Nontrivial Lower Bounds for the p-adic Valuation of Certain Rational Numbers

Bakir Farhi Laboratoire de Mathématiques Appliquées Faculté des Sciences Exactes Université de Bejaia 06000 Bejaia Algeria

                  bakir.farhi@univ-bejaia.dz

Abstract We show that the p-adic valuation (where p is a given prime number) of certain rational numbers is unusually large. This generalizes very recent results of the author and of Dubickas, which are both related to the special case p = 2. The crucial point for obtaining our main result is the fact that the p-adic valuation of the rational numbers in question is unbounded from above. We confirm this fact by three different methods; the first two are elementary, while the third one relies on p-adic analysis.

1 Introduction and notation

Throughout this paper, we let ℕ denote the set of positive integers and ℕ_{0} := ℕ∪{0} denote the set of non-negative integers. For x ∈ ℝ, we let ⌊x⌋ denote the integer part of x. For a given prime number p and a given non-zero rational number r, we let ν_{p}(r) denote the usual p-adic valuation of r; if in addition r is positive then we let log_{p}(r) denote its logarithm to the base p (i.e., log_{p}(r) := \frac{log r}{log p}). Next, the least common multiple of given positive integers u_{1}, u_{2}, …, u_{n} (n ∈ ℕ) is denoted by lcm(u_{1}, u_{2}, …, u_{n}) . In several places of this paper, we will use the estimate ν_{p}(n) ≤ log_{p}(n) (for primes p and all n ∈ ℕ). We also often use the formula ν_{p} (lcm(1, 2, …, n)) = ⌊log_{p}(n)⌋ (for primes p and all n ∈ ℕ). At the end of the paper, we need to use the p-adic logarithm function which we denote by L_{p} (to differentiate from the notation log_{p}, which is reserved to denote the logarithm to the base p). With the usual notation ℚ_{p} for the field of p-adic numbers, ℂ_{p} for the field of the p-adic complex numbers, and | · |_{p} for the usual p-adic absolute value on ℂ_{p}, recall that L_{p} can be defined by

−L_{p}(1 − x) := ∑_{k=1}^{+∞} \frac{x^{k}}{k} (∀x ∈ ℂ_{p}, |x|_{p} < 1).

(See [5]). The fundamental property of L_{p} is that it satisfies the functional equation

L_{p}(uv) = L_{p}(u) + L_{p}(v)

(for all u, v ∈ ℂ_{p}, with |u − 1|_{p} < 1 and |v − 1|_{p} < 1).

The author [2, 3] has obtained nontrivial lower bounds for the 2-adic valuation of the rational numbers of the form ∑_{k=1}^{n} \frac{2^{k}}{k} (n ∈ ℕ). The stronger one is

ν_{2} (∑_{k=1}^{n} \frac{2^{k}}{k} ) ≥ n − ⌊log_{2}(n)⌋  (∀n ∈ ℕ).  (1)

The author [3] has also posed the problem of generalizing Eq. (1) to other prime numbers p other than p = 2. Dubickas [1] has found arguments for improving and optimizing Eq. (1) by relying solely on the fact that the sequence (ν_{2} (∑_{k=1}^{n} \frac{2^{k}}{k} ))_{n≥1} is unbounded from above.

However, he did not establish a method to prove this fact without relying on Eq. (1). The main result in [1] states that

ν_{2} (∑_{k=1}^{n} \frac{2^{k}}{k} ) ≥ (n + 1) − log_{2}(n + 1),  (2)

for all n ∈ ℕ, with equality if and only if n = 2^{α} − 1 (α ∈ ℕ).

The goal of this paper is twofold. On the one hand, we expand and improve the arguments in [1] to establish a general result providing to us nontrivial lower bounds for the p-adic valuation of a sum of rational numbers under some conditions (see Theorem 2). On the other hand, we solve the problem posed in [3] by generalizing Eqs. (1) and (2) to other prime numbers. More precisely, we show (in different ways) that for all prime numbers p and all non-multiple integers a of p, the sequence

(ν_{p} (∑_{k=1}^{n} (\frac{1}{a^{k}} + \frac{1}{(p − a)^{k}} ) \frac{p^{k}}{k} ))_{n≥1}  (3)

is unbounded from above. Then, by using our general theorem 2, we derive an optimal lower bound for the sequence in Eq. (3). It must be noted that the crucial point of the unboundedness from above of the sequence in Eq. (3) is established by three methods. The first two are elementary and effective while the third one relies on the p-adic analysis and it is ineffective; precisely, it uses the function L_{p} described above. Personally, we consider that the deep reason why the sequence in Eq. (3) is unbounded from above is rather given by the third method.

2 The results and the proofs

Our main result is the following:

Theorem 1. Let p be a prime number, and let a be an integer that is not a multiple of p. Then

ν_{p} (∑_{k=1}^{n} (\frac{1}{a^{k}} + \frac{1}{(p − a)^{k}} ) \frac{p^{k}}{k} ) ≥ (n + 1) − log_{p} (\frac{n + 1}{2})  (4)

for all positive integers n. In addition, this inequality becomes an equality if and only if n has the form n = 2p^{α} − 1 (α ∈ ℕ_{0}).

Note that Theorem 1 generalizes the recent results of the author [2, 3] and Dubickas [1], which are both related to the particular case p = 2. In particular, if we take p = 2 and a = 1 in Theorem 1, we exactly obtain (after some obvious simplifications) the main result of [1], stating that

ν_{2} (∑_{k=1}^{n} \frac{2^{k}}{k} ) ≥ (n + 1) − log_{2}(n + 1) (∀n ∈ ℕ),

with equality if and only if n has the form (2^{α} − 1) (α ∈ ℕ).

The proof of Theorem 1 is based in part on the following result, which can be useful in other situations for establishing a lower bound on the p-adic valuation of certain sums of rational numbers. It must be also noted that the result below is obtained by generalizing the arguments in [1].

Theorem 2. Let p be a fixed prime number and (r_{n})_{n≥1} be a sequence of rational numbers such that the sequence (ν_{p} (∑_{k=1}^{n} r_{k}))_{n≥1} is unbounded from above. Let also (ℓ_{k})_{k≥2} be an increasing real sequence satisfying the property

ℓ_{k} ≤ ν_{p}(r_{k})  (∀k ≥ 2).  (5)

Then we have

ν_{p} (∑_{k=1}^{n} r_{k}) ≥ min_{k≥n+1} ν_{p}(r_{k}) ≥ ℓ_{n+1}  (6)

for all positive integers n. In addition, the inequality ν_{p} (∑_{k=1}^{n} r_{k}) ≥ ℓ_{n+1} becomes an equality if and only if we have

ν_{p}(r_{n+1}) = ℓ_{n+1}.  (7)

Our main result (i.e., Theorem 1) is proven in two steps. In the first one, we suppose (in the situation of Theorem 1) that the sequence (ν_{p} (∑_{k=1}^{n} (\frac{1}{a^{k}} + \frac{1}{(p−a)^{k}} ) \frac{p^{k}}{k} ))_{n≥1} is un-

bounded from above, and we apply for it Theorem 2 to establish the lower bound (4) and characterize the n’s for which it is attained. In the second one, we focus on proving the unboundedness from above of the sequence in question. This is accomplished using three different methods. The first method relies on two identities, one combinatorial and the other arithmetic. The second method utilizes a specific functional equation and Taylor polynomials. Lastly, the third method employs p-adic analysis, specifically the p-adic logarithm function.

Let us begin by proving Theorem 2.

Proof of Theorem 2. Let n be a fixed positive integer. Let us show the first inequality of (6). Since, by hypothesis, the sequence (ν_{p} (∑_{k=1}^{N} r_{k}))_{N≥1} is unbounded from above then

there exists m ∈ ℕ, with m > n, such that

ν_{p} (∑_{k=1}^{m} r_{k}) > ν_{p} (∑_{k=1}^{n} r_{k}) .

Then, by using the elementary properties of the p-adic valuation, we have on the one hand

ν_{p} ( ∑_{k=n+1}^{m} r_{k}) = ν_{p} (∑_{k=1}^{m} r_{k} − ∑_{k=1}^{n} r_{k}) = min (ν_{p} (∑_{k=1}^{m} r_{k}) , ν_{p} (∑_{k=1}^{n} r_{k})) = ν_{p} (∑_{k=1}^{n} r_{k}) ,

and on the other hand

ν_{p} ( ∑_{k=n+1}^{m} r_{k}) ≥ min_{n+1≤k≤m} ν_{p}(r_{k}) ≥ min_{k≥n+1} ν_{p}(r_{k}).

By comparing these two results, we deduce that

ν_{p} (∑_{k=1}^{n} r_{k}) ≥ min_{k≥n+1} ν_{p}(r_{k}),

which is the first inequality of (6). The second inequality of (6) is immediately derived from its first inequality, combined with the properties of the sequence (ℓ_{k})_{k≥2}. Indeed, we have

ν_{p} (∑_{k=1}^{n} r_{k}) ≥ min_{k≥n+1} ν_{p}(r_{k})  (by the first inequality of (6))
≥ min_{k≥n+1} ℓ_{k} (by using (5))
= ℓ_{n+1} (since (ℓ_{k})_{k} is increasing by hypothesis),

confirming the second inequality of (6).

Now, let us prove the second part of Theorem 2. If ν_{p} (∑_{k=1}^{n} r_{k}) = ℓ_{n+1} then we have (according to Eq. (6), proved above) min_{k≥n+1} ν_{p}(r_{k}) = ℓ_{n+1}. Then, using Eq. (5) and the increase of (ℓ_{k})_{k}, we have

ℓ_{n+1} = min_{k≥n+1} ν_{p}(r_{k}) = min (ν_{p}(r_{n+1}), min_{k≥n+2} ν_{p}(r_{k})) ≥ min (ν_{p}(r_{n+1}), min_{k≥n+2} ℓ_{k})
= min (ν_{p}(r_{n+1}), ℓ_{n+2}) .

But since ℓ_{n+1} < ℓ_{n+2}, we must have ν_{p}(r_{n+1}) ≤ ℓ_{n+1}, implying (according to Eq. (5)) that ν_{p}(r_{n+1}) = ℓ_{n+1}, as required. Conversely, suppose that ν_{p}(r_{n+1}) = ℓ_{n+1} and let us show that ν_{p} (∑_{k=1}^{n} r_{k}) = ℓ_{n+1}. If ν_{p} (∑_{k=1}^{n} r_{k}) ≠ ℓ_{n+1}, we must have (in view of Eq. (6))

ν_{p} (∑_{k=1}^{n} r_{k}) > ℓ_{n+1} = ν_{p}(r_{n+1});

consequently, we get

ν_{p} (∑_{k=1}^{n+1} r_{k}) = min (ν_{p} (∑_{k=1}^{n} r_{k}) , ν_{p}(r_{n+1})) = ν_{p}(r_{n+1}) = ℓ_{n+1} < ℓ_{n+2},

contradicting Eq. (6) (applied for the positive integer (n + 1) instead of n). Hence

ν_{p} (∑_{k=1}^{n} r_{k}) = ℓ_{n+1},

as required. This confirms the second part of Theorem 2 and completes this proof.

Next, we have the following fundamental result:

Theorem 3. Let p be a prime number and a be an integer that is not a multiple of p. Then the sequence

(ν_{p} (∑_{k=1}^{n} (\frac{1}{a^{k}} + \frac{1}{(p − a)^{k}} ) \frac{p^{k}}{k} ))_{n≥1}

is unbounded from above.

Admitting Theorem 3 for the moment, our main result is obtained as an application of Theorem 2.

Proof of Theorem 1 by admitting Theorem 3. Let us put ourselves in the situation of Theorem 1. We apply Theorem 2 with r_{k} := (\frac{1}{a^{k}} + \frac{1}{(p−a)^{k}} ) \frac{p^{k}}{k} (∀k ∈ ℕ) and ℓ_{k} := k − log_{p} (\frac{k}{2}) (∀k ≥ 2). The unboundedness from above of the sequence (ν_{p} (∑_{k=1}^{n} r_{k}))_{n≥1} is guaranteed by Theorem 3 (admitted for the moment). Next, the increase of the sequence (ℓ_{k})_{k≥2} can be derived from the increase of the function x ↦ x − log_{p} (\frac{x}{2}) on the interval [2, +∞). Finally, we have for

ν_{p}(r_{k}) = ν_{p} ((\frac{1}{a^{k}} + \frac{1}{(p − a)^{k}} ) \frac{p^{k}}{k} )

= ν_{p} (\frac{a^{k} + (p − a)^{k}}{(a(p − a))^{k}} · \frac{p^{k}}{k} )

= ν_{p} (a^{k} + (p − a)^{k}) + k − ν_{p}(k)  (8)

for all integers k (since a is coprime with p). If k is even, we use ν_{p} (a^{k} + (p − a)^{k}) ≥ ν_{p}(2) (for p > 2, this is obvious and for p = 2, observe that a^{k} + (p − a)^{k} is even). So, we obtain

ν_{p}(r_{k}) ≥ k − ν_{p} (\frac{k}{2}) ≥ k − log_{p} (\frac{k}{2}) = ℓ_{k}

(because k/2 is a positive integer if k is even). However, if k is odd, we use ν_{p} (a^{k} + (p − a)^{k}) ≥ 1 (since a^{k} + (p − a)^{k} ≡ a^{k} + (−a)^{k} (mod p) ≡ 0 (mod p)). So, we again obtain

ν_{p}(r_{k}) ≥ 1 + k − ν_{p}(k) ≥ log_{p}(2) + k − log_{p}(k) = k − log_{p} (\frac{k}{2}) = ℓ_{k}.

Consequently, we have ν_{p}(r_{k}) ≥ ℓ_{k} for all integers k ≥ 2. So, all the hypothesis of Theorem 2 are satisfied; thus we can apply it for our situation. Applying the first part of Theorem 2, we get

ν_{p} (∑_{k=1}^{n} (\frac{1}{a^{k}} + \frac{1}{(p − a)^{k}} ) \frac{p^{k}}{k} ) ≥ min_{k≥n+1} ν_{p} ((\frac{1}{a^{k}} + \frac{1}{(p − a)^{k}} ) \frac{p^{k}}{k} ) ≥ (n + 1) − log_{p} (\frac{n + 1}{2})

for all positive integers n, thus confirming Inequality (4) of Theorem 1. Next, for a given positive integer n, the second part of Theorem 2 tells us that (4) becomes an equality if and only if we have

ν_{p}(r_{n+1}) = (n + 1) − log_{p} (\frac{n + 1}{2}) ,

which is equivalent (by using Eq. (8) and simplifying) to

ν_{p} (a^{n+1} + (p − a)^{n+1}) − ν_{p}(n + 1) = − log_{p} (\frac{n + 1}{2}) .  (9)

So, it remains to prove that Eq. (9) holds if and only if n has the form n = 2p^{α} − 1 (α ∈ ℕ_{0}). Let us prove this last fact.

• Suppose that Eq. (9) holds. Then, we have

log_{p} (\frac{n + 1}{2}) = ν_{p}(n + 1) − ν_{p} (a^{n+1} + (p − a)^{n+1}) ∈ ℤ.

But since log_{p} (\frac{n+1}{2}) ≥ 0, we have even log_{p} (\frac{n+1}{2}) ∈ ℕ_{0}. By setting α := log_{p} (\frac{n+1}{2}) ∈ ℕ_{0}, we get n = 2p^{α} − 1, as required.

• Conversely, suppose that n = 2p^{α} − 1 for some α ∈ ℕ_{0}. Then we have ν_{p}(n + 1) = ν_{p}(2) + α and log_{p} (\frac{n+1}{2}) = α. So Eq. (9) is equivalent to

ν_{p} (a^{n+1} + (p − a)^{n+1}) = ν_{p}(2).  (10)

To confirm Eq. (10), we distinguish two cases:

Case 1: (If p = 2). In this case, since a is coprime with p then a and (p − a) are both odd, implying that a^{2} ≡ 1 (mod 4) and (p−a)^{2} ≡ 1 (mod 4). Then, because n+1 = 2p^{α} is even, we have also a^{n+1} ≡ 1 (mod 4) and (p−a)^{n+1} ≡ 1 (mod 4); thus a^{n+1} + (p − a)^{n+1} ≡ 2 (mod 4), implying that ν_{p} (a^{n+1} + (p − a)^{n+1}) = 1 = ν_{p}(2).

Case 2: (If p is odd). In this case, because n + 1 = 2p^{α} is even, we have a^{n+1} + (p − a)^{n+1} ≡ a^{n+1} + (−a)^{n+1} (mod p) ≡ 2a^{n+1} (mod p) ≢ 0 (mod p) (since p is assumed odd and a is coprime with p). Thus ν_{p} (a^{n+1} + (p − a)^{n+1}) = 0 = ν_{p}(2). Consequently, Formula (10) is confirmed in all cases. This completes the proof of the second part of Theorem 1 and completes this proof.

The rest of the paper is now devoted to proving Theorem 3. We achieve this by three different methods.

2.1 The first method

We rely on two identities. The first one (due to Mansour [6]) is combinatorial and states that

∑_{k=0}^{n} \frac{x^{k}y^{n−k}}{\binom{n}{k}} = \frac{n + 1}{(x + y) (\frac{1}{x} + \frac{1}{y})^{n+1}} ∑_{k=1}^{n+1} \frac{(x^{k} + y^{k}) (\frac{1}{x} + \frac{1}{y})^{k}}{k}  (11)

(for x, y ∈ ℝ^{∗}, with x + y ≠ 0, and n ∈ ℕ_{0}). While the second one (due to the author [4]) is arithmetic and states that

lcm (\binom{n}{0}, \binom{n}{1}, …, \binom{n}{n}) = \frac{lcm (1, 2, …, n, n + 1)}{n + 1}  (12)

(for all n ∈ ℕ_{0}).

Using Eqs. (11) and (12), we are now ready to prove Theorem 3. Let p be a prime number and a be an integer non-multiple of p. By applying Eq. (11) for x = a and y = p − a and replacing n by (n − 1) (where n ∈ ℕ), we get (after simplifying and rearranging)

∑_{k=1}^{n} (\frac{1}{a^{k}} + \frac{1}{(p − a)^{k}} ) \frac{p^{k}}{k} = \frac{p^{n+1}}{n (a(p − a))^{n}} ∑_{k=0}^{n−1} \frac{a^{k}(p − a)^{n−1−k}}{\binom{n−1}{k}} .  (13)

On the other hand, for all n ∈ ℕ, we have (according to Eq. (12))

1 = \frac{n}{lcm(1, 2, …, n)}lcm (\binom{n − 1}{0}, \binom{n − 1}{1}, …, \binom{n − 1}{n − 1}) .  (14)

Then, for a given n ∈ ℕ, by multiplying Eqs. (13) and (14), we obtain

∑_{k=1}^{n} (\frac{1}{a^{k}} + \frac{1}{(p − a)^{k}} ) \frac{p^{k}}{k} = \frac{p^{n+1}}{(a(p − a))^{n} lcm(1, 2, …, n)}

× lcm (\binom{n − 1}{0}, \binom{n − 1}{1}, …, \binom{n − 1}{n − 1}) ∑_{k=0}^{n−1} \frac{a^{k}(p − a)^{n−1−k}}{\binom{n−1}{k}} .

But since the rational number

lcm (\binom{n − 1}{0}, \binom{n − 1}{1}, …, \binom{n − 1}{n − 1}) ∑_{k=0}^{n−1} \frac{a^{k}(p − a)^{n−1−k}}{\binom{n−1}{k}}

is obviously an integer, we derive from the last identity that

ν_{p} (∑_{k=1}^{n} (\frac{1}{a^{k}} + \frac{1}{(p − a)^{k}} ) \frac{p^{k}}{k} ) ≥ ν_{p} (\frac{p^{n+1}}{(a(p − a))^{n} lcm(1, 2, …, n)})

= n + 1 − ν_{p} (lcm(1, 2, …, n)) (since a is not a multiple of p)

= n + 1 − ⌊log_{p}(n)⌋ ,

confirming the unboundedness from above of the sequence (ν_{p} (∑_{k=1}^{n} (\frac{1}{a^{k}} + \frac{1}{(p−a)^{k}} ) \frac{p^{k}}{k} ))_{n≥1} (since n + 1 − ⌊log_{p}(n)⌋ → +∞ as n → +∞). □

2.2 The second method

Let p be a prime number and a be an integer non-multiple of p. For a given n ∈ ℕ, consider the rational function R_{n} defined by

R_{n}(X) := ∑_{k=1}^{n} (\frac{1}{a^{k}} + \frac{1}{(X − a)^{k}} ) \frac{X^{k}}{k} = ∑_{k=1}^{n} \frac{(\frac{X}{a} )k}{k} + ∑_{k=1}^{n} \frac{(\frac{X}{X−a})k}{k} .

Also consider the real function f defined at the neighborhood of 0 by

f(X) := − log (1 − X) ,

which satisfies the functional equation

f (\frac{X}{a} ) + f (\frac{X}{X − a}) = 0  (15)

and whose the n^{th} degree Taylor polynomial at 0 is ∑_{k=1}^{n} \frac{X^{k}}{k}.

On the one hand, according to the well-known properties of Taylor polynomials, the n^{th} degree Taylor polynomial of the function X ↦^{g} f (\frac{X}{a} ) + f (\frac{X}{X−a}) at 0 is the same with the n^{th} degree Taylor polynomial of

∑_{k=1}^{n} \frac{(\frac{X}{a} )k}{k} + ∑_{k=1}^{n} \frac{(\frac{X}{X−a})k}{k} = R_{n}(X).

But on the other hand, in view of Eq. (15), this n^{th} degree Taylor polynomial of g at 0 is zero. Comparing these two results, we deduce that the multiplicity of 0 in R_{n} is at least (n + 1). Consequently, R_{n}(X) can be written as

R_{n}(X) = X^{n+1} · \frac{U_{n}(X)}{a^{n}(X − a)^{n}lcm(1, 2, …, n)},

where U_{n} ∈ ℤ[X]. In particular, we have

R_{n}(p) = p^{n+1} · \frac{U_{n}(p)}{a^{n}(p − a)^{n}lcm(1, 2, …, n)}.

Next, because U_{n}(p) ∈ ℤ (since U_{n} ∈ ℤ[X]) and a is not a multiple of p, then by taking the p-adic valuations in the two sides of the last identity, we derive that

ν_{p} (R_{n}(p)) ≥ n + 1 − ν_{p} (lcm(1, 2, …, n)) = n + 1 − ⌊log_{p}(n)⌋ ,

implying that the sequence (ν_{p} (R_{n}(p)))_{n≥1} is unbounded from above, as required by Theorem 3. □

Remark 4. Curiously, the two previous methods give the same upper bound

ν_{p} (∑_{k=1}^{n} (\frac{1}{a^{k}} + \frac{1}{(p − a)^{k}} ) \frac{p^{k}}{k} ) ≥ n + 1 − ⌊log_{p}(n)⌋ .

Furthermore, this last estimate is remarkably very close to the optimal one of Theorem 1.

In the third method below, we will show the unboundedness of the sequence in Theorem 3 without providing any estimate!

2.3 The third method

Let p be a prime number and a be an integer non-multiple of p. For all n ∈ ℕ, set

r_{n} := (\frac{1}{a^{n}} + \frac{1}{(p − a)^{n}} ) \frac{p^{n}}{n} and s_{n} := ∑_{k=1}^{n} r_{k}.

The property we have to show is that the sequence (ν_{p}(s_{n}))_{n≥1} is unbounded from above; in other words, we have that lim sup_{n→+∞} ν_{p}(s_{n}) = +∞. So, if we show the stronger property lim_{n→+∞} ν_{p}(s_{n}) = +∞, then we are done. To do so, observe that

lim_{n→+∞} ν_{p}(s_{n}) = +∞ ⟺ lim_{n→+∞} |s_{n}|_{p} = 0
⟺ lim_{n→+∞} s_{n} = 0 (in the p-adic sense)
⟺ ∑_{k=1}^{+∞} r_{k} = 0 (in the p-adic sense).

Consequently, it suffices to show that

∑_{k=1}^{+∞} (\frac{1}{a^{k}} + \frac{1}{(p − a)^{k}} ) \frac{p^{k}}{k} = 0  (16)

(in the p-adic sense). Let us show Eq. (16). By using the p-adic logarithm function (recalled in §1), we have

∑_{k=1}^{+∞} (\frac{1}{a^{k}} + \frac{1}{(p − a)^{k}} ) \frac{p^{k}}{k} = ∑_{k=1}^{+∞} \frac{(\frac{p}{a})k}{k} + ∑_{k=1}^{+∞} \frac{(\frac{p}{p−a})k}{k}

= −L_{p} (1 − \frac{p}{a}) − L_{p} (1 − \frac{p}{p − a})

= − [L_{p} (\frac{a − p}{a}) + L_{p} (\frac{−a}{p − a})]

= −L_{p} (\frac{a − p}{a} · \frac{−a}{p − a})

= −L_{p}(1) = 0,

as required. The unboundedness from above of the sequence (ν_{p}(s_{n}))_{n≥1} follows. □

3 Acknowledgment

The author is grateful to the anonymous reviewers for their insightful and constructive comments.

References

[1] A. Dubickas, On a sequence of integers with unusual divisibility by a power of 2, to appear in Miskolc Math. Notes.

[2] B. Farhi, On a curious integer sequence, preprint, 2022. Available at https://arxiv.

   org/abs/2204.10136.

[3] B. Farhi, The integrality of the Genocchi numbers obtained through a new identity and other results, Notes Number Theory Discrete Math. 28 (2022), 749–757.

[4] B. Farhi, An identity involving the least common multiple of binomial coefficients and its application, Amer. Math. Monthly 116 (2009), 836–839.

[5] N. Koblitz, p-adic numbers, p-adic analysis, and zeta-functions, 2nd ed., Springer-Verlag, 1984.

[6] T. Mansour, Combinatorial identities and inverse binomial coefficients, Adv. Appl. Math 28 (2002), 196–202.

2020 Mathematics Subject Classification: Primary 11B83; Secondary 11A41, 05A10, 05A19, 11B65. Keywords: p-adic valuation, binomial coefficient, combinatorial identity, p-adic analysis.

(Concerned with sequences \underline{A002944} and \underline{A087910}.)

Received March 11 2023; revised versions received March 16 2023; June 11 2023. Published in Journal of Integer Sequences, June 12 2023.

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