frac{(ℓ_{1}−ℓ_{2})p^{b}}{m} ⌋!, which can be evaluated by using Theorem 7. We consider the case b ≥ a in Theorem 9 and the other case in Theorem 10.

Theorem 9. Let p be a prime, let a be a nonnegative integer, and let b, m, ℓ_{1}, ℓ_{2} be positive integers satisfying b ≥ a and ℓ_{1}p^{b} − ℓ_{2}p^{a} > 0. Assume that p ≡ ±1 (mod m). Then the following statements hold.

(i) If p ≡ 1 (mod m), then

ν_{p} (⌊\frac{ℓ_{1}p^{b} − ℓ_{2}p^{a}}{m} ⌋!) = \frac{(ℓ_{1}p^{b−a} − ℓ_{2})(p^{a} − 1)}{m(p − 1)} − a \{\frac{ℓ_{1} − ℓ_{2}}{m} \} + ν_{p} (⌊\frac{ℓ_{1}p^{b−a} − ℓ_{2}}{m} ⌋!) .

(ii) If p ≡ −1 (mod m) and a ≡ b (mod 2), then

ν_{p} (⌊\frac{ℓ_{1}p^{b} − ℓ_{2}p^{a}}{m} ⌋!) = \frac{(ℓ_{1}p^{b−a} − ℓ_{2})(p^{a} − 1)}{m(p − 1)} − \{\frac{ℓ_{1} − ℓ_{2}}{m} \} [a ≡ 1 (mod 2)]

− ⌊\frac{a}{2}⌋ [ℓ_{1} ≢ ℓ_{2} (mod m)] + ν_{p} (⌊\frac{ℓ_{1}p^{b−a} − ℓ_{2}}{m} ⌋!) .

(iii) If p ≡ −1 (mod m) and a ≢ b (mod 2), then

ν_{p} (⌊\frac{ℓ_{1}p^{b} − ℓ_{2}p^{a}}{m} ⌋!) = \frac{(ℓ_{1}p^{b−a} − ℓ_{2})(p^{a} − 1)}{m(p − 1)} − \{−\frac{ℓ_{1} + ℓ_{2}}{m} \} [a ≡ 1 (mod 2)]

− ⌊\frac{a}{2}⌋ [ℓ_{1} ≢ −ℓ_{2} (mod m)] + ν_{p} (⌊\frac{ℓ_{1}p^{b−a} − ℓ_{2}}{m} ⌋!) .

We remark that if m = 1, the expressions in each case of this theorem are equal.

Proof. The result is easily checked when a = 0, and as discussed above, if b = a, then the result can be verified using Theorem 7. So we assume throughout that a ≥ 1 and b > a. Similar to the proof of Theorem 7, we use Lemmas 4(i), 4(vi), and 5 repeatedly without reference. Then, as for (3), we obtain

ν_{p} (⌊\frac{ℓ_{1}p^{b} − ℓ_{2}p^{a}}{m} ⌋!) = ∑_{j=1}^{a} ⌊\frac{ℓ_{1}p^{b−j} − ℓ_{2}p^{a−j}}{m} ⌋ + ∑_{j=a+1}^{∞} ⌊\frac{ℓ_{1}p^{b−j} − ℓ_{2}p^{a−j}}{m} ⌋

= ∑_{j=1}^{a} ⌊\frac{ℓ_{1}p^{b−j} − ℓ_{2}p^{a−j}}{m} ⌋ + ν_{p} (⌊\frac{ℓ_{1}p^{b−a} − ℓ_{2}}{m} ⌋!) .  (6)

We see that when m = 1, (6) becomes

ν_{p} ((ℓ_{1}p^{b} − ℓ_{2}p^{a})!) = \frac{(ℓ_{1}p^{b−a} − ℓ_{2})(p^{a} − 1)}{p − 1} + ν_{p} ((ℓ_{1}p^{b−a} − ℓ_{2})!) .

So assume throughout that m ≥ 2. We begin with the proof of (i). Suppose that p ≡ 1 (mod m). For each 1 ≤ j ≤ a, we have

⌊\frac{ℓ_{1}p^{b−j} − ℓ_{2}p^{a−j}}{m} ⌋ = ⌊\frac{ℓ_{1}p^{b−j} − ℓ_{1}}{m} − \frac{ℓ_{2}p^{a−j} − ℓ_{2}}{m} + \frac{ℓ_{1} − ℓ_{2}}{m} ⌋

= \frac{ℓ_{1}p^{b−j} − ℓ_{2}p^{a−j}}{m} − \frac{ℓ_{1} − ℓ_{2}}{m} + ⌊\frac{ℓ_{1} − ℓ_{2}}{m} ⌋ .

Then the sum ∑_{j=1}^{a} ⌊\frac{ℓ_{1}p^{b−j}−ℓ_{2}p^{a−j}}{m} ⌋ appearing in (6) is equal to

\frac{ℓ_{1}}{m} ∑_{1≤j≤a} p^{b−j} − \frac{ℓ_{2}}{m} ∑_{1≤j≤a} p^{a−j} − a (\frac{ℓ_{1} − ℓ_{2}}{m} − ⌊\frac{ℓ_{1} − ℓ_{2}}{m} ⌋)

= \frac{ℓ_{1}}{m} (\frac{p^{b−a}(p^{a} − 1)}{p − 1}) − \frac{ℓ_{2}}{m} (\frac{p^{a} − 1}{p − 1}) − a \{\frac{ℓ_{1} − ℓ_{2}}{m} \}

= \frac{(ℓ_{1}p^{b−a} − ℓ_{2})(p^{a} − 1)}{m(p − 1)} − a \{\frac{ℓ_{1} − ℓ_{2}}{m} \} .

This proves (i). So from this point on, we assume that p ≡ −1 (mod m). For each 1 ≤ j ≤ a, we have

⌊\frac{ℓ_{1}p^{b−j} − ℓ_{2}p^{a−j}}{m} ⌋ = ⌊\frac{ℓ_{1}p^{b−j} − (−1)^{b−j}ℓ_{1}}{m} − \frac{ℓ_{2}p^{a−j} − (−1)^{a−j}ℓ_{2}}{m} + \frac{(−1)^{b−j}ℓ_{1} − (−1)^{a−j}ℓ_{2}}{m} ⌋

= \frac{ℓ_{1}p^{b−j} − ℓ_{2}p^{a−j}}{m} − \frac{(−1)^{b−j}ℓ_{1} − (−1)^{a−j}ℓ_{2}}{m} + ⌊\frac{(−1)^{b−j}ℓ_{1} − (−1)^{a−j}ℓ_{2}}{m} ⌋

= \begin{cases}\frac{ℓ_{1}p^{b−j}−ℓ_{2}p^{a−j}}{m} − \frac{(−1)^{b−j}(ℓ_{1}−ℓ_{2})}{m} + ⌊\frac{(−1)^{b−j}(ℓ_{1}−ℓ_{2})}{m}⌋ , & if a ≡ b (mod 2); \\ \frac{ℓ_{1}p^{b−j}−ℓ_{2}p^{a−j}}{m} − \frac{(−1)^{b−j}(ℓ_{1}+ℓ_{2})}{m} + ⌊\frac{(−1)^{b−j}(ℓ_{1}+ℓ_{2})}{m}⌋ , & if a ≢ b (mod 2). \end{cases}

Case 1. a ≡ b (mod 2). Then the sum ∑_{j=1}^{a} ⌊\frac{ℓ_{1}p^{b−j}−ℓ_{2}p^{a−j}}{m} ⌋ appearing in (6) is equal to

\frac{ℓ_{1}}{m} ∑_{1≤j≤a} p^{b−j} − \frac{ℓ_{2}}{m} ∑_{1≤j≤a} p^{a−j} − (\frac{ℓ_{1} − ℓ_{2}}{m} ) ∑_{1≤j≤a}(−1)^{b−j} + ∑_{1≤j≤a} ⌊\frac{(−1)^{b−j}(ℓ_{1} − ℓ_{2})}{m}⌋

= \frac{(ℓ_{1}p^{b−a} − ℓ_{2})(p^{a} − 1)}{m(p − 1)} − (\frac{ℓ_{1} − ℓ_{2}}{m} ) ∑_{1≤j≤a}(−1)^{b−j} + ∑_{1≤j≤a} ⌊\frac{(−1)^{b−j}(ℓ_{1} − ℓ_{2})}{m}⌋ .  (7)

Observe that

∑_{1≤j≤a}(−1)^{b−j} = \begin{cases}0, & if a is even; \\ 1, & if a is odd. \end{cases}

So we have

(\frac{ℓ_{1} − ℓ_{2}}{m} ) ∑_{1≤j≤a}(−1)^{b−j} = (\frac{ℓ_{1} − ℓ_{2}}{m} ) [a ≡ 1 (mod 2)].

It remains to calculate the last term in (7). If ℓ_{1} ≡ ℓ_{2} (mod m), then we obtain by Lemma 4(iii) that

∑_{1≤j≤a} ⌊\frac{(−1)^{b−j}(ℓ_{1} − ℓ_{2})}{m}⌋ = \begin{cases}0, & if a is even; \\ ⌊\frac{ℓ_{1}−ℓ_{2}}{m} ⌋ , & if a is odd; \end{cases}

= ⌊\frac{ℓ_{1} − ℓ_{2}}{m} ⌋ [a ≡ 1 (mod 2)].

Similarly, if ℓ_{1} ≢ ℓ_{2} (mod m), then we obtain by Lemma 4(iii) that

∑_{1≤j≤a} ⌊\frac{(−1)^{b−j}(ℓ_{1} − ℓ_{2})}{m}⌋ = \begin{cases}−\frac{a}{2}, & if a is even; \\ ⌊\frac{ℓ_{1}−ℓ_{2}}{m} ⌋ − \frac{a−1}{2}, & if a is odd; \end{cases}

= ⌊\frac{ℓ_{1} − ℓ_{2}}{m} ⌋ [a ≡ 1 (mod 2)] − ⌊\frac{a}{2}⌋ .

In any case,

∑_{1≤j≤a} ⌊\frac{(−1)^{b−j}(ℓ_{1} − ℓ_{2})}{m}⌋ = ⌊\frac{ℓ_{1} − ℓ_{2}}{m} ⌋ [a ≡ 1 (mod 2)] − ⌊\frac{a}{2}⌋ [ℓ_{1} ≢ ℓ_{2} (mod m)].

Therefore (7) is equal to

\frac{(ℓ_{1}p^{b−a} − ℓ_{2})(p^{a} − 1)}{m(p − 1)} − (\frac{ℓ_{1} − ℓ_{2}}{m} ) [a ≡ 1 (mod 2)] + ⌊\frac{ℓ_{1} − ℓ_{2}}{m} ⌋ [a ≡ 1 (mod 2)]

− ⌊\frac{a}{2}⌋ [ℓ_{1} ≢ ℓ_{2} (mod m)]

= \frac{(ℓ_{1}p^{b−a} − ℓ_{2})(p^{a} − 1)}{m(p − 1)} − \{\frac{ℓ_{1} − ℓ_{2}}{m} \} [a ≡ 1 (mod 2)] − ⌊\frac{a}{2}⌋ [ℓ_{1} ≢ ℓ_{2} (mod m)].

This proves (ii). Next we prove (iii).

Case 2. a ≢ b (mod 2). Similar to Case 1, the sum ∑_{j=1}^{a} ⌊\frac{ℓ_{1}p^{b−j}−ℓ_{2}p^{a−j}}{m} ⌋ appearing in (6) is equal to

\frac{(ℓ_{1}p^{b−a} − ℓ_{2})(p^{a} − 1)}{m(p − 1)} − (\frac{ℓ_{1} + ℓ_{2}}{m} ) ∑_{1≤j≤a}(−1)^{b−j} + ∑_{1≤j≤a} ⌊\frac{(−1)^{b−j}(ℓ_{1} + ℓ_{2})}{m}⌋

= \frac{(ℓ_{1}p^{b−a} − ℓ_{2})(p^{a} − 1)}{m(p − 1)} + (\frac{ℓ_{1} + ℓ_{2}}{m} ) [a ≡ 1 (mod 2)] + ∑_{1≤j≤a} ⌊\frac{(−1)^{b−j}(ℓ_{1} + ℓ_{2})}{m}⌋ . (8)

If ℓ_{1} ≡ −ℓ_{2} (mod m), then we obtain by Lemma 4(iii) that

∑_{1≤j≤a} ⌊\frac{(−1)^{b−j}(ℓ_{1} + ℓ_{2})}{m}⌋ = \begin{cases}0, & if a is even; \\ ⌊−\frac{ℓ_{1}+ℓ_{2}}{m} ⌋ , & if a is odd; \end{cases}

= ⌊−\frac{ℓ_{1} + ℓ_{2}}{m} ⌋ [a ≡ 1 (mod 2)].

Similarly, if ℓ_{1} ≢ −ℓ_{2} (mod m), then we obtain by Lemma 4(iii) that

∑_{1≤j≤a} ⌊\frac{(−1)^{b−j}(ℓ_{1} + ℓ_{2})}{m}⌋ = \begin{cases}−\frac{a}{2}, & if a is even; \\ ⌊−\frac{ℓ_{1}+ℓ_{2}}{m} ⌋ − \frac{a−1}{2}, & if a is odd; \end{cases}

= ⌊−\frac{ℓ_{1} + ℓ_{2}}{m} ⌋ [a ≡ 1 (mod 2)] − ⌊\frac{a}{2}⌋ .

In any case, ∑_{1≤j≤a} ⌊\frac{(−1)^{b−j}(ℓ_{1}+ℓ_{2})}{m}⌋ = ⌊−\frac{ℓ_{1}+ℓ_{2}}{m} ⌋ [a ≡ 1 (mod 2)] − ⌊\frac{a}{2}⌋ [ℓ_{1} ≢ −ℓ_{2} (mod m)]. Therefore (8) is equal to

\frac{(ℓ_{1}p^{b−a} − ℓ_{2})(p^{a} − 1)}{m(p − 1)} + (\frac{ℓ_{1} + ℓ_{2}}{m} ) [a ≡ 1 (mod 2)] + ⌊−\frac{ℓ_{1} + ℓ_{2}}{m} ⌋ [a ≡ 1 (mod 2)]

− ⌊\frac{a}{2}⌋ [ℓ_{1} ≢ −ℓ_{2} (mod m)]

= \frac{(ℓ_{1}p^{b−a} − ℓ_{2})(p^{a} − 1)}{m(p − 1)} − \{−\frac{ℓ_{1} + ℓ_{2}}{m} \} [a ≡ 1 (mod 2)] − ⌊\frac{a}{2}⌋ [ℓ_{1} ≢ −ℓ_{2} (mod m)].

This completes the proof.

Next we replace the assumption b ≥ a in Theorem 9 by b < a. The calculation follows from the same idea so we skip the details of the proof. Although we do not use it in this article, it may be useful for future reference. So we record it in the next theorem.

Theorem 10. Let p be a prime, let b be a nonnegative integer, and let a, m, ℓ_{1}, ℓ_{2} be positive integers satisfying b < a and ℓ_{1}p^{b} − ℓ_{2}p^{a} > 0. Assume that p ≡ ±1 (mod m). Then the following statements hold.

(i) If p ≡ 1 (mod m), then

ν_{p} (⌊\frac{ℓ_{1}p^{b} − ℓ_{2}p^{a}}{m} ⌋!) = \frac{(ℓ_{1} − ℓ_{2}p^{a−b})(p^{b} − 1)}{m(p − 1)} − b \{\frac{ℓ_{1} − ℓ_{2}}{m} \} + ν_{p} (⌊\frac{ℓ_{1} − ℓ_{2}p^{a−b}}{m} ⌋!) .

(ii) If p ≡ −1 (mod m) and a ≡ b (mod 2), then

ν_{p} (⌊\frac{ℓ_{1}p^{b} − ℓ_{2}p^{a}}{m} ⌋!) = \frac{(ℓ_{1} − ℓ_{2}p^{a−b})(p^{b} − 1)}{m(p − 1)} − \{\frac{ℓ_{1} − ℓ_{2}}{m} \} [b ≡ 1 (mod 2)]

− ⌊\frac{b}{2}⌋ [ℓ_{1} ≢ ℓ_{2} (mod m)] + ν_{p} (⌊\frac{ℓ_{1} − ℓ_{2}p^{a−b}}{m} ⌋!) .

(iii) If p ≡ −1 (mod m) and a ≢ b (mod 2), then

ν_{p} (⌊\frac{ℓ_{1}p^{b} − ℓ_{2}p^{a}}{m} ⌋!) = \frac{(ℓ_{1} − ℓ_{2}p^{a−b})(p^{b} − 1)}{m(p − 1)} − \{\frac{ℓ_{1} + ℓ_{2}}{m} \} [b ≡ 1 (mod 2)]

− ⌊\frac{b}{2}⌋ [ℓ_{1} ≢ ℓ_{2} (mod m)] + ν_{p} (⌊\frac{ℓ_{1} − ℓ_{2}p^{a−b}}{m} ⌋!) .

Proof. We begin by writing ν_{p} (⌊\frac{ℓ_{1}p^{b}−ℓ_{2}p^{a}}{m} ⌋!) as

∑_{j=1}^{b} ⌊\frac{ℓ_{1}p^{b−j} − ℓ_{2}p^{a−j}}{m} ⌋ + ∑_{j=b+1}^{∞} ⌊\frac{ℓ_{1}p^{b−j} − ℓ_{2}p^{a−j}}{m} ⌋ .

The second sum above is ν_{p} (⌊\frac{ℓ_{1}−ℓ_{2}p^{a−b}}{m} ⌋!). The first sum can be evaluated in the same way as in Theorem 9. We leave the details to the reader.

When we put more restrictions on the range of ℓ_{1} and ℓ_{2}, the expression ν_{p} (⌊\frac{ℓ_{1}p^{b−a}−ℓ_{2}}{m} ⌋!) appearing in Theorems 9 and 10 can be evaluated further. Nevertheless, since we do not need it in our application, we do not give them here. In the future, we plan to put it in the second author’s Researchgate account. So the interested reader can find it there.

4 The p-adic valuations of Fibonomial coefficients

Recall that the binomial coefficients \binom{m}{k} is defined by

\binom{m}{k} = \begin{cases}\frac{m!}{k!(m−k)!}, & if 0 ≤ k ≤ m; \\ 0, & if k < 0 or k > m. \end{cases}

A classical result of Kummer states that for 0 ≤ k ≤ m, ν_{p} (\binom{m}{k}) is equal to the number of carries when we add k and m − k in base p. From this, it is not difficult to show that for all primes p and positive integers k, b, a with b ≥ a, we have

ν_{p} (\binom{p^{b}}{p^{a}}) = b − a, or more generally, ν_{p} (\binom{p^{a}}{k}) = a − ν_{p}(k).

Knuth and Wilf [8] also obtain the result analogous to that of Kummer for a C-nomial coefficient. However, our purpose is to obtain ν_{p} (\binom{m}{k}_{F} ) is an explicit form. So we first express ν_{p} (\binom{m}{k}_{F} ) in terms of the p-adic valuation of some binomial coefficients in Theorem 11. Then we write it in a form which is easy to use in Corollary 12. Then we apply it to obtain the p-adic valuation of Fibonomial coefficients of the form \binom{ℓ_{1}p^{b}}{ℓ_{2}p^{a}}_{F}.

Theorem 11. Let 0 ≤ k ≤ m be integers. Then the following statements hold.

(i) Let m′ = ⌊\frac{m}{6}⌋, k′ = ⌊\frac{k}{6}⌋, and let r = m mod 6 and s = k mod 6 be the least nonnegative residues of m and k modulo 6, respectively. Then

ν_{2} (\binom{m}{k}_{F} ) = ν_{2} (\binom{m′}{k′}) + ⌊\frac{r + 3}{6}⌋ − ⌊\frac{r − s + 3}{6}⌋ − ⌊\frac{s + 3}{6}⌋ − 3 ⌊\frac{r − s}{6}⌋

+ [r < s]ν_{2} (⌊\frac{m − k + 6}{6}⌋) .

(ii) ν_{5} (\binom{m}{k}_{F} ) = ν_{5} (\binom{m}{k}).

(iii) Suppose that p is a prime, p ≠ 2, and p ≠ 5. Let m′ = ⌊\frac{m}{z(p)}⌋, k′ = ⌊\frac{k}{z(p)}⌋, and let r = m mod z(p), and s = k mod z(p) be the least nonnegative residues of m and k modulo z(p), respectively. Then

ν_{p} (\binom{m}{k}_{F} ) = ν_{p} (\binom{m′}{k′}) + [r < s] (ν_{p} (⌊\frac{m − k + z(p)}{z(p)}⌋) + ν_{p}(F_{z(p)})) .

Proof. We will use Lemmas 4(i) and 5 repeatedly without reference. In addition, it is useful to recall that for every a, b ∈ ℕ, ν_{p}(ab) = ν_{p}(a)+ν_{p}(b) and if b | a, then ν_{p}(\frac{a}{b}) = ν_{p}(a)−ν_{p}(b). Since the formulas to prove clearly hold when k = 0 or m, we assume m ≥ 2 and 1 ≤ k < m.

By Lemma 2, we obtain, for every ℓ ≥ 1,

ν_{2}(F_{1}F_{2}F_{3} ⋯ F_{ℓ}) = ∑_{\begin{matrix}1≤n≤ℓ \\ n ≡ 3 (mod 6) \end{matrix}} ν_{2}(F_{n}) + ∑_{\begin{matrix}1≤n≤ℓ \\ n ≡ 0 (mod 6) \end{matrix}} ν_{2}(F_{n})

= ∑_{\begin{matrix}1≤n≤ℓ \\ n ≡ 3 (mod 6) \end{matrix}} 1 + ∑_{\begin{matrix}1≤n≤ℓ \\ n ≡ 0 (mod 6) \end{matrix}}(ν_{2}(n) + 2)

= ⌊\frac{ℓ + 3}{6}⌋ + 2 ⌊\frac{ℓ}{6}⌋ + ∑_{1≤j≤ \frac{ℓ}{6}} ν_{2}(6j)

= ⌊\frac{ℓ + 3}{6}⌋ + 3 ⌊\frac{ℓ}{6}⌋ + ∑_{1≤j≤ \frac{ℓ}{6}} ν_{2}(j)

= ⌊\frac{ℓ + 3}{6}⌋ + 3 ⌊\frac{ℓ}{6}⌋ + ν_{2} (⌊\frac{ℓ}{6}⌋!) .  (9)

Then we obtain from the definition of \binom{m}{k}_{F} and from (9) that

ν_{2} (\binom{m}{k}_{F} ) = ν_{2}(F_{1}F_{2} ⋯ F_{m}) − ν_{2}(F_{1}F_{2} ⋯ F_{m−k}) − ν_{2}(F_{1}F_{2} ⋯ F_{k})

= (⌊\frac{m + 3}{6}⌋ − ⌊\frac{m − k + 3}{6}⌋ − ⌊\frac{k + 3}{6}⌋) + 3 (⌊\frac{m}{6}⌋ − ⌊\frac{m − k}{6} ⌋ − ⌊\frac{k}{6}⌋)

+ ν_{2} (⌊\frac{m}{6}⌋!) − ν_{2} (⌊\frac{m − k}{6} ⌋!) − ν_{2} (⌊\frac{k}{6}⌋!) .  (10)

The expression in the first parenthesis in (10) is equal to

⌊\frac{m − r}{6} + \frac{r + 3}{6}⌋ − ⌊\frac{(m − r) − (k − s)}{6} + \frac{r − s + 3}{6}⌋ − ⌊\frac{k − s}{6} + \frac{s + 3}{6}⌋

= \frac{m − r}{6} + ⌊\frac{r + 3}{6}⌋ − \frac{(m − r) − (k − s)}{6} − ⌊\frac{r − s + 3}{6}⌋ − \frac{k − s}{6} − ⌊\frac{s + 3}{6}⌋

= ⌊\frac{r + 3}{6}⌋ − ⌊\frac{r − s + 3}{6}⌋ − ⌊\frac{s + 3}{6}⌋ .

Similarly, the expression in the second parenthesis is

3 (⌊\frac{r}{6}⌋ − ⌊\frac{r − s}{6}⌋ − ⌊\frac{s}{6}⌋) = −3 ⌊\frac{r − s}{6}⌋ .

Therefore (10) becomes

ν_{2} (\binom{m}{k}_{F} ) = ⌊\frac{r + 3}{6}⌋ − ⌊\frac{r − s + 3}{6}⌋ − ⌊\frac{s + 3}{6}⌋ − 3 ⌊\frac{r − s}{6}⌋ + ν_{2} (\frac{⌊x + y⌋!}{⌊x⌋!⌊y⌋!}) (11)

where x = \frac{m−k}{6} and y = \frac{k}{6}. By Lemma 4(v), we see that

\frac{⌊x + y⌋!}{⌊x⌋!⌊y⌋!} = \begin{cases}\binom{⌊x+y⌋}{⌊y⌋}, & if \{x\} + \{y\} < 1; \\ \binom{⌊x+y⌋}{⌊y⌋}(⌊x⌋ + 1), & if \{x\} + \{y\} ≥ 1; \end{cases}

= \begin{cases}\binom{m′}{k′}, & if \{x\} + \{y\} < 1; \\ \binom{m′}{k′} (⌊\frac{m−k+6}{6}⌋) , & if \{x\} + \{y\} ≥ 1. \end{cases}

By Lemma 4(ii), we obtain

\{x\} = \{\frac{(m − r) − (k − s)}{6} + \frac{r − s}{6}\} = \{\frac{r − s}{6}\} and \{y\} = \{\frac{k − s}{6} + \frac{s}{6}\} = \frac{s}{6}.

If r ≥ s, then \{x\} + \{y\} = \{\frac{r−s}{6}\} + \frac{s}{6} = \frac{r−s}{6} + \frac{s}{6} = \frac{r}{6} < 1. If r < s, then we obtain by Lemma 4(iv) that \{x\} + \{y\} = \{−\frac{s−r}{6}\} + \frac{s}{6} = 1 − \frac{s−r}{6} + \frac{s}{6} = 1 + \frac{r}{6} ≥ 1. Therefore

\frac{⌊x + y⌋!}{⌊x⌋!⌊y⌋!} = \begin{cases}\binom{m′}{k′}, & if r ≥ s; \\ \binom{m′}{k′} (⌊\frac{m−k+6}{6}⌋) , & if r < s. \end{cases}  (12)

Substituting (12) in (11), we obtain part (i) of this theorem. The calculation in parts (ii) and (iii) are similar, so we give fewer details than given in part (i). By Lemma 2, for every ℓ ≥ 1, we have

ν_{5}(F_{1}F_{2} ⋯ F_{ℓ}) = ∑_{1≤n≤ℓ} ν_{5}(F_{n}) = ∑_{1≤n≤ℓ} ν_{5}(n) = ν_{5}(ℓ!),

which implies

ν_{5} (\binom{m}{k}_{F} ) = ν_{5}(m!) − ν_{5}(k!) − ν_{5}((m − k)!) = ν_{5} (\binom{m}{k}) .

For (iii), we apply Lemmas 2 and 1(iv) to obtain

ν_{p}(F_{1}F_{2} ⋯ F_{ℓ}) = ∑_{\begin{matrix}1≤n≤ℓ \\ n ≡ 0 (mod z(p)) \end{matrix}} ν_{p}(F_{n}) = ∑_{\begin{matrix}1≤n≤ℓ \\ n ≡ 0 (mod z(p)) \end{matrix}}(ν_{p}(n) + ν_{p}(F_{z(p)}))

= ∑_{1≤k≤ \frac{ℓ}{z(p)}} ν_{p}(kz(p)) + ⌊\frac{ℓ}{z(p)}⌋ ν_{p}(F_{z(p)})

= ν_{p} (⌊\frac{ℓ}{z(p)}⌋!) + ⌊\frac{ℓ}{z(p)}⌋ ν_{p}(F_{z(p)}).

As in part (i), the above implies that

ν_{p} (\binom{m}{k}_{F} ) = ν_{p} (\frac{⌊x + y⌋!}{⌊x⌋!⌊y⌋!}) + (⌊x + y⌋ − ⌊x⌋ − ⌊y⌋)ν_{p}(F_{z(p)}),  (13)

where x = \frac{m−k}{z(p)} and y = \frac{k}{z(p)}. In addition, if r ≥ s, then \{x\} + \{y\} < 1 and if r < s, then {x} + {y} ≥ 1. Therefore (13) can be simplified to the desired result. This completes the proof.

By Theorem 11(ii), we see that the 5-adic valuations of Fibonomial and binomial coefficients are the same. So we focus our investigation only on the p-adic valuations of Fibonomial coefficients when p ≠ 5. Calculating r and s in Theorem 11(i) in every case and writing Theorem 11(iii) in another form, we obtain the following corollary.

Corollary 12. Let m, k, r, and s be as in Theorem 11. Let

A_{2} = ν_{2} (⌊\frac{m}{6}⌋!) − ν_{2} (⌊\frac{k}{6}⌋!) − ν_{2} (⌊\frac{m − k}{6} ⌋!) ,

and for each prime p ≠ 2, 5, let A_{p} = ν_{p} (⌊\frac{m}{z(p)}⌋!) − ν_{p} (⌊\frac{k}{z(p)}⌋!) − ν_{p} (⌊\frac{m−k}{z(p)}⌋!). Then the following statements hold.

(i) ν_{2} (\binom{m}{k}_{F} ) = \begin{cases}A_{2}, & if r ≥ s and (r, s) ≠ (3, 1), (3, 2), (4, 2); \\ A_{2} + 1, & if (r, s) = (3, 1), (3, 2), (4, 2); \\ A_{2} + 3, & if r < s and (r, s) ≠ (0, 3), (1, 3), (2, 3), \\ & (1, 4), (2, 4), (2, 5); \\ A_{2} + 2, & if (r, s) = (0, 3), (1, 3), (2, 3), (1, 4), (2, 4), \\ & (2, 5). \end{cases}

(ii) For p ≠ 2, 5, we have

ν_{p} (\binom{m}{k}_{F} ) = \begin{cases}A_{p}, & if r ≥ s; \\ A_{p} + ν_{p}(F_{z(p)}), & if r < s. \end{cases}

Proof. For (i), we have 0 ≤ r ≤ 5 and 0 ≤ s ≤ 5, so we can directly consider every case and reduce Theorem 11(i) to the result in this corollary. In addition, (ii) follows directly from (13).

In a series of papers (see [11] and references therein), Marques and Trojovský obtain a formula for ν_{p} (\binom{p^{b}}{p^{a}}_{F} ) only when b = a + 1. Then Ballot [2] extends it to any case b > a.

Corollary 12 enables us to compute ν_{p} (\binom{ℓ_{1}p^{b}}{ℓ_{2}p^{a}}_{F} ). We illustrate this in the next theorem.

Theorem 13. Let a, b, ℓ_{1}, and ℓ_{2} be positive integers and b ≥ a. Let p ≠ 5 be a prime. Assume that ℓ_{1}p^{b} > ℓ_{2}p^{a} and let m_{p} = ⌊\frac{ℓ_{1}p^{b−a}}{z(p)} ⌋ and k_{p} = ⌊\frac{ℓ_{2}}{z(p)}⌋. Then the following statements hold.

(i) If a ≡ b (mod 2), then ν_{2} (\binom{ℓ_{1}2^{b}}{ℓ_{2}2^{a}}_{F} ) is equal to

ν_{2} (\binom{m_{2}}{k_{2}}) ,  if ℓ_{1} ≡ ℓ_{2} (mod 3) or ℓ_{2} ≡ 0 (mod 3);
a + 2 + ν_{2} (m_{2} − k_{2}) + ν_{2} (\binom{m_{2}}{k_{2}}) ,  if ℓ_{1} ≡ 0 (mod 3) and ℓ_{2} ≢ 0 (mod 3);
⌈\frac{a}{2}⌉ + 1 + ν_{2} (m_{2} − k_{2}) + ν_{2} (\binom{m_{2}}{k_{2}}) , if ℓ_{1} ≡ 1 (mod 3) and ℓ_{2} ≡ 2 (mod 3);
⌈\frac{a+1}{2}⌉ + ν_{2} (\binom{m_{2}}{k_{2}}) ,  if ℓ_{1} ≡ 2 (mod 3) and ℓ_{2} ≡ 1 (mod 3),

and if a ≢ b (mod 2), then ν_{2} (\binom{ℓ_{1}2^{b}}{ℓ_{2}2^{a}}_{F} ) is equal to

ν_{2} (\binom{m_{2}}{k_{2}}) ,  if ℓ_{1} ≡ −ℓ_{2} (mod 3) or ℓ_{2} ≡ 0 (mod 3);
a + 2 + ν_{2} (m_{2} − k_{2}) + ν_{2} (\binom{m_{2}}{k_{2}}) ,  if ℓ_{1} ≡ 0 (mod 3) and ℓ_{2} ≢ 0 (mod 3);
⌈\frac{a+1}{2}⌉ + ν_{2} (\binom{m_{2}}{k_{2}}) ,  if ℓ_{1} ≡ 1 (mod 3) and ℓ_{2} ≡ 1 (mod 3);
⌈\frac{a}{2}⌉ + 1 + ν_{2} (m_{2} − k_{2}) + ν_{2} (\binom{m_{2}}{k_{2}}) , if ℓ_{1} ≡ 2 (mod 3) and ℓ_{2} ≡ 2 (mod 3).

(ii) Let p ≠ 5 be an odd prime and let r = ℓ_{1}p^{b} mod z(p) and s = ℓ_{2}p^{a} mod z(p). If

p ≡ ±1 (mod 5), then

ν_{p} (\binom{ℓ_{1}p^{b}}{ℓ_{2}p^{a}}_{F} ) = [r < s] (a + ν_{p} (m_{p} − k_{p}) + ν_{p}(F_{z(p)})) + ν_{p} (\binom{m_{p}}{k_{p}}) ,

and if p ≡ ±2 (mod 5), then ν_{p} (\binom{ℓ_{1}p^{b}}{ℓ_{2}p^{a}}_{F} ) is equal to

ν_{p} (\binom{m_{p}}{k_{p}}) ,  if r = s or ℓ_{2} ≡ 0 (mod z(p));
a + ν_{p}(F_{z(p)}) + ν_{p} (m_{p} − k_{p}) + ν_{p} (\binom{m_{p}}{k_{p}}) , if ℓ_{1} ≡ 0 (mod z(p)) and

ℓ_{2} ≢ 0 (mod z(p));

\frac{a}{2} + ν_{p} (\binom{m_{p}}{k_{p}}) ,  if r > s, ℓ_{1}, ℓ_{2} ≢ 0 (mod z(p)),

and a is even;

\frac{a}{2} + ν_{p}(F_{z(p)}) + ν_{p} (m_{p} − k_{p}) + ν_{p} (\binom{m_{p}}{k_{p}}) , if r < s, ℓ_{1}, ℓ_{2} ≢ 0 (mod z(p)),

and a is even;

\frac{a+1}{2} + ν_{p} (m_{p} − k_{p}) + ν_{p} (\binom{m_{p}}{k_{p}}) ,  if r > s, ℓ_{1}, ℓ_{2} ≢ 0 (mod z(p)),

and a is odd;

\frac{a−1}{2} + ν_{p}(F_{z(p)}) + ν_{p} (\binom{m_{p}}{k_{p}}) ,  if r < s, ℓ_{1}, ℓ_{2} ≢ 0 (mod z(p)),

and a is odd.

Remark 14. In the proof of this theorem, we also show that the condition r = s in Theorem 13(ii) is equivalent to ℓ_{1} ≡ ℓ_{2} − 2ℓ_{2}[a ≢ b (mod 2)] (mod z(p)). It seems more natural to write r = s in the statement of the theorem, but it is more convenient in the proof to use the condition ℓ_{1} ≡ ℓ_{2} − 2ℓ_{2}[a ≢ b (mod 2)] (mod z(p)).

Proof of Theorem 13. We apply Corollary 12 to calculate ν_{2} (\binom{ℓ_{1}2^{b}}{ℓ_{2}2^{a}}_{F} ) with m = ℓ_{1}2^{b}, k = ℓ_{2}2^{a}, r = ℓ_{1}2^{b} mod 6, and s = ℓ_{2}2^{a} mod 6. For convenience, we also let r′ = ℓ_{1} mod 3, and s′ = ℓ_{2} mod 3. Therefore A_{2} given in Corollary 12 is

A_{2} = ν_{2} (⌊\frac{ℓ_{1}2^{b−1}}{3} ⌋!) − ν_{2} (⌊\frac{ℓ_{2}2^{a−1}}{3} ⌋!) − ν_{2} (⌊\frac{ℓ_{1}2^{b−1} − ℓ_{2}2^{a−1}}{3} ⌋!) .  (14)

By Corollary 8, the first term on the right-hand side of (14) is equal to

\frac{ℓ_{1}(2^{b−1} − 1)}{3} − ⌊\frac{b − 1}{2}⌋ [ℓ_{1} ≢ 0 (mod 3)] − \{\frac{ℓ_{1}}{3} \} [b ≡ 0 (mod 2)] + ν_{2} (⌊\frac{ℓ_{1}}{3} ⌋!)

= \frac{ℓ_{1}(2^{b−1} − 1)}{3} − ⌊\frac{b − 1}{2}⌋ [r′ ≠ 0] − \frac{r′}{3} [b ≡ 0 (mod 2)] + ν_{2} (⌊\frac{ℓ_{1}}{3} ⌋!) .  (15)

Similarly, the second term is

\frac{ℓ_{2}(2^{a−1} − 1)}{3} − ⌊\frac{a − 1}{2}⌋ [s′ ≠ 0] − \frac{s′}{3} [a ≡ 0 (mod 2)] + ν_{2} (⌊\frac{ℓ_{2}}{3} ⌋!) .  (16)

To evaluate the third term on the right-hand side of (14), we divide the proof into two cases according to the parity of a and b.

Case 1. a ≡ b (mod 2). Observe that ℓ_{1} ≡ ℓ_{2} (mod 3) if and only if r′ = s′. In addition, \{\frac{ℓ_{1}−ℓ_{2}}{3} \} = \{\frac{r′−s′}{3} \} and ⌊\frac{r′−s′}{3} ⌋ = −[r′ < s′]. Then by Theorem 9, the third term on the right-hand side of (14) is equal to

\frac{(ℓ_{1}2^{b−a} − ℓ_{2})(2^{a−1} − 1)}{3} − \{\frac{r′ − s′}{3} \} [a ≡ 0 (mod 2)] − ⌊\frac{a − 1}{2}⌋ [r′ ≠ s′] + ν_{2} (⌊\frac{ℓ_{1}2^{b−a} − ℓ_{2}}{3} ⌋!)

= \frac{(ℓ_{1}2^{b−a} − ℓ_{2})(2^{a−1} − 1)}{3} − (\frac{r′ − s′}{3} + [r′ < s′]) [a ≡ 0 (mod 2)] − ⌊\frac{a − 1}{2}⌋ [r′ ≠ s′]

+ ν_{2} (⌊\frac{ℓ_{1}2^{b−a} − ℓ_{2}}{3} ⌋!) .  (17)

Recall that m_{2} = ⌊\frac{ℓ_{1}2^{b−a}}{3} ⌋ and k_{2} = ⌊\frac{ℓ_{2}}{3} ⌋. Since b − a is even, 2^{b−a} ≡ 1 (mod 3) and we obtain by Lemma 6 that

⌊\frac{ℓ_{1}2^{b−a} − ℓ_{2}}{3} ⌋ = m_{2} − k_{2} − [r′ < s′].

Therefore ν_{2} (⌊\frac{ℓ_{1}2^{b−a}−ℓ_{2}}{3} ⌋!) is equal to

ν_{2}(m_{2}!) − [r′ < s′]ν_{2}(m_{2} − k_{2}) − ν_{2} (\frac{m_{2}!}{(m_{2} − k_{2})!}) .

By Corollary 8, ν_{2}(m_{2}!) is equal to

\frac{ℓ_{1}(2^{b−a} − 1)}{3} − \frac{b − a}{2}[r′ ≠ 0] + ν_{2} (⌊\frac{ℓ_{1}}{3} ⌋!) .

We substitute the value of ν_{2} (⌊\frac{ℓ_{1}2^{b−a}−ℓ_{2}}{3} ⌋!) in (17) and then substitute (15), (16), and (17) in (14) to obtain A_{2}. We see that there are some cancellations. For instance,

\frac{r′}{3} ([a ≡ 0 (mod 2)] − [b ≡ 0 (mod 2)]) = 0.

and

ν_{2} (\frac{m_{2}!}{(m_{2} − k_{2})!}) − ν_{2} (⌊\frac{ℓ_{2}}{3} ⌋!) = ν_{2} (\binom{m_{2}}{k_{2}})

Then we obtain

A_{2} = − ⌊\frac{b − 1}{2}⌋ [r′ ≠ 0] + ⌊\frac{a − 1}{2}⌋ [s′ ≠ 0] + [r′ < s′][a ≡ 0 (mod 2)]
+ ⌊\frac{a − 1}{2}⌋ [r′ ≠ s′] + \frac{b − a}{2}[r′ ≠ 0] + [r′ < s′]ν_{2}(m_{2} − k_{2}) + ν_{2} (\binom{m_{2}}{k_{2}}) .  (18)

Next we divide the calculation of A_{2} into 4 cases:

• Case 1.1. ℓ_{1} ≡ ℓ_{2} (mod 3) or ℓ_{2} ≡ 0 (mod 3),

• Case 1.2. ℓ_{1} ≡ 0 (mod 3) and ℓ_{2} ≢ 0 (mod 3),

• Case 1.3. ℓ_{1} ≡ 1 (mod 3) and ℓ_{2} ≡ 2 (mod 3),

• Case 1.4. ℓ_{1} ≡ 2 (mod 3) and ℓ_{2} ≡ 1 (mod 3). Since the calculation in each case is similar, we only show the details in Case 1.1 and Case 1.2. So assume that ℓ_{1} ≡ ℓ_{2} (mod 3). Then r′ = s′ and (18) becomes

A_{2} = − ⌊\frac{b − 1}{2}⌋ [r′ ≠ 0] + ⌊\frac{a − 1}{2}⌋ [r′ ≠ 0] + \frac{b − a}{2}[r′ ≠ 0] + ν_{2} (\binom{m_{2}}{k_{2}})

Since − ⌊\frac{b−1}{2}⌋ + ⌊\frac{a−1}{2}⌋ + \frac{b−a}{2} = 0, we see that A_{2} = ν_{2} (\binom{m_{2}}{k_{2}}). Next if ℓ_{2} ≡ 0 (mod 3), then s′ = 0 and the same calculation leads to A_{2} = ν_{2} (\binom{m_{2}}{k_{2}}). Next assume that ℓ_{1} ≡ 0 (mod 3) and ℓ_{2} ≢ 0 (mod 3). Then r′ = 0, s′ ≠ 0, and (18) becomes

A_{2} = ⌊\frac{a − 1}{2}⌋ + [a ≡ 0 (mod 2)] + ⌊\frac{a − 1}{2}⌋ + ν_{2}(m_{2} − k_{2}) + ν_{2} (\binom{m_{2}}{k_{2}}) .

Observing that the sum of the first three terms above is equal to a − 1, we obtain A_{2} = a − 1 + ν_{2}(m_{2} − k_{2}) + ν_{2} (\binom{m_{2}}{k_{2}}). The other cases are similar. Therefore A_{2} is

ν_{2} (\binom{m_{2}}{k_{2}}) ,  if ℓ_{1} ≡ ℓ_{2} (mod 3) or ℓ_{2} ≡ 0 (mod 3);
a − 1 + ν_{2} (m_{2} − k_{2}) + ν_{2} (\binom{m_{2}}{k_{2}}) , if ℓ_{1} ≡ 0 (mod 3) and ℓ_{2} ≢ 0 (mod 3);
⌊\frac{a}{2}⌋ + ν_{2} (m_{2} − k_{2}) + ν_{2} (\binom{m_{2}}{k_{2}}) , if ℓ_{1} ≡ 1 (mod 3) and ℓ_{2} ≡ 2 (mod 3);
⌊\frac{a−1}{2}⌋ + ν_{2} (\binom{m_{2}}{k_{2}}) ,  if ℓ_{1} ≡ 2 (mod 3) and ℓ_{2} ≡ 1 (mod 3).

Recall that r = ℓ_{1}2^{b} mod 6 and s = ℓ_{2}2^{a} mod 6. Therefore

r = \begin{cases}0, & if ℓ_{1} ≡ 0 (mod 3); \\ 2, & if b is even and ℓ_{1} ≡ 2 (mod 3) or if b is odd and ℓ_{1} ≡ 1 (mod 3); \\ 4 & if b is even and ℓ_{1} ≡ 1 (mod 3) or if b is odd and ℓ_{1} ≡ 2 (mod 3), \end{cases}

and

s = \begin{cases}0, & if ℓ_{2} ≡ 0 (mod 3); \\ 2, & if a is even and ℓ_{2} ≡ 2 (mod 3) or if a is odd and ℓ_{2} ≡ 1 (mod 3); \\ 4 & if a is even and ℓ_{2} ≡ 1 (mod 3) or if a is odd and ℓ_{2} ≡ 2 (mod 3). \end{cases}

To obtain the formula for ν_{2} (\binom{ℓ_{1}2^{b}}{ℓ_{2}2^{a}}_{F} ), we divide the calculation into 4 cases: Case 1.1 to Case 1.4 as before. Then we consider the values of r and s in each case, and substitute A_{2} in Corollary 12. This leads to the desired result. Since the calculation in each case is similar, we only give the details in Case 1.3. In this case, A_{2} = ⌊\frac{a}{2}⌋ + ν_{2}(m_{2} − k_{2}) + ν_{2} (\binom{m_{2}}{k_{2}}), (r, s) = (2, 4) if a and b are odd, and (r, s) = (4, 2) if a and b are even. By Corollary 12, we obtain

ν_{2} (\binom{ℓ_{1}2^{b}}{ℓ_{2}2^{a}}_{F} ) = \begin{cases}A_{2} + 2, & if a and b are odd; \\ A_{2} + 1, & if a and b are even, \end{cases}

= ⌈\frac{a}{2}⌉ + 1 + ν_{2}(m_{2} − k_{2}) + ν_{2} (\binom{m_{2}}{k_{2}}) ,

as required. The other cases are similar. Case 2. a ≢ b (mod 2). The calculation in this case is similar to Case 1, so we omit some details. By Theorem 9, the third term on the right-hand side of (14) is equal to

\frac{(ℓ_{1}2^{b−a} − ℓ_{2})(2^{a−1} − 1)}{3} − \{−\frac{r′ + s′}{3} \} [a ≡ 0 (mod 2)] − ⌊\frac{a − 1}{2}⌋ [ℓ_{1} ≢ −ℓ_{2} (mod 3)]

+ ν_{2} (⌊\frac{ℓ_{1}2^{b−a} − ℓ_{2}}{3} ⌋!) .  (19)

Since b − a is odd, ℓ_{1}2^{b−a} ≡ −r′ (mod 3) and we obtain by Lemma 6 that

⌊\frac{ℓ_{1}2^{b−a} − ℓ_{2}}{3} ⌋ = m_{2} − k_{2} − B

where B = [(r′, s′) ∈ \{(0, 1), (0, 2), (2, 2)\}]. Similar to Case 1, ν_{2} (⌊\frac{ℓ_{1}2^{b−a}−ℓ_{2}}{3} ⌋!) is

ν_{2} (m_{2}!) − Bν_{2} (m_{2} − k_{2}) − ν_{2} (\frac{m_{2}!}{(m_{2} − k_{2})!}) .

Then we evaluate ν_{2}(m_{2}!) by Corollary 8, and substitute all of these in (14) to obtain that A_{2} is equal to

(\frac{b − a − 1}{2} − ⌊\frac{b − 1}{2}⌋) [r′ ≠ 0] − \frac{r′}{3} [b ≡ 0 (mod 2)] + ⌊\frac{a − 1}{2}⌋ [s′ ≠ 0]

+ (\frac{s′}{3} + \{−\frac{r′ + s′}{3} \}) [a ≡ 0 (mod 2)] + ⌊\frac{a − 1}{2}⌋ [ℓ_{1} ≢ −ℓ_{2} (mod 3)] + \frac{r′}{3}

+ Bν_{2} (m_{2} − k_{2}) + ν_{2} (\binom{m_{2}}{k_{2}}) .

Then we divide the calculation into 4 cases and obtain that A_{2} is

ν_{2} (\binom{m_{2}}{k_{2}}) ,  if ℓ_{1} ≡ −ℓ_{2} (mod 3) or ℓ_{2} ≡ 0 (mod 3);
a − 1 + ν_{2} (m_{2} − k_{2}) + ν_{2} (\binom{m_{2}}{k_{2}}) , if ℓ_{1} ≡ 0 (mod 3) and ℓ_{2} ≢ 0 (mod 3);
⌊\frac{a−1}{2}⌋ + ν_{2} (\binom{m_{2}}{k_{2}}) ,  if ℓ_{1} ≡ 1 (mod 3) and ℓ_{2} ≡ 1 (mod 3);
⌊\frac{a}{2}⌋ + ν_{2} (m_{2} − k_{2}) + ν_{2} (\binom{m_{2}}{k_{2}}) , if ℓ_{1} ≡ 2 (mod 3) and ℓ_{2} ≡ 2 (mod 3).

We illustrate the calculation of A_{2} above only for the case ℓ_{2} ≡ 0 (mod 3) since the other cases are similar. So suppose ℓ_{2} ≡ 0 (mod 3). So s′ = 0. If r′ = 0, then it is easy to see that A_{2} is equal to ν_{2} (\binom{m_{2}}{k_{2}}). So assume that r′ ≠ 0. Then A_{2} is equal to x + y + ν_{2} (\binom{m_{2}}{k_{2}}), where

x = \frac{b − a − 1}{2} − ⌊\frac{b − 1}{2}⌋ + ⌊\frac{a − 1}{2}⌋ = \begin{cases}0, & if a is odd; \\ −1, & if a is even, \end{cases}

y = \frac{−r′}{3} [b ≡ 0 (mod 2)] + \{−\frac{r′}{3} \} [a ≡ 0 (mod 2)] + \frac{r′}{3} = \begin{cases}0, & if a is odd; \\ 1, & if a is even. \end{cases}

Therefore A_{2} = ν_{2} (\binom{m_{2}}{k_{2}}), as required. As in Case 1, we divide the calculation of ν_{2} (\binom{ℓ_{1}2^{b}}{ℓ_{2}2^{a}}_{F} ) into 4 cases according to the value of A_{2}, which leads to the desired result. This proves (i).

For (ii), we apply Corollary 12 with m = ℓ_{1}p^{b} and k = ℓ_{2}p^{a}. For convenience, we let r′ = ℓ_{1} mod z(p) and s′ = ℓ_{2} mod z(p). The calculation of this part is similar to that of part (i), so we omit some details. We have

A_{p} = ν_{p} (⌊\frac{ℓ_{1}p^{b}}{z(p)}⌋!) − ν_{p} (⌊\frac{ℓ_{2}p^{a}}{z(p)}⌋!) − ν_{p} (⌊\frac{ℓ_{1}p^{b} − ℓ_{2}p^{a}}{z(p)} ⌋!) .  (20)

Case 1. p ≡ ±1 (mod 5). Then by Lemma 1(iii), p ≡ 1 (mod z(p)). By Corollary 8, the first term on the right-hand side of (20) is equal to

\frac{ℓ_{1}(p^{b} − 1)}{z(p)(p − 1)} + ν_{p} (⌊\frac{ℓ_{1}}{z(p)}⌋!) − ⌊\frac{b}{2}⌋ [r′ ≠ 0] − \frac{r′}{z(p)}[b ≡ 1 (mod 2)] + ⌊\frac{b}{2}⌋ [r′ ≠ 0] (1 − \frac{2r′}{z(p)})

= \frac{ℓ_{1}(p^{b} − 1)}{z(p)(p − 1)} − \frac{br′}{z(p)} + ν_{p} (⌊\frac{ℓ_{1}}{z(p)}⌋!) ,

and similarly, the second term is

\frac{ℓ_{2}(p^{a} − 1)}{z(p)(p − 1)} − \frac{as′}{z(p)} + ν_{p} (⌊\frac{ℓ_{2}}{z(p)}⌋!) .

By Theorem 9, the third term is

\frac{(ℓ_{1}p^{b−a} − ℓ_{2})(p^{a} − 1)}{z(p)(p − 1)} − a (\frac{r′ − s′}{z(p)} + [r′ < s′]) + ν_{p} (⌊\frac{ℓ_{1}p^{b−a} − ℓ_{2}}{z(p)} ⌋!) .

Since p ≡ 1 (mod z(p)), we obtain by Lemma 6 that

⌊\frac{ℓ_{1}p^{b−a} − ℓ_{2}}{z(p)} ⌋ = m_{p} − k_{p} − [r′ < s′].

Therefore ν_{p} (⌊\frac{ℓ_{1}p^{b−a}−ℓ_{2}}{z(p)} ⌋!) is equal to

ν_{p} (m_{p}!) − [r′ < s′]ν_{p} (m_{p} − k_{p}) − ν_{p} (\frac{m_{p}!}{(m_{p} − k_{p})!}) .

As usual, the first term above can be evaluated by Corollary 8 and is equal to

\frac{ℓ_{1}(p^{b−a} − 1)}{z(p)(p − 1)} − (b − a)\frac{r′}{z(p)} + ν_{p} (⌊\frac{ℓ_{1}}{z(p)}⌋!) .

We substitute all of these in (20) to obtain

A_{p} = [r′ < s′] (a + ν_{p} (m_{p} − k_{p})) + ν_{p} (\binom{m_{p}}{k_{p}}) .

Since p ≡ 1 (mod z(p)), r = r′ and s = s′. Substituting A_{p} and applying Corollary 12, we obtain the desired result.

Case 2. p ≡ ±2 (mod 5). Then by Lemma 1(iii), p ≡ −1 (mod z(p)). By Corollary 8, the first term on the right-hand side of (20) is equal to

\frac{ℓ_{1}(p^{b} − 1)}{z(p)(p − 1)} − ⌊\frac{b}{2}⌋ [r′ ≠ 0] − \frac{r′}{z(p)}[b ≡ 1 (mod 2)] + ν_{p} (⌊\frac{ℓ_{1}}{z(p)}⌋!) .

Similarly, the second term is

\frac{ℓ_{2}(p^{a} − 1)}{z(p)(p − 1)} − ⌊\frac{a}{2}⌋ [s′ ≠ 0] − \frac{s′}{z(p)}[a ≡ 1 (mod 2)] + ν_{p} (⌊\frac{ℓ_{2}}{z(p)}⌋!) .

For the third term, we divide the proof into two cases according to the parity of a and b.

Case 2.1. a ≡ b (mod 2). Then by Theorem 9, the third term on the right-hand side of (20) is equal to

\frac{(ℓ_{1}p^{b−a} − ℓ_{2})(p^{a} − 1)}{z(p)(p − 1)} − (\frac{r′ − s′}{z(p)} + [r′ < s′]) [a ≡ 1 (mod 2)] − ⌊\frac{a}{2}⌋ [r′ ≠ s′]

+ ν_{p} (⌊\frac{ℓ_{1}p^{b−a} − ℓ_{2}}{z(p)} ⌋!) .

As in Case 1, we apply Lemma 6 to write

⌊\frac{ℓ_{1}p^{b−a} − ℓ_{2}}{z(p)} ⌋ = m_{p} − k_{p} − [r′ < s′],

and then use Corollary 8 to show that ν_{p} (⌊\frac{ℓ_{1}p^{b−a}−ℓ_{2}}{z(p)} ⌋!) is equal to

\frac{ℓ_{1}(p^{b−a} − 1)}{z(p)(p − 1)} − \frac{b − a}{2}[r′ ≠ 0] + ν_{p} (⌊\frac{ℓ_{1}}{z(p)}⌋!) − [r′ < s′]ν_{p} (m_{p} − k_{p}) − ν_{p} (\frac{m_{p}!}{(m_{p} − k_{p})!}) .

Substituting all of these in (20), we see that A_{p} is equal to

− ⌊\frac{b}{2}⌋ [r′ ≠ 0] + ⌊\frac{a}{2}⌋ [s′ ≠ 0] + [r′ < s′][b ≡ 1 (mod 2)] + ⌊\frac{a}{2}⌋ [r′ ≠ s′] + \frac{b − a}{2}[r′ ≠ 0]

+ [r′ < s′]ν_{p} (m_{p} − k_{p}) + ν_{p} (\binom{m_{p}}{k_{p}})

= \begin{cases}ν_{p} (\binom{m_{p}}{k_{p}}) , & if ℓ_{1} ≡ ℓ_{2} (mod z(p)) or ℓ_{2} ≡ 0 (mod z(p)); \\ a + ν_{p} (m_{p} − k_{p}) + ν_{p} (\binom{m_{p}}{k_{p}}) , & if ℓ_{1} ≡ 0 (mod z(p)) and ℓ_{2} ≢ 0 (mod z(p)); \\ ⌊\frac{a}{2}⌋ + ν_{p} (\binom{m_{p}}{k_{p}}) , & if ℓ_{1}, ℓ_{2} ≢ 0 (mod z(p)) and r′ > s′; \\ ⌈\frac{a}{2}⌉ + ν_{p} (m_{p} − k_{p}) + ν_{p} (\binom{m_{p}}{k_{p}}) , & if ℓ_{1}, ℓ_{2} ≢ 0 (mod z(p)) and r′ < s′. \end{cases}

Recall that r = ℓ_{1}p^{b} mod z(p) and s = ℓ_{2}p^{a} mod z(p). If a and b are even, then p^{b} ≡ p^{a} ≡ 1 (mod z(p)), r = r′, and s = s′, and we can obtain ν_{p} (\binom{ℓ_{1}p^{b}}{ℓ_{2}p^{a}}_{F} ) by substituting A_{p} in Corollary 12. Suppose a and b are odd. Then r ≡ −r′ (mod z(p)) and s ≡ −s′ (mod z(p)) and thus when r and s are both nonzero or are both zero, we have

r ≥ s if and only if r′ ≤ s′.

Similar to the above, we can obtain ν_{p} (\binom{ℓ_{1}p^{b}}{ℓ_{2}p^{a}}_{F} ) by the substitution of A_{p} in Corollary 12. We see that ν_{p} (\binom{ℓ_{1}p^{b}}{ℓ_{2}p^{a}}) is equal to

ν_{p} (\binom{m_{p}}{k_{p}}) ,  if ℓ_{1} ≡ ℓ_{2} (mod z(p)) or ℓ_{2} ≡ 0 (mod z(p));
a + ν_{p}(F_{z(p)}) + ν_{p} (m_{p} − k_{p}) + ν_{p} (\binom{m_{p}}{k_{p}}) , if ℓ_{1} ≡ 0 (mod z(p)) and ℓ_{2} ≢ 0 (mod z(p));
\frac{a}{2} + ν_{p} (\binom{m_{p}}{k_{p}}) ,  if r > s, ℓ_{1}, ℓ_{2} ≢ 0 (mod z(p)), and a is even;
\frac{a}{2} + ν_{p}(F_{z(p)}) + ν_{p} (m_{p} − k_{p}) + ν_{p} (\binom{m_{p}}{k_{p}}) , if r < s, ℓ_{1}, ℓ_{2} ≢ 0 (mod z(p)), and a is even;
\frac{a+1}{2} + ν_{p} (m_{p} − k_{p}) + ν_{p} (\binom{m_{p}}{k_{p}}) ,  if r > s, ℓ_{1}, ℓ_{2} ≢ 0 (mod z(p)), and a is odd;
\frac{a−1}{2} + ν_{p}(F_{z(p)}) + ν_{p} (\binom{m_{p}}{k_{p}}) ,  if r < s, ℓ_{1}, ℓ_{2} ≢ 0 (mod z(p)), and a is odd.

Since a ≡ b (mod 2), we see that p^{a} ≡ p^{b} (mod z(p)) and therefore

ℓ_{1} ≡ ℓ_{2} (mod z(p)) ⇔ r = s  (21)

So the condition ℓ_{1} ≡ ℓ_{2} (mod z(p)) can be replaced by r = s. Case 2.2. a ≢ b (mod 2). The calculation in this case is similar to that given before. So we skip some details. By Theorem 9, the third term on the right-hand side of (20) is equal to

\frac{(ℓ_{1}p^{b−a} − ℓ_{2})(p^{a} − 1)}{z(p)(p − 1)}−[a ≡ 1 (mod 2)]B_{1}−⌊\frac{a}{2}⌋ [ℓ_{1} ≢ −ℓ_{2} (mod z(p))]+ν_{p} (⌊\frac{ℓ_{1}p^{b−a} − ℓ_{2}}{z(p)} ⌋!) ,

where B_{1} = \{−\frac{r′+s′}{z(p)} \} = −\frac{r′+s′}{z(p)} + [r′ + s′ > 0] + [r′ + s′ > z(p)]. Since p ≡ −1 (mod z(p)), we obtain by Lemma 6 and a straightforward verification that

⌊\frac{ℓ_{1}p^{b−a} − ℓ_{2}}{z(p)} ⌋ = m_{p} − k_{p} − ε,

where ε = [−r′ mod z(p) < s′] = [r′ = 0 and s′ ≠ 0] + [r′ + s′ > z(p)]. Then by Corollary 8, ν_{p} (⌊\frac{ℓ_{1}p^{b−a}−ℓ_{2}}{z(p)} ⌋!) is equal to

\frac{ℓ_{1}(p^{b−a} − 1)}{z(p)(p − 1)} − ⌊\frac{b − a}{2}⌋ [r′ ≠ 0] − \frac{r′}{z(p)} + ν_{p} (⌊\frac{ℓ_{1}}{z(p)}⌋!) − B_{2} − ν_{p} (\frac{m_{p}!}{(m_{p} − k_{p})!}) ,

where B_{2} = εν_{p} (m_{p} − k_{p}). Since a ≢ b (mod 2), [b ≡ 1 (mod 2)] = 1 − [a ≡ 1 (mod 2)] and ⌊\frac{b−a}{2}⌋ + ⌊\frac{a}{2}⌋ − ⌊\frac{b}{2}⌋ + [a ≡ 1 (mod 2)]. We substitute all of these in (20) to obtain that A_{p} is equal to

(⌊\frac{b − a}{2}⌋ − ⌊\frac{b}{2}⌋) [r′ ≠ 0] + ⌊\frac{a}{2}⌋ ([s′ ≠ 0] + [ℓ_{1} ≢ −ℓ_{2} (mod z(p))])

+ ([r′ + s′ > 0] + [r′ + s′ > z(p)]) [a ≡ 1 (mod 2)] + B_{2} + ν_{p} (\binom{m_{p}}{k_{p}})

= \begin{cases}ν_{p} (\binom{m_{p}}{k_{p}}) , & if ℓ_{1} ≡ −ℓ_{2} (mod z(p)) or ℓ_{2} ≡ 0 (mod z(p)); \\ a + ν_{p} (m_{p} − k_{p}) + ν_{p} (\binom{m_{p}}{k_{p}}) , & if ℓ_{1} ≡ 0 (mod z(p)) and ℓ_{2} ≢ 0 (mod z(p)); \\ ⌊\frac{a}{2}⌋ + ν_{p} (\binom{m_{p}}{k_{p}}) , & if ℓ_{1}, ℓ_{2} ≢ 0 (mod z(p)) and r′ + s′ < z(p); \\ ⌈\frac{a}{2}⌉ + ν_{p} (m_{p} − k_{p}) + ν_{p} (\binom{m_{p}}{k_{p}}) , & if ℓ_{1}, ℓ_{2} ≢ 0 (mod z(p)) and r′ + s′ > z(p). \end{cases}

Recall that r = ℓ_{1}p^{b} mod z(p) and s = ℓ_{2}p^{a} mod z(p). Suppose that a is odd and b is even. Then r = r′ and s ≡ −s′ (mod z(p)). Moreover, if s′ ≠ 0, then s = z(p) − s′ and thus

r < s ⇔ r′ + s′ < z(p) and r > s ⇔ r′ + s′ > z(p).

Similarly, if a is even and b is odd, then r ≡ −r′ (mod z(p)) and s = s′, and for r′ ≠ 0, we have

r < s ⇔ r′ + s′ > z(p) and r > s ⇔ r′ + s′ < z(p).

From the above observation and the substitution of A_{p} in Corollary 12, we see that ν_{p} (\binom{ℓ_{1}p^{b}}{ℓ_{2}p^{a}}_{F} ) is equal to

ν_{p} (\binom{m_{p}}{k_{p}}) ,  if ℓ_{1} ≡ −ℓ_{2} (mod z(p)) or ℓ_{2} ≡ 0 (mod z(p));
a + ν_{p}(F_{z(p)}) + ν_{p} (m_{p} − k_{p}) + ν_{p} (\binom{m_{p}}{k_{p}}) , if ℓ_{1} ≡ 0 (mod z(p)) and ℓ_{2} ≢ 0 (mod z(p));
\frac{a+1}{2} + ν_{p} (m_{p} − k_{p}) + ν_{p} (\binom{m_{p}}{k_{p}}) ,  if r > s, ℓ_{1}, ℓ_{2} ≢ 0 (mod z(p)), and a is odd;
\frac{a−1}{2} + ν_{p}(F_{z(p)}) + ν_{p} (\binom{m_{p}}{k_{p}}) ,  if r < s, ℓ_{1}, ℓ_{2} ≢ 0 (mod z(p)), and a is odd;
\frac{a}{2} + ν_{p} (\binom{m_{p}}{k_{p}}) ,  if r > s, ℓ_{1}, ℓ_{2} ≢ 0 (mod z(p)), and a is even;
\frac{a}{2} + ν_{p}(F_{z(p)}) + ν_{p} (m_{p} − k_{p}) + ν_{p} (\binom{m_{p}}{k_{p}}) , if r < s, ℓ_{1}, ℓ_{2} ≢ 0 (mod z(p)), and a is even.

Since a ≢ b (mod 2), we see that p^{a} ≡ −p^{b} (mod z(p)) and therefore

ℓ_{1} ≡ −ℓ_{2} (mod z(p)) ⇔ r = s.

Combining this with (21), we conclude that

ℓ_{1} ≡ ℓ_{2} − 2ℓ_{2}[a ≢ b (mod 2)] (mod z(p)) ⇔ r = s.

This completes the proof.

5 Examples

In this last section, we give several examples to show applications of our main results. We also recall from Remark 14 that the condition r = s in Theorem 13(ii) can be replaced by ℓ_{1} ≡ ℓ_{2} − 2ℓ_{2}[a ≢ b (mod 2)] (mod z(p)). In the calculation given in this section, we will use this observation without further reference.

Example 15. Let a, b, and ℓ be positive integers and b ≥ a. We assert that for ℓ ≢ 0 (mod 3), we have

ν_{2} (\binom{ℓ · 2^{b}}{2^{a}}_{F} ) = ⌈\frac{a + 1}{2}⌉ (ε_{1}ε_{2} + ε′_{1}ε′_{2}) ,  (22)

where ε_{1} = [ℓ ≡ 2 (mod 3)], ε_{2} = [a ≡ b (mod 2)], ε′_{1} = [ℓ ≡ 1 (mod 3)], and ε′_{2} = [a ≢ b (mod 2)]. In addition, if ℓ ≡ 0 (mod 3), then

ν_{2} (\binom{ℓ · 2^{b}}{2^{a}}_{F} ) = b + 2 + ν_{2}(ℓ).  (23)

Proof. We apply Theorem 13 to verify our assertion. Here m_{2} = ⌊\frac{ℓ·2^{b−a}}{3} ⌋ and k_{2} = ⌊\frac{1}{3}⌋ = 0. So we immediately obtain the following: if a ≡ b (mod 2), then

ν_{2} (\binom{ℓ · 2^{b}}{2^{a}}_{F} ) = \begin{cases}0, & if ℓ ≡ 1 (mod 3); \\ a + 2 + ν_{2}(m_{2}), & if ℓ ≡ 0 (mod 3); \\ ⌈\frac{a+1}{2}⌉ , & if ℓ ≡ 2 (mod 3), \end{cases}

and if a ≢ b (mod 2), then

ν_{2} (\binom{ℓ · 2^{b}}{2^{a}}_{F} ) = \begin{cases}0, & if ℓ ≡ 1 (mod 3); \\ a + 2 + ν_{2}(m_{2}), & if ℓ ≡ 0 (mod 3); \\ ⌈\frac{a+1}{2}⌉ , & if ℓ ≡ 1 (mod 3). \end{cases}

This proves (22). If ℓ ≡ 0 (mod 3), then m_{2} = \frac{ℓ}{3} · 2^{b−a} and ν_{2}(m_{2}) is equal to

ν_{2}(m_{2}) = ν_{2} (ℓ) + ν_{2}(2^{b−a}) − ν_{2}(3) = b − a + ν_{2} (ℓ) ,

which implies (23).

Example 16. Substituting ℓ = 1 in Example 15, we see that

ν_{2} (\binom{2^{b}}{2^{a}}_{F} ) = ⌈\frac{a + 1}{2}⌉ [a ≢ b (mod 2)]

= \begin{cases}0, & if a ≡ b (mod 2); \\ ⌈\frac{a+1}{2}⌉ , & if a ≢ b (mod 2). \end{cases}  (24)

Our example also implies that (24) still holds for the 2-adic valuations of \binom{2^{b+2c}}{2^{a}}_{F}, \binom{7·2^{b}}{2^{a}}_{F}, \binom{5·2^{b+1}}{2^{a}}_{F}, \binom{13·2^{b}}{2^{a}}_{F}, etc.

Example 17. Let a, b, and ℓ be positive integers, b ≥ a, and p a prime distinct from 2 and 5. If p ≡ ±1 (mod 5), then

ν_{p} (\binom{ℓp^{b}}{p^{a}}_{F} ) = (b + ν_{p}(F_{z(p)}) + ν_{p}(ℓ)) [ℓ ≡ 0 (mod z(p))],

and if p ≡ ±2 (mod 5), then

ν_{p} (\binom{ℓp^{b}}{p^{a}}_{F} ) = \begin{cases}0, & if ℓ ≡ 1 − 2ε (mod z(p)); \\ b + ν_{p}(F_{z(p)}) + ν_{p}(ℓ), & if ℓ ≡ 0 (mod z(p)); \\ \frac{a}{2}, & if ℓ ≢ 0, 1 − 2ε (mod z(p)) and a is even; \\ \frac{a−1}{2} + ν_{p}(F_{z(p)}), & if ℓ ≢ 0, 1 − 2ε (mod z(p)) and a is odd, \end{cases}

where ε = [a ≢ b (mod 2)].

Proof. Similar to Example 15, we verify this by applying Theorem 13. Here m_{p} = ⌊\frac{ℓp^{b−a}}{z(p)} ⌋,

k_{p} = ⌊\frac{1}{z(p)}⌋ = 0, r = ℓp^{b} mod z(p), and s = p^{a} mod z(p). We first assume that p ≡ ±1 (mod 5). Then by Lemma 1, we have p ≡ 1 (mod z(p)). Therefore s = 1, r ≡ ℓ (mod z(p)), and

ν_{p} (\binom{ℓp^{b}}{p^{a}}_{F} ) = (a + ν_{p}(m_{p}) + ν_{p}(F_{z(p)})) [ℓ ≡ 0 (mod z(p))].

Similarly, if p ≡ ±2 (mod 5) and a ≡ b (mod 2), then we obtain by Lemma 1 and Theorem 13 that

ν_{p} (\binom{ℓp^{b}}{p^{a}}_{F} ) = \begin{cases}0, & if ℓ ≡ 1 (mod z(p)); \\ a + ν_{p}(m_{p}) + ν_{p}(F_{z(p)}), & if ℓ ≡ 0 (mod z(p)); \\ \frac{a}{2}, & if ℓ ≢ mod 0, 1z(p) and a is even; \\ \frac{a−1}{2} + ν_{p}(F_{z(p)}), & if ℓ ≢ 0, 1 (mod z(p)) and a is odd. \end{cases}

In addition, if p ≡ ±2 (mod 5) and a ≢ b (mod 2), then

ν_{p} (\binom{ℓp^{b}}{p^{a}}_{F} ) = \begin{cases}0, & if ℓ ≡ −1 (mod z(p)); \\ a + ν_{p}(m_{p}) + ν_{p}(F_{z(p)}), & if ℓ ≡ 0 (mod z(p)); \\ \frac{a}{2}, & if ℓ ≢ 0, −1 (mod z(p)) and a is even; \\ \frac{a−1}{2} + ν_{p}(F_{z(p)}), & if ℓ ≢ 0, −1 (mod z(p)) and a is odd. \end{cases}

It remains to calculate ν_{p}(m_{p}) when ℓ ≡ 0 (mod z(p)). In this case, we have

ν_{p}(m_{p}) = ν_{p} (\frac{ℓp^{b−a}}{z(p)} ) = ν_{p}(ℓ) + ν_{p}(p^{b−a}) − ν_{p}(z(p)) = b − a + ν_{p}(ℓ).

This implies the desired result.

Example 18. Substituting ℓ = 1 in Example 17, we see that for p ≠ 2, 5, we have

ν_{p} (\binom{p^{b}}{p^{a}}_{F} ) = \begin{cases}0, & if p ≡ ±1 (mod 5) or a ≡ b (mod 2); \\ \frac{a}{2}, & if p ≡ ±2 (mod 5), a ≢ b (mod 2), and a is even; \\ \frac{a−1}{2} + ν_{p}(F_{z(p)}), & if p ≡ ±2 (mod 5), a ≢ b (mod 2), and a is odd. \end{cases} (25)

Our example also implies that (25) still holds for the p-adic valuations of \binom{p^{b+2c}}{p^{a}}_{F} and

\binom{(z(p)+1)·p^{b}}{p^{a}}_{F}. Similarly, for p ≠ 2, 5, we have

ν_{p} (\binom{2p^{b}}{p^{a}}_{F} ) = \begin{cases}0, & if p ≡ ±1 (mod 5); \\ \frac{a}{2}, & if p ≡ ±2 (mod 5) and a is even; \\ \frac{a−1}{2} + ν_{p}(F_{z(p)}), & if p ≡ ±2 (mod 5) and a is odd. \end{cases}  (26)

In addition, (26) also holds when \binom{2p^{b}}{p^{a}}_{F} is replaced by \binom{ℓp^{b}}{p^{a}}_{F} for ℓ ≢ 0, ±1 (mod z(p)) and

p ≠ 2, 5. Furthermore, replacing \binom{2p^{b}}{p^{a}}_{F} by \binom{(z(p)−1)p^{b}}{p^{a}}_{F}, the formula becomes

0,  if p ≡ ±1 (mod 5) or a ≢ b (mod 2);
\frac{a}{2},  if p ≡ ±2 (mod 5), a ≡ b (mod 2), and a is even;
\frac{a−1}{2} + ν_{p}(F_{z(p)}), if p ≡ ±2 (mod 5), a ≡ b (mod 2), and a is odd.

Example 19. We know that the 5-adic valuations of Fibonomial coefficients are the same as those of binomial coefficients. For example, by Theorem 11(ii) and Kummer’s theorem, we obtain

ν_{5} (\binom{ℓ · 5^{b}}{5^{a}}_{F} ) = ν_{5} (\binom{ℓ · 5^{b}}{5^{a}}) = b − a + ν_{5}(ℓ),

for every a, b, ℓ ∈ ℕ with b ≥ a. Similarly, ν_{5} (\binom{5^{b}}{ℓ·5^{a}}_{F} ) = b − a − ν_{5}(ℓ) for every a, b, ℓ ∈ ℕ such that 5^{b} > ℓ · 5^{a}.

Example 20. Let a, b, and ℓ be positive integers and 2^{b} > ℓ · 2^{a}. Let m_{2} = ⌊\frac{2^{b−a}}{3} ⌋ and

k_{2} = ⌊\frac{ℓ}{3}⌋. Then

ν_{2} (\binom{2^{b}}{ℓ · 2^{a}}_{F} ) = ν_{2} (\binom{m_{2}}{k_{2}}) + (⌈\frac{a + 2}{2}⌉ + ν_{2}(m_{2} − k_{2})) ε_{1}ε_{2} + ⌈\frac{a + 1}{2}⌉ ε_{3}ε_{4}, (27)

where ε_{1} = [a ≡ b (mod 2)], ε_{2} = [ℓ ≡ 2 (mod 3)], ε_{3} = [a ≢ b (mod 2)], and ε_{4} = [ℓ ≡ 1 (mod 3)].

Proof. Similar to Example 15, this follows from the application of Theorem 13. So we leave the details to the reader.

Example 21. Let k ≥ 2. We observe that

⌊\frac{2^{k}}{3} ⌋ = \begin{cases}\frac{2^{k}−1}{3}, & if k is even; \\ \frac{2(2^{k−1}−1)}{3}, & if k is odd, \end{cases}

which implies,

ν_{2} (⌊\frac{2^{k}}{3} ⌋) = [k ≡ 1 (mod 2)].  (28)

By a similar reason, we also see that for k ≥ 3,

ν_{2} (⌊\frac{2^{k}}{3} ⌋ − 1) = 2[k ≡ 0 (mod 2)].  (29)

From (27), (28), and (29), we obtain the following results:

(i) if b − a ≥ 2, then ν_{2} (\binom{2^{b}}{3·2^{a}}_{F} ) = [a ≢ b (mod 2)], (ii) if b − a ≥ 3, then ν_{2} (\binom{2^{b}}{5·2^{a}}_{F} ) is equal to

[a ≢ b (mod 2)] + (⌈\frac{a + 2}{2}⌉ + 2[a ≡ b (mod 2)]) [a ≡ b (mod 2)]

= 1 + ⌈\frac{a + 4}{2}⌉ [a ≡ b (mod 2)],

(iii) if b − a ≥ 3, then ν_{2} (\binom{2^{b}}{6·2^{a}}_{F} ) = [a ≡ b (mod 2)], (iv) if b − a ≥ 4, then ν_{2} (\binom{2^{b}}{7·2^{a}}_{F} ) = [a ≡ b (mod 2)] + ⌈\frac{a+1}{2}⌉ [a ≢ b (mod 2)]. Example 22. Let p ≠ 5 be an odd prime and let a, b, and ℓ be positive integers, p^{b} > ℓp^{a}, m_{p} = ⌊\frac{p^{b−a}}{z(p)} ⌋, and k_{p} = ⌊\frac{ℓ}{z(p)}⌋. Then the following statements hold.

(i) If p ≡ ±1 (mod 5), then

ν_{p} (\binom{p^{b}}{ℓp^{a}}_{F} ) = (a + ν_{p} (m_{p} − k_{p}) + ν_{p}(F_{z(p)})) [ℓ ≢ 0, 1 (mod z(p))] + ν_{p} (\binom{m_{p}}{k_{p}}) ,

(ii) If p ≡ ±2 (mod 5), then ν_{p} (\binom{p^{b}}{ℓp^{a}}_{F} ) is equal to

ν_{p} (\binom{m_{p}}{k_{p}})+ε_{1}ε_{2}ε_{5} (⌈\frac{a}{2}⌉ + ν_{p}(m_{p} − k_{p}) + ε_{3}ν_{p}(F_{z}(p)))+ε_{1}ε_{4}(1−ε_{5}) (⌊\frac{a}{2}⌋ + ε_{3}ν_{p}(F_{z}(p)))

(30)

where ε_{1} = [ℓ ≢ 0 (mod z(p))], ε_{2} = [ℓ ≢ 1 (mod z(p))], ε_{3} = [b ≡ 0 (mod 2)],

ε_{4} = [ℓ ≢ −1 (mod z(p))], and ε_{5} = [a ≡ b (mod 2)].

Proof. Similar to Example 17, this follows from the application of Lemma 1 and Theorem 13. Since (i) is easily verified, we only give the proof of (ii). The calculation is done in two cases. If p ≡ ±2 (mod 5) and a ≡ b (mod 2), then ν_{p} (\binom{p^{b}}{ℓp^{a}}_{F} ) is equal to

ν_{p} (\binom{m_{p}}{k_{p}}) ,  if ℓ ≡ 0, 1 (mod z(p));
\frac{a}{2} + ν_{p}(F_{z(p)}) + ν_{p}(m_{p} − k_{p}) + ν_{p} (\binom{m_{p}}{k_{p}}) , if ℓ ≢ 0, 1 (mod z(p)) and a is even;
\frac{a+1}{2} + ν_{p}(m_{p} − k_{p}) + ν_{p} (\binom{m_{p}}{k_{p}}) ,  if ℓ ≢ 0, 1 (mod z(p)) and a is odd,

= ν_{p} (\binom{m_{p}}{k_{p}}) + ε_{1}ε_{2} (⌈\frac{a}{2}⌉ + ν_{p}(m_{p} − k_{p}) + ε_{3}ν_{p}(F_{z(p)})) ,

where ε_{1} = [ℓ ≢ 0 (mod z(p))], ε_{2} = [ℓ ≢ 1 (mod z(p))], and ε_{3} = [b ≡ 0 (mod 2)]. If p ≡ ±2 (mod 5) and a ≢ b (mod 2), then

ν_{p} (\binom{p^{b}}{ℓp^{a}}_{F} ) = \begin{cases}ν_{p} (\binom{m_{p}}{k_{p}}) , & if ℓ ≡ 0, −1 (mod z(p)); \\ \frac{a}{2} + ν_{p} (\binom{m_{p}}{k_{p}}) , & if ℓ ≢ 0, −1 (mod z(p)) and a is even; \\ \frac{a−1}{2} + ν_{p}(F_{z(p)}) + ν_{p} (\binom{m_{p}}{k_{p}}) , & if ℓ ≢ 0, −1 (mod z(p)) and a is odd, \end{cases}

= ν_{p} (\binom{m_{p}}{k_{p}}) + ε_{1}ε_{4} (⌊\frac{a}{2}⌋ + ε_{3}ν_{p}(F_{z(p)})) ,

where ε_{1}, ε_{2}, ε_{3} are as above and ε_{4} = [ℓ ≢ −1 (mod z(p))]. Let ε_{5} = [a ≡ b (mod 2)]. Then both cases can be combined to obtain (ii).

Example 23. Let k ≥ 2. We observe that z(7) = 8 and

⌊\frac{7^{k}}{8} ⌋ = \begin{cases}\frac{7^{k}−1}{8}, & if k is even; \\ \frac{7(7^{k−1}−1)}{8}, & if k is odd. \end{cases}

Therefore

ν_{7} (⌊\frac{7^{k}}{8} ⌋) = [k ≡ 1 (mod 2)] and ν_{7} (⌊\frac{7^{k}}{8} ⌋ − 1) = 0.  (31)

From (30) and (31), we obtain the following results:

(i) if b − a ≥ 2, then ν_{7} (\binom{7^{b}}{8·7^{a}}_{F} ) = [a ≢ b (mod 2)],

(ii) if b − a ≥ 2, then ν_{7} (\binom{7^{b}}{9·7^{a}}_{F} ) = (⌊\frac{a+2}{2}⌋ + [b ≡ 0 (mod 2)]) [a ≢ b (mod 2)],

(iii) if b − a ≥ 2, then ν_{7} (\binom{7^{b}}{15·7^{a}}_{F} ) is equal to

[a ≢ b (mod 2)] + (⌈\frac{a}{2}⌉ + [b ≡ 0 (mod 2)]) [a ≡ b (mod 2)].

To keep this article not too lengthy, we plan to give more applications of our main results in the next article.

6 Acknowledgments

We are very grateful to the anonymous referee for his/her careful reading, kind words of praise, and many valuable suggestions which greatly improve the presentation of this article. Phakhinkon Phunphayap receives a scholarship from Science Achievement Scholarship of Thailand(SAST). Prapanpong Pongsriiam receives financial support jointly from The Thailand Research Fund and Faculty of Science Silpakorn University, grant number RSA5980040.

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2010 Mathematics Subject Classification: Primary 11B39; Secondary 11B65. Keywords: Fibonacci number, binomial coefficient, Fibonomial coefficient, p-adic valuation, p-adic order, divisibility.

(Concerned with sequences \underline{A000045}, \underline{A003267}, \underline{A010048}, and \underline{A055870}.)

Received October 1 2017; revised version received March 7 2018. Published in Journal of Integer Sequences, March 8 2018.

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