The Arithmetic Partial Derivative

Brad Emmons and Xiao Xiao Department of Mathematics Utica University Utica, NY 13502 USA

                   bemmons@utica.edu
                   xixiao@utica.edu

Abstract

The arithmetic partial derivative (with respect to a prime p) is a function from the set of integers that sends p to 1 and satisfies the Leibniz rule. In this paper, we prove that the p-adic valuation of the sequence of higher order partial derivatives is eventually periodic. We also prove a criterion to determine when an integer has integral anti-partial derivatives. As an application, we show that there are infinitely many integers with exactly n integral anti-partial derivatives for every nonnegative integer n.

1 Introduction

Let p be a prime number. We recall the p-adic valuation ord_{p} : ℤ → ℕ ∪ {+∞}. For x ∈ ℤ, we have

ord_{p}(x) = \begin{cases}max\{ℓ ∈ ℕ : p^{ℓ} | x\}, & if x ≠ 0; \\ +∞, & if x = 0. \end{cases}

In other words, if x ≠ 0, then ord_{p}(x) is the largest integer ℓ such that p^{ℓ} | x. We sometimes denote this by p^{ℓ} || x.

The arithmetic derivative of a nonzero integer x is a function D : ℤ\{0} → ℤ defined as

D(x) = x ∑_{p|x} \frac{ord_{p}(x)}{p}.

If x = p^{k} is a prime power, then

D(p^{k}) = p^{k} · \frac{k}{p} = kp^{k−1}.

This shows that D mimics the power rule for differentiable functions on prime powers. One can check that D satisfies the Leibniz rule D(xy) = xD(y) + yD(x), and sends every prime to 1. By the Leibniz rule, we are able to extend D uniquely to ℤ by defining D(0) = 0.

Ufnarovski and Åhlander [6] made several conjectures related to this arithmetic derivative, two of which we will investigate here. The first conjecture relates to the dynamical system by considering higher order arithmetic derivatives of an integer. We denote by D^{i}(x) the i-th arithmetic derivative of x for i ≥ 1.

We note that the first conjecture that we would like to address is not correct as stated. The original conjecture [6, Conjecture 2] states that for every positive integer x, exactly one of the following can happen: either D^{i}(x) = 0 for sufficiently large i, lim_{i→+∞} D^{i}(x) = +∞, or

x = p^{p} for some prime p. The fact that D(1647082) = 823543 = 7^{7} means that it is not quite true. We modify it slightly here to the following conjecture.

Conjecture 1. For every positive integer x, exactly one of the following can happen: either D^{i}(x) = 0 or p^{p} for some prime p for sufficiently large i, or lim_{i→+∞} D^{i}(x) = +∞.

The second conjecture relates to the number of anti-derivatives of an integer. An integer y is said to be an (integral) anti-derivative of x if D(y) = x. Unlike anti-derivatives of functions, x has only finitely many anti-derivatives [6, Corollary 3] for every integer x > 1. We denote by i(x) the number of anti-derivatives of an integer x.

Conjecture 2. For each nonnegative integer n, the number of solutions to the equation i(x) = n is infinite.

Part of what makes investigating the function D(x) challenging is that the derivative of an integer x will often have prime factors that are not factors of x itself. For instance, if x = 72 = 2^{3} 3^{2}, then D(x) = 156 = 2^{2} 3^{1} 13, and we see that while the multiplicities of the prime factors 2 and 3 each decreased by 1, the derivative D(x) has a factor of 13. In particular, when investigating the dynamical system D^{i}(x), we would have to keep track of new prime factors that may arise following each successive differentiation. If x has more than one prime factor, predicting new prime factors of D(x) relies on the ability of predicting prime factors of a + b when knowing the prime factors of a and b. A difficult conjecture that tries to tackle a question in this direction is the famous abc conjecture. There is still ongoing debate in the mathematics community about whether abc conjecture has been proved [5]. Loosely speaking, the abc conjecture implies that when two numbers a and b are divisible by large powers of small primes, then a+b tends to be divisible by small powers of large primes. Because of difficulties like these, we start by investigating these conjectures in the case of arithmetic partial derivatives with respect to a prime p, which was formally introduced by Kovič [3].

For nonzero x ∈ ℤ, the (arithmetic) partial derivative (with respect to p) is defined as

D_{p}(x) = \frac{x}{p} · ord_{p}(x).

One can check that D_{p} satisfies the Leibniz rule D_{p}(xy) = D_{p}(x)y + xD_{p}(y). And to extend the definition of D_{p} to 0, we observe that for every nonzero x ∈ ℤ we have

D_{p}(0) = D_{p}(0 · x) = D_{p}(0) · x + 0 · D_{p}(x) = D_{p}(0) · x.

This implies that D_{p}(0) = 0. If p does not divide x, then D_{p}(x) = 0. For every integer x, the arithmetic derivative of x is a finite sum of arithmetic partial derivative of x, that is,

D(x) = ∑_{p|x} D_{p}(x).

The main advantage that we have in exploring the dynamical system D_{p}^{i}(x) over D^{i}(x) is that now we do not need to keep track of new primes that may arise when we differentiate. This is because if x = ap^{ℓ}, where ℓ = ord_{p}(x), then D_{p}(x) = aD_{p}(p^{ℓ}), and so upon repeated differentiation, we only need to keep track of ord_{p}(x) and its partial derivatives with respect to p. Any new prime factors that may arise when differentiating get absorbed into the coefficient.

Because we do not have to deal with new primes occurring in our dynamical system, we are able to prove a stronger version of Conjecture 1 for the arithmetic partial derivative. Theorem 8 provides an exact prescription of the ord_{p} sequence of x when ord_{p}(x) ≥ p, and as a corollary we prove Theorem 10, which is analogous to Conjecture 1; see Section 2 for the definition of ord_{p} sequence. We go on to prove in Theorem 11 that given any possible ord_{p} sequence as prescribed in Theorem 8, we are able to construct an integer x with this ord_{p} sequence.

For each integer x, we say that an integer y is an (integral) anti-partial derivative of x if D_{p}(y) = x. Let i_{p}(x) be the number of anti-partial derivatives of x. Theorem 21 and Example 13 together show that for each nonnegative integer n, the number of solutions to the equation i_{p}(x) = n is infinite, which is analogous to Conjecture 2. We think the natural next step is to study analogous conjectures for the arithmetic subderivative D_{S} defined by Haukkanen, Merikoski, and Tossavainen [2] in the following way:

D_{S}(x) := x ∑_{p∈S} \frac{ord_{p}(x)}{p} = ∑_{p∈S} D_{p}(x),

where S is a finite set of primes.

The definition of arithmetic derivative and arithmetic partial derivative can be generalized to rational numbers [6, 1]. Haukkanen, Merikoski, and Tossavainen [1, Conjecture 27] conjecture that for every nonzero rational number x, there are at most four rational anti-partial derivatives if p = 2, and at most two rational anti-partial derivatives if p > 2. As a result of Theorem 21, we prove that this conjecture is false; see Remark 22. This also confirms Pandey and Saxena’s earlier observations [4] that [1, Conjecture 27] might be false.

2 Period of the ord_{p} sequence

We fix a prime p. For x ∈ ℤ, we refer to the sequence x, D_{p}(x), D_{p}^{2}(x), … as the D_{p} sequence of x, and the sequence ord_{p}(x), ord_{p}(D_{p}(x)), ord_{p}(D_{p}^{2}(x)), … as the ord_{p} sequence of x. In this section, we will study properties of ord_{p} sequences and show that they are eventually periodic. For the sake of completeness, we first give the definition of eventually periodic.

Definition 3. Given a sequence x_{1}, x_{2}, …, we say that it is eventually periodic if there exist integers m, L ≥ 1 such that for all 0 ≤ j ≤ L − 1 and i ≥ 1, we have x_{m+j+iL} = x_{m+j}. If a sequence is eventually periodic with m = 1, then we say the sequence is periodic. The smallest possible L is called the period of the sequence.

If the D_{p} sequence of x is periodic (resp., eventually periodic), then the ord_{p} sequence of x is periodic (resp., eventually periodic), but not vice versa.

Lemma 4. Let x ∈ ℤ. If x = 0, then the D_{p} sequence of x (and thus the ord_{p} sequence of x) is periodic with period 1. If ord_{p}(x) < p, then the D_{p} sequence of x (and thus the ord_{p} sequence of x) is eventually periodic with period 1.

Proof. If x = 0, then D_{p}(x) = 0 and thus D_{p}^{i}(x) = 0 for all i ≥ 0. Therefore the D_{p} sequence of x is periodic of period 1.

Suppose x ≠ 0. We write x = ap^{ℓ} where p ∤ a and ord_{p}(x) = ℓ < p. We can compute that D_{p}^{ℓ}(x) = aℓ!. Since 0 ≤ ℓ < p, we know that p ∤ aℓ!. This implies that D_{p}^{ℓ+1}(x) = 0 and thus D_{p}^{i}(x) = 0 for all i ≥ ℓ + 1. Therefore the D_{p} sequence of x is eventually periodic of period 1.

Suppose ord_{p}(D_{p}^{i}(x)) ≠ +∞ for all i ≥ 1. To understand how each term in ord_{p} sequence changes, we study the sequence of increments (of consecutive terms) of the ord_{p} sequence

ord_{p}(D_{p}(x)) − ord_{p}(x), ord_{p}(D_{p}^{2}(x)) − ord_{p}(D_{p}(x)), ord_{p}(D_{p}^{3}(x)) − ord_{p}(D_{p}^{2}(x)), …

and we call this the inc_{p} sequence of x. Recall that for integers x ∈ ℤ, we let x mod p denote the unique integer between 0 and p − 1 that is congruent to x modulo p. We denote ⌊x⌋_{p} := x − (x mod p). In other words, ⌊x⌋_{p} is the largest multiple of p that is less than or equal to x.

We are ready to handle the case ord_{p}(x) ≥ p.

Lemma 5. Let x = ap^{bp^{k}} be a nonzero integer with p ∤ ab and k ≥ 1. Define k′ := (k − 1 mod p) + 1. The first k′ terms of the inc_{p} sequence of x are

(k − 1, \underbrace{−1, −1, …, −1}_{(k−1 mod p) copies}).

Proof. Since D_{p}(x) = abp^{bp^{k}+k−1} and p ∤ ab, we know ord_{p}(D_{p}(x)) = bp^{k} + k − 1. If k′ ≥ 2, then D_{p}^{2}(x) = ab(bp^{k} + k − 1)p^{bp^{k}+k−2}. Since p ∤ ab(bp^{k} + k − 1), we know ord_{p}(D_{p}^{2}(x)) = bp^{k} + k − 2. Following this pattern, we know that when 1 ≤ i ≤ k′, the (i + 1)-th term of the D_{p} sequence is

ab ∏_{j=1}^{i−1}(bp^{k} + k − j)p^{bp^{k}+k−i}

because p ∤ ab ∏_{j=1}^{i−1}(bp^{k} + k − j) for 1 ≤ i ≤ k′. The first k′ + 1 terms of the ord_{p} sequence are

(bp^{k}, bp^{k} + k − 1, bp^{k} + k − 2, …, bp^{k} + k − k′).

Therefore the first k′ terms of the inc_{p} sequence of x are

(k − 1, \underbrace{−1, −1, …, −1}_{(k−1 mod p) copies}).

Corollary 6. Let x = ap^{bp^{k}} be a nonzero integer with p ∤ ab and 1 ≤ k ≤ p. The ord_{p} sequence and the inc_{p} sequence of x is periodic of period k.

Proof. If 1 ≤ k ≤ p, then k′ = (k − 1 mod p) + 1 = k − 1 + 1 = k. The first k + 1 terms of the ord_{p} sequence are

(bp^{k}, bp^{k} + k − 1, bp^{k} + k − 2, …, bp^{k}).

It is now clear that the ord_{p} sequence and the inc_{p} sequence of x is periodic of period k.

Definition 7. For integers k ≥ 1, we call the following sequence

S_{k} := (k − 1, \underbrace{−1, −1, …, −1}_{(k−1 mod p) copies})

the k-segment.

For integers ℓ ≥ p, we define ℓ_{1} := ord_{p}(⌊ℓ⌋_{p}) and for i ≥ 2

ℓ_{i} := \begin{cases}ord_{p}(⌊ℓ_{i−1} − 1⌋_{p}), & if ℓ_{i−1} < +∞; \\ +∞, & if ℓ_{i−1} = +∞. \end{cases}  (1)

It is clear that if ℓ_{i} < +∞, then ℓ_{i+1} < log_{p}(ℓ_{i}). For every integer ℓ ≥ p, there exists a unique positive integer N = N(ℓ) such that 1 ≤ ℓ_{N} ≤ p, and ℓ_{i} = +∞ for all i > N.

Theorem 8. Let x = ap^{ℓ} be a nonzero integer with ℓ ≥ p and p ∤ a.The inc_{p} sequence of x is of the form

(\underbrace{−1, −1, …, −1}_{(ℓ mod p) copies}, S_{ℓ_{1}}, S_{ℓ_{2}}, S_{ℓ_{3}}, …, S_{ℓ_{N}} , S_{ℓ_{N}} , S_{ℓ_{N}} , …).

As a result, the ord_{p} sequence and the inc_{p} sequence of x are eventually periodic of period ℓ_{N}.

Proof. Let ℓ′_{0} = ℓ mod p. Then the first ℓ′_{0} + 1 terms of the D_{p} sequence of x are

(ap^{ℓ}, aℓp^{ℓ−1}, aℓ(ℓ − 1)p^{ℓ−2}, …, a ∏_{i=1}^{ℓ′_{0}}(ℓ − i)p^{ℓ−ℓ′_{0}}).

The first ℓ′_{0} + 1 terms of the ord_{p} sequence of x are

(ℓ, ℓ − 1, ℓ − 2, …, ℓ − ℓ′_{0}).

Hence the first ℓ′_{0} terms of the inc_{p} sequence of x are

(\underbrace{−1, −1, …, −1}_{ℓ′_{0} copies}).

Note that ℓ − ℓ′_{0} = ⌊ℓ⌋_{p} and ℓ_{1} = ord_{p}(⌊ℓ⌋_{p}), we can write D_{p}^{ℓ′_{0}}(x) = a_{0}p^{b_{0}p^{ℓ_{1}}} with p ∤ a_{0}b_{0}. We can compute D_{p}^{ℓ′_{0}+1}(x) = a_{0}b_{0}p^{b_{0}p^{ℓ_{1}}+ℓ_{1}−1}. Note that b_{0}p^{ℓ_{1}} + ℓ_{1} − 1 mod p = ℓ_{1} − 1 mod p.

Let ℓ′_{1} := (ℓ_{1} − 1) mod p. The (ℓ′_{0} + 1)-th term to the (ℓ′_{0} + ℓ′_{1} + 2)-th term of the D_{p} sequence of x are

(a_{0}p^{b_{0}p^{ℓ_{1}}}, a_{0}b_{0}p^{b_{0}p^{ℓ_{1}}+ℓ_{1}−1}, …, a_{0}b_{0} ∏_{i=1}^{ℓ′_{1}}(b_{0}p^{ℓ_{1}} + ℓ_{1} − i)p^{b_{0}p^{ℓ_{1}}+ℓ_{1}−1−ℓ′_{1}}).

The (ℓ′_{0} + 1)-th term to the (ℓ′_{0} + ℓ′_{1} + 2)-th term of the ord_{p} sequence of x are

(b_{0}p^{ℓ_{1}}, b_{0}p^{ℓ_{1}} + ℓ_{1} − 1, …, b_{0}p^{ℓ_{1}} + ℓ_{1} − 1 − ℓ′_{1}).

The next ℓ′_{1} + 1 terms of the inc_{p} sequence of x is the ℓ_{1}-segment

(ℓ_{1} − 1, \underbrace{−1, −1, …, −1}_{ℓ′_{1} copies}) = S_{ℓ_{1}}.

As ℓ_{1} − 1 − ℓ′_{1} = ⌊ℓ_{1} − 1⌋_{p} and ℓ_{2} = ord_{p}(⌊ℓ_{1} − 1⌋_{p}), we can write D_{p}^{ℓ′_{0}+ℓ′_{1}+1}(x) = a_{1}p^{b_{1}p^{ℓ_{2}}} with p ∤ a_{1}b_{1}. There exists a positive integer N such that 1 ≤ ℓ_{N} ≤ p. Using induction, we know that the initial terms of the inc_{p} sequence of x is of the form

(\underbrace{−1, −1, …, −1}_{(ℓ mod p) copies}, S_{ℓ_{1}}, S_{ℓ_{2}}, S_{ℓ_{3}}, …, S_{ℓ_{N}}).

Corollary 6 implies that if a_{N−1}p^{b_{N−1}p^{ℓ_{N}}}, where p ∤ a_{N−1}b_{N−1}, is a term in the D_{p} sequence of x, then S_{ℓ_{N}} will appear repeatedly in the inc_{p} sequence of x. This completes the proof of the theorem.

We summarize our results of Lemma 4 and Theorem 8 into the following theorem.

Theorem 9. The ord_{p} sequence of any integer x is eventually periodic of period at most p.

As an application of Theorem 8, we can prove the following theorem that has a similar structure as Conjecture 1.

Theorem 10. For any x ∈ ℤ, exactly one of the following three cases will happen:

(i) D_{p}^{i}(x) = 0 for sufficiently large i;

(ii) D_{p}^{i}(x) = ap^{p} where gcd(a, p) = 1 for sufficiently large i;

(iii) lim_{i→+∞} D_{p}^{i}(x) = ±∞.

Proof. Suppose x = 0 or 0 ≤ ord_{p}(x) < p. Lemma 4 implies that D_{p}^{i}(x) = 0 for sufficiently large i.

Suppose ord_{p}(x) ≥ p. Theorem 8 implies that the ord_{p} sequence of x is eventually periodic of period ℓ_{N}. Let x_{1} = ap^{bp^{k}} be the first term of the periodic cycle where 1 ≤ k ≤ p and p ∤ ab. If b = k = 1, then D_{p}(x_{1}) = D_{p}(ap^{p}) = ap^{p} = x_{1}. We know that D_{p}^{i}(x) = ap^{p} for sufficiently large i. Suppose b > 1 or k > 1. If x_{1} > 0, then

x_{1} < D_{p}(x_{1}) < D_{p}^{2}(x_{1}) < ⋯ < D_{p}^{ℓ_{N_{0}}−1}(x_{1}) < D_{p}^{ℓ_{N_{0}}}(x_{1}) = αx_{1},

for some α > 1. Therefore

lim_{i→+∞} D_{p}^{i}(x) = lim_{i→+∞} α^{i}x_{1} = +∞.

Similarly, if x_{1} < 0, then

x_{1} > D_{p}(x_{1}) > D_{p}^{2}(x_{1}) > ⋯ > D_{p}^{ℓ_{N_{0}}−1}(x_{1}) > D_{p}^{ℓ_{N_{0}}}(x_{1}) = αx_{1},

for some α > 1. Therefore

lim_{i→+∞} D_{p}^{i}(x) = lim_{i→+∞} α^{i}x_{1} = −∞.

Theorem 8 gives the form of the inc_{p} sequence of any integer x such that ord_{p}(x) ≥ p. It is a sequence of nonnegative integers that rapidly decreases until it is less than p, after which it is constant, where the first term determines all subsequent terms. This sequence can be interrupted by strings of −1’s, where the length of the string is equal to the previous nonnegative integer modulo p.

The question remains about whether all sequences of the form

(\underbrace{−1, −1, …, −1}_{(ℓ mod p) copies}, S_{ℓ_{1}}, S_{ℓ_{2}}, S_{ℓ_{3}}, …, S_{ℓ_{N}} , S_{ℓ_{N}} , S_{ℓ_{N}} , …)

has an x such that this is its inc_{p} sequence. Since the initial ℓ_{1} determines all the subsequent ℓ_{i}’s, the answer is no. But we do have some control over the length of the interrupting −1’s and the number of jumps N.

Theorem 11. Let N > 0 be an integer. For N + 1 integers 0 ≤ i_{0}, i_{1}, i_{2}, …, i_{N} ≤ p − 1, there exists a positive integer ℓ such that the inc_{p} sequence of x = p^{ℓ} is of the form

(\underbrace{−1, −1, …, −1}_{i_{0} copies}, S_{ℓ_{1}}, S_{ℓ_{2}}, S_{ℓ_{3}}, …, S_{ℓ_{N}} , S_{ℓ_{N}} , S_{ℓ_{N}} , …),

such that for all 1 ≤ j ≤ N, S_{ℓ_{j}} is the ℓ_{j}-segment with i_{j} consecutive copies of −1.

Proof. Given a sequence of N + 1 nonnegative integers 0 ≤ i_{0}, i_{1}, i_{2}, …, i_{N} ≤ p − 1, we construct a sequence k_{N}, k_{N−1}, …, k_{1}, k_{0} as follows. Let k_{N} = i_{N} + 1 and k_{j} = p^{k_{j+1}} + i_{j} + 1 for 0 ≤ j ≤ N − 1. Let ℓ := k_{0} − 1 and x = p^{k_{0}−1}. We first show that ℓ_{j} = k_{j} for all 1 ≤ j ≤ N. By definition of ℓ_{1}, we can compute

ℓ_{1} = ord_{p}(⌊ℓ⌋_{p}) = ord_{p}(⌊k_{0} − 1⌋_{p}) = ord_{p}(⌊p^{k_{1}} + i_{0}⌋_{p}) = ord_{p}(p^{k_{1}}) = k_{1}.

Suppose for 2 ≤ j ≤ m − 1 < N, we have ℓ_{j} = k_{j}. Then we have

ℓ_{m} = ord_{p}(⌊ℓ_{m−1} − 1⌋_{p}) = ord_{p}(⌊k_{m−1} − 1⌋_{p}) = ord_{p}(⌊p^{k_{m}} + i_{m−1}⌋_{p}) = ord_{p}(p^{k_{m}}) = k_{m}.

It remains to show that the inc_{p} sequence of x is of the desired form. As k_{0} − 1 = p^{k_{1}} + i_{0}, it is clear that the first i_{0} terms of the inc_{p} sequence of x are −1 and ord_{p}(D_{p}^{i_{0}}(x)) = p^{k_{1}}. Suppose D_{p}^{i_{0}}(x) = αp^{p^{k_{1}}}. Set k′_{1} := (k_{1} − 1 mod p) + 1 = (i_{1} mod p) + 1 = i_{1} + 1. By Lemma 5, we know that the next k′_{1} terms of the inc_{p} sequence is a k_{1}-segment S_{k_{1}} = S_{ℓ_{1}}. The number of consecutive −1 in this k_{1}-segment S_{k_{1}} is equal to k′_{1} − 1 = i_{1} as desired. The proof is complete by induction.

For a nonzero integer x = ap^{ℓ} with p ∤ a and ℓ ≥ 1, Theorem 8 implies that the ord_{p} sequence increases (not necessarily consecutively) exactly N − 1 times before it reaches the periodic cycle. Therefore it is impossible for the ord_{p}-sequence to continue to increase indefinitely. However, we see that the number of jumps can be made arbitrarily large. And as the following corollary tells us, we can even prescribe these jumps to be consecutive.

Corollary 12. For every integer N ≥ 1, there exists x ∈ ℤ such that

ord_{p}(x) < ord_{p}(D_{p}(x)) < ⋯ < ord_{p}(D_{p}^{N}(x)).

Proof. Use Theorem 11 by choosing i_{0} = i_{1} = ⋯ = i_{N} = 0.

3 Anti-partial derivatives

Not every integer has an anti-partial derivative with respect to p. We would like to have a criterion to determine when an integer does have an anti-partial derivative (Question 1). If we know an integer has an anti-partial derivative, a natural extension of Question 1 is how many anti-partial derivatives can it have (Question 3)? In order to answer that, it would be useful to know when two integers have the same partial derivatives (Question 2). Furthermore, if we know that there exists at least one integer that has exactly n anti-partial derivatives, then we can also ask how many integers there are that have exactly n anti-partial derivatives (Question 4). In this section, we will attempt to answer these four questions:

1. When does an integer have an anti-partial derivative?

2. When do two integers have the same partial derivatives?

3. How many anti-partial derivatives can an integer have?

4. For a positive integer n, how many integers have exactly n anti-partial derivatives? For a nonzero integer x, if ord_{p}(x) > 0, then there exist unique integers a ≠ 0, b > 0, and k ≥ 0 with p ∤ ab such that x = ap^{bp^{k}}. We call this the standard form of x.

3.1 When does an integer have an anti-partial derivative?

We start this subsection with an example of a family of integers that does not have an anti-partial derivative. Example 13. Suppose y = a_{0}p^{p−1} for some integer a_{0} where p ∤ a_{0}. If D_{p}(x) = y, then it is clear that x ≠ 0 and ord_{p}(x) > 0. Let x = ap^{bp^{k}} be in standard form such that D_{p}(x) = a_{0}p^{p−1}. Then abp^{bp^{k}+k−1} = a_{0}p^{p−1}. This implies that bp^{k} + k = p. As b ≥ 1, we know that k = 0. Therefore b = p and this contradicts to p ∤ ab. Therefore a_{0}p^{p−1} does not have an (integral) anti-partial derivative (with respect to p).

The main goal of this subsection is to determine when an integer y has an anti-partial derivative. We first consider two simple cases.

• If y = 0, then D_{p}(a) = 0 for all a ∈ ℤ with p ∤ a.

• If ord_{p}(y) = 0, then D_{p}(x) = y if and only if x = yp. Now we assume that ord_{p}(y) > 0. If D_{p}(x) = y, then ord_{p}(x) > 0. Lemma 14. Let y = a_{0}p^{ℓ_{0}} where p ∤ a_{0} and ℓ_{0} > 0. For x = ap^{bp^{k}} ∈ ℤ in standard form, we have D_{p}(x) = y if and only if

ab = a_{0},  (2)
bp^{k} + k − 1 = ℓ_{0}.  (3)

Proof. Since D_{p}(x) = abp^{bp^{k}+k−1}, it is clear that abp^{bp^{k}+k−1} = a_{0}p^{ℓ_{0}} if and only if (2) and (3) hold.

Using Lemma 14, we can show that every nonzero integer has finitely many anti-partial derivatives. This is already known as a result of [1, Theorem 1].

Corollary 15. For each nonzero y ∈ ℤ, the set {x ∈ ℤ : D_{p}(x) = y} is finite (possibly empty).

Proof. If ord_{p}(y) = 0, then D_{p}(x) = y if and only if x = yp. If ord_{p}(y) > 0, then Eq. (3) implies that k ≤ log_{p}(ℓ_{0}). Since b is determined by k by (3) and a is determined by b by (2), there are only finitely many a, b, k such that (2) and (3) hold.

Corollary 16. For two nonzero integers x_{1} and x_{2} with ord_{p}(x_{1}), ord_{p}(x_{2}) > 0, let x_{1} = a_{1}p^{b_{1}p^{k_{1}}} and x_{2} = a_{2}p^{b_{2}p^{k_{2}}} be in standard form such that D_{p}(x_{1}) = D_{p}(x_{2}). The following three statements are equivalent.

(i) k_{1} < k_{2}  (ii) b_{1}p^{k_{1}} > b_{2}p^{k_{2}}  (iii) b_{1} > b_{2}

If D_{p}(x_{1}) = D_{p}(x_{2}) > 0, then the above three statements are further equivalent to the following two statements.

(iv) a_{1} < a_{2} (v) x_{1} < x_{2}

Proof. If D_{p}(x_{1}) = D_{p}(x_{2}), then we have

a_{1}b_{1} = a_{2}b_{2}  (4)
b_{1}p^{k_{1}} + k_{1} − 1 = b_{2}p^{k_{2}} + k_{2} − 1,  (5)

Eq. (5) implies that (i) and (ii) are equivalent. If (i) and (ii) hold, then (iii) holds. If (iii) holds, then Eq. (5) implies that (i) holds.

Now suppose D_{p}(x_{1}) = D_{p}(x_{2}) > 0. Hence a_{1}, a_{2}, x_{1}, x_{2} > 0. Eq. (4) implies that (iii) and (iv) are equivalent. It remains to show that (v) is equivalent to (ii). Note that

\frac{x_{1}}{x_{2}} = \frac{a_{1}}{a_{2}} · \frac{p^{b_{1}p^{k_{1}}}}{p^{b_{2}p^{k_{2}}}} = \frac{b_{2}}{b_{1}} p^{b_{1}p^{k_{1}}−b_{2}p^{k_{2}}} = \frac{b_{2}}{b_{1}} p^{k_{2}−k_{1}} = \frac{b_{2}p^{k_{2}}}{b_{1}p^{k_{1}}} .

Therefore (v) is equivalent to (ii).

If D_{p}(x_{1}) = D_{p}(x_{2}) < 0, then (iv) and (v) of Corollary 16 need to be replaced by a_{1} > a_{2} and x_{1} > x_{2} respectively.

3.2 When do two integers have the same partial derivatives?

Anti-partial derivatives do not exist uniquely. For example

D_{p}(p^{p^{p+1}+p}) = D_{p}((p^{p} + 1)p^{p^{p+1}}).

The main goal of this subsection is to determine when two integers have the same partial derivatives. If D_{p}(x) = 0, then x = 0 or ord_{p}(x) = 0. This means that the set of anti-partial derivatives of 0 is

{x ∈ ℤ : p ∤ x} ∪ {0}.

Now we suppose that ord_{p}(x) > 0 and let y = D_{p}(x). By Corollary 16, there is an element x_{0} in the set of all anti-partial derivative of y with the smallest value k. We call x_{0} the primitive anti-partial derivative of y and denote x_{0} = a_{0}p^{b_{0}p^{k_{0}}} in the standard form.

Theorem 17. Let x_{0} = a_{0}p^{b_{0}p^{k_{0}}} ∈ ℤ be the standard form of the primitive anti-partial derivative of y = D_{p}(x_{0}). Define

C = C(x_{0}) := \{c ∈ [0, b_{0}) ∩ ℤ : p^{p^{k_{0}}c} || (b_{0} − c), \frac{b_{0} − c}{p^{p^{k_{0}}c}} | a_{0}b_{0}\}.

There is a one-to-one correspondence between C and the set of all anti-partial derivatives of y. Furthermore, if x = ap^{bp^{k}} (in its standard form) is an anti-partial derivative of y, then there exists a unique c ∈ C such that

k = p^{k_{0}}c + k_{0}, b = \frac{b_{0} − c}{p^{k−k_{0}}} = \frac{b_{0} − c}{p^{p^{k_{0}}c}} ∈ ℤ, a = \frac{a_{0}b_{0}}{b} ∈ ℤ.

Proof. We show that a anti-partial derivative x = ap^{bp^{k}} of y is associated with a unique c ∈ C. If x = x_{0}, then we associated x with c = 0. If x ≠ x_{0}, since x_{0} is the primitive anti-partial derivative, we know that k > k_{0}. Then p^{k_{0}}(b_{0} − bp^{k−k_{0}}) = k − k_{0}. Since p ∤ bb_{0}, we know that p ∤ (b_{0} − bp^{k−k_{0}}). Therefore p^{k_{0}} || (k − k_{0}). Let k − k_{0} = p^{k_{0}}c where c > 0 and p ∤ c. By plugging k − k_{0} = p^{k_{0}}c into p^{k_{0}}(b_{0} − bp^{k−k_{0}}) = k − k_{0}, we get b_{0} − bp^{k−k_{0}} = c. Since bp^{k−k_{0}} > 0, we know that c < b_{0}. Since p ∤ b, we know that p^{k−k_{0}} || (b_{0} − c). Since a ∈ ℤ, we also have b | a_{0}b_{0}, that is, \frac{b_{0}−c}{p^{p^{k_{0}} c}} | a_{0}b_{0}.

Then we show that for each c ∈ C, we can define a unique x = ap^{bp^{k}} such that D_{p}(x) = y. Since c < b_{0} and p^{p^{k_{0}}c} || (b_{0} − c), we know that b > 0 is an integer and ord_{p}(b) = 0. Since b | a_{0}b_{0}, we know that a ∈ ℤ and ord_{p}(a) = 0 as well. We can compute

bp^{k} + k − 1 = \frac{b_{0} − c}{p^{k−k_{0}}}p^{k} + k − 1 = (b_{0} − c)p^{k_{0}} + p^{k_{0}}c + k_{0} − 1 = b_{0}p^{k_{0}} + k_{0} − 1.

Therefore

D_{p}(x) = abp^{bp^{k}+k−1} = a_{0}b_{0}p^{b_{0}p^{k_{0}}+k_{0}−1} = D_{p}(x_{0}) = y.

3.3 How many anti-partial derivatives can an integer have?

By Corollary 15, we know that every nonzero integer can only have finitely many anti-partial derivatives. This partially answers Question 3, but we can further ask for a given positive integer n, is there an integer x with exactly n anti-partial derivatives? The answer is yes. In order to prove that, we will construct k_{0} such that x_{0} is the primitive anti-partial derivative of D_{p}(x_{0}) and then construct a_{0} and b_{0} so that D_{p}(x_{0}) has exactly n anti-partial derivatives. We first show that for certain values of k_{0}, the quantity x_{0} is always the primitive anti-partial derivative of D_{p}(x_{0}) no matter how a_{0} and b_{0} are defined as long as p ∤ a_{0}b_{0}.

Lemma 18. Fix an integer m ≥ 2 and let k_{0} := p+p^{2}+p^{3}+⋯+p^{m}. For integers a_{0}, b_{0} with b_{0} > 0 such that p ∤ a_{0}b_{0}, let x_{0} := a_{0}p^{b_{0}p^{k_{0}}}. Then x_{0} is the primitive anti-partial derivative of D_{p}(x_{0}).

Proof. Suppose x_{0} is not the primitive anti-partial derivative of D_{p}(x_{0}). Let x = ap^{bp^{k}} be an anti-partial derivative of D_{p}(x_{0}) such that 0 ≤ k < k_{0}. Since D_{p}(x) = D_{p}(x_{0}), we get bp^{k} +k = b_{0}p^{k_{0}} +k_{0}. This means that p^{k} || k_{0} −k. It is clear that k ≠ 0 because p || k_{0} −0. It is also clear that k ≠ 1 because p ∤ k_{0} − 1. Suppose k ≥ 2. If p^{k} || k_{0} − ℓ for some ℓ ∈ ℤ, then ℓ ≥ p+p^{2} +⋯+p^{k−1} > k. Therefore there does not exist an anti-partial derivative x = ap^{bp^{k}} with k < k_{0}. This means that x_{0} is the primitive anti-partial derivative of D_{p}(x_{0}).

We now construct b_{0} such that there are exactly n − 1 different possible values of c ∈ (0, b_{0}) ∩ ℤ such that p^{p^{k_{0}}c} || (b_{0} − c) for k_{0} ≥ 0. This means that the set C(x_{0}) (as defined in Theorem 17) has at most n elements (including 0).

Lemma 19. Fix integers n > 0 and k_{0} ≥ 0. Let c_{1} = 0. For 2 ≤ i ≤ n + 1, let c_{i} := p^{p^{k_{0}}c_{i−1}} + c_{i−1} and b_{0} := c_{n+1}. We have

{c ∈ (0, b_{0}) ∩ ℤ : p^{p^{k_{0}}c} || b_{0} − c} = {c_{2}, …, c_{n}}.

Proof. First, we prove ⊇. By the definition of c_{j}, we know that c_{j+1} > c_{j} and p^{p^{k_{0}}c_{j}} || c_{j+1}−c_{j} for all 1 ≤ j ≤ n. For 1 ≤ i ≤ n, we have

b_{0} − c_{i} = ∑_{j=i}^{n}(c_{j+1} − c_{j}) = ∑_{j=i}^{n} p^{p^{k_{0}}c_{j}}.

Therefore p^{p^{k_{0}}c_{i}} || b_{0} − c_{i}.

Second, we prove ⊆. Let c ∈ (0, b_{0}) be an integer such that p^{p^{k_{0}}c} || b_{0} − c. If c ∉ {c_{2}, …, c_{n}}, then there exists a unique 3 ≤ i ≤ n + 1 such that c_{i−1} < c < c_{i}. As

b_{0} − c = ∑_{j=i}^{n}(c_{j+1} − c_{j}) + (c_{i} − c) = ∑_{j=i}^{n} p^{p^{k_{0}}c_{j}} + (c_{i} − c).

Since c < c_{j} for all i ≤ j ≤ n, we have p^{p^{k_{0}}c} | p^{p^{k_{0}}c_{j}}. Since p^{p^{k_{0}}c} || b_{0} − c, we know that p^{p^{k_{0}}c} || c_{i} − c. Let ℓ > 0 with p ∤ ℓ such that c_{i} = ℓp^{p^{k_{0}}c} + c. By definition of c_{i}, we know that c_{i} = p^{p^{k_{0}}c_{i−1}} + c_{i−1}. This means that

ℓp^{p^{k_{0}}c} + c = p^{p^{k_{0}}c_{i−1}} + c_{i−1}.

Since c_{i−1} < c, we know that ℓp^{p^{k_{0}}c} < p^{p^{k_{0}}c_{i−1}}. But this is not possible as ℓ ≥ 1 and c > c_{i−1}. Hence there does not exist an integer c ∈ (0, b_{0}) such that p^{p^{k_{0}}c} || b_{0} − c and c ∉ {c_{2}, …, c_{n}}. This completes the proof of ⊆.

In order for D_{p}(x_{0}) to have exactly n anti-partial derivatives, we want to construct a_{0} so that

\frac{b_{0} − c}{p^{p^{k_{0}}c}} | a_{0}b_{0}

for every 0 < c < b_{0} satisfying p^{p^{k_{0}}c} || b_{0} − c. This can be achieved by the next lemma.

Lemma 20. Fix integers n > 0 and k_{0} ≥ 0. Let c_{1}, c_{2}, …, c_{n+1} = b_{0} be defined as in Lemma 19. For each 2 ≤ i ≤ n, let a_{i} := (b_{0} − c_{i})/p^{p^{k_{0}}c_{i}} and let a_{0} := ∏_{j=2}^{n} a_{j}. Then p ∤ a_{0} and the set C(x_{0}) (as defined in Theorem 17) has exactly n elements.

Proof. Lemma 19 shows that C(x_{0}) ⊆ {c_{1}, c_{2}, …, c_{n}}. For each 2 ≤ i ≤ n, we know that (b_{0} − c_{i})/p^{p^{k_{0}}c_{i}} = a_{i} divides a_{0} by the definition of a_{0}. Hence it divides a_{0}b_{0}. Since p ∤ a_{i} for 2 ≤ i ≤ n, we know that p ∤ a_{0}. Therefore C(x_{0}) = {c_{1}, c_{2}, …, c_{n}}.

Now we are ready to give a fuller answer to Question 3 as well as an answer to Question 4 at the same time.

Theorem 21. For each positive integer n, there are infinitely many integers x_{0} such that D_{p}(x_{0}) has exactly n anti-partial derivatives.

Proof. Lemma 18 implies that there are infinitely many k_{0} such that x_{0} is the primitive anti-partial derivative of D_{p}(x_{0}). Lemmas 19 and 20 imply that for a given k_{0}, we can construct a_{0} and b_{0} so that x_{0} = a_{0}p^{b_{0}p^{k_{0}}} is the primitive anti-partial derivative of D_{p}(x_{0}) and D_{p}(x_{0}) has exactly n anti-partial derivatives. Therefore, for each positive integer n, there exists infinitely many integers x_{0} = a_{0}p^{b_{0}p^{k_{0}}} such that D_{p}(x_{0}) has exactly n anti-partial derivatives.

Remark 22. For an nteger x_{0}, let D_{ℤ,p}(x_{0}) (resp., D_{ℚ,p}(x_{0})) be the set of all integral (resp., rational) anti-partial derivatives of D_{p}(x_{0}). It is clear that D_{ℤ}(x_{0}) ⊂ D_{ℚ}(x_{0}). Therefore Theorem 21 implies that [1, Conjecture 27] is false.

Theorem 21 remains true even if we allow rational anti-partial derivatives. We give a brief explanation as follows. Let x_{0} := a_{0}p^{b_{0}p^{k_{0}}} be defined as in Lemma 18, so x_{0} is the primitive anti-partial derivative of D_{p}(x_{0}) (even among all the rational anti-partial derivatives) with b_{0} > 0. Let x ≠ x_{0} be a rational anti-partial derivative of D_{p}(x_{0}). We can write x = ap^{bp^{k}} where 0 ≠ a ∈ ℚ, 0 ≠ b ∈ ℤ, k ∈ ℤ, k ≥ 0, and ord_{p}(ab) ≠ 0. Since x_{0} is the primitive anti-partial derivative of D_{p}(x_{0}), we have k > k_{0}. As D_{p}(x) = D_{p}(x_{0}), we have b_{0}p^{k_{0}} + k_{0} − 1 = bp^{k} + k − 1, or p^{k_{0}}(b_{0} − bp^{k−k_{0}}) = k − k_{0}. If b < 0, then k − k_{0} < −bp^{k−k_{0}} < b_{0} − bp^{k−k_{0}} < p^{k_{0}}(b_{0} − bp^{k−k_{0}}). This is a contradiction, so b > 0. This is saying that if x_{0} is the primitive rational anti-partial derivative of D_{p}(x_{0}) with ord_{p}(x_{0}) > 0, then every other rational anti-partial derivative x of D_{p}(x_{0}) also satisfies ord_{p}(x) > 0. In this case, there is a one-to-one correspondence between the set of all rational anti-partial derivatives of D_{p}(x_{0}) and the following set

C_{ℚ}(x_{0}) := {c ∈ [0, b_{0}) ∩ ℤ : p^{p^{k_{0}}c} || (b_{0} − c)}.

If we compare C_{ℚ}(x_{0}) with C(x_{0}) defined in Theorem 17, we have dropped the extra condition \frac{b_{0}−c}{p^{p^{k_{0}} c}} | a_{0}b_{0}. This is because we no longer need a = a_{0}b_{0}/b to be an integer. Therefore Lemma 19 still applies, that is, we can find b_{0} such that C_{ℚ}(x_{0}) has exactly n elements for every positive integer n. Hence, for every positive integer n, there are infinitely many integers x_{0} such that D_{p}(x_{0}) has exactly n rational anti-partial derivatives.

4 Acknowledgments

The authors would like to thank the anonymous referee for carefully reading the manuscript and the editor-in-chief for many useful comments and suggestions.

References

[1] P. Haukkanen, J. K. Merikoski, and T. Tossavainen, On arithmetic partial differential equations, J. Integer Sequences 19 (2016), Article 16.8.6.

[2] P. Haukkanen, J. K. Merikoski, and T. Tossavainen, Arithmetic subderivative and Leibniz-additive functions, Ann. Math. Inform. 50 (2019), 145–157.

[3] J. Kovič, The arithmetic derivative and antiderivative, J. Integer Sequences 15 (2012), Article 12.3.8.

[4] R. K. Pandey and R. Saxena, On some conjectures about arithmetic partial differential equations, J. Integer Sequences 20 (2017), Article 17.5.2.

[5] P. Scholze and J. Stix, Why abc is still a conjecture, 2018. Available at https://www.

   math.uni-bonn.de/people/scholze/WhyABCisStillaConjecture.pdf.

[6] V. Ufnarovski and B. Åhlander, How to differentiate a number, J. Integer Sequences 6 (2003), Article 03.3.4.

2010 Mathematics Subject Classification: Primary 11A25; Secondary 11A41, 11Y55. Keywords: arithmetic derivative, arithmetic partial derivative.

(Concerned with sequence \underline{A003415}.)

Received January 29 2022; revised version received June 1 2022. Published in Journal of Integer Sequences, June 4 2022.

Return to Journal of Integer Sequences home page.