The Arithmetic Derivative and Antiderivative

Jurij Kovič Institute of Mathematics, Physics, and Mechanics University of Ljubljana, Slovenia

                 amadea.kovic@siol.net

Abstract

The notion of the arithmetic derivative, a function sending each prime to 1 and satisfying the Leibnitz rule, is extended to the case of complex numbers with rational real and imaginary parts. Some constraints on the solutions to some arithmetic differential equations are found. The homogeneous arithmetic differential equation of the k-th order is studied. The factorization structure of the antiderivatives of natural numbers is presented. Arithmetic partial derivatives are defined and some arithmetic partial differential equations are solved.

1 Introduction: Basic concepts

The goal of this paper is: (i) to review what is known about the arithmetic derivative [1, 2, 3], (ii) to define some related new concepts, and (iii) to prove some new results, mostly related to the conjectures formulated in [1, 3].

1.1 Definition of the arithmetic derivative

Barbeau [1] defined the arithmetic derivative as the function D : ℤ → ℤ, defined by the rules:

D(1) = D(0) = 0 D(p) = 1 for any prime p ∈ ℙ := {2, 3, 5, 7, …, p_{i}, …}. D(ab) = D(a)b + aD(b) for any a, b ∈ ℕ (the Leibnitz rule) D(−n) = −D(n) for any n ∈ ℕ. Ufnarovski and Åhlander [3] extended D to rational numbers by the rule:

D(\frac{a}{b}) = (\frac{a}{b})′ = \frac{a′b − ab′}{b^{2}}

They also defined D for real numbers of the form x = ∏_{i=1}^{k} p_{i}^{x_{i}}, where p_{i} are different primes and x_{i} ∈ ℚ, by the additional rule:

D(x) = x′ = x ∑_{i=1}^{k} \frac{x_{i}}{p_{i}} .

They showed that the definition of D can be extended to all real numbers and to the arbitrary unique factorization domain.

I extend the definition of the arithmetic derivative to the numbers from the unit circle in the complex plane 𝔼 = {e^{iφ}, φ ∈ [0, 2π]} and to the Gaussian integers ℤ[i] = {a+bi, a, b ∈ ℤ}.

It is not possible to extend the definition of the arithmetic derivative to the Gaussian integers by the demand D(q) = q′ = 0 for all irreducible Gaussian integers q (the analog of primes). If one wants the Leibnitz rule to still remain valid, one easily gets a contradiction: 2 = (1 + i)(1 − i), hence D(2) = 1 · (1 − i) + (1 + i) · 1 = 2, but D(2) = 1 because 2 is a prime.

However, one can extend D to ℚ[i] = {a + bi, a, b ∈ ℚ} using the polar decomposition of complex numbers as follows:

Proposition 1. i) For any φ = (m/n)π, where m, n ∈ ℚ and n ≠ 0, the equation D(e^{iφ}) = 0 holds.

ii) Let 𝔼 = {z ∈ ℂ; |z| = 1} be the unit circle in the complex plane. There are uncountably many functions D : 𝔼 → 𝔼 satisfying the condition D(zw) = D(z)w+zD(w) for all z, w ∈ 𝔼.

iii) If one defines the arithmetic derivative on 𝔼 as follows: D(e^{iφ}) = 0 for all e^{iφ} ∈ 𝔼, then the definition of the arithmetic derivative can be uniquely extended to ℚ[i] so that the Leibnitz rule and the quotient rule remain valid.

Proof. i) Take any φ = (m/n)π, where m, n ∈ ℤ and n ≠ 0. Then e^{inφ} = ±1, hence D(e^{inφ}) = 0. The Leibnitz rule implies D(e^{inφ}) = D(e^{iφ})(e^{iφ})^{n−1} and because (e^{iφ})^{n−1} ≠ 0 it must be that D(e^{iφ}) = 0.

ii) Sketch of the proof: One can easily see that there is a 1-1 correspondence between the functions D : 𝔼 → ℂ satisfying the Leibnitz rule D(zw) = D(z)W + zD(w) and the additive functions L : ℝ → ℂ with a period of 2π (the correspondence is given by the formulas L(s) := \frac{D(e^{is})}{e^{is}} and D(e^{is}) := L(s)e^{is}).

It is also easy to see that for any additive function (satisfying Cauchy’s functional equation L(s + t) = L(s) + L(t)) it is: L(qx) = q · L(x) for all q ∈ ℚ and x ∈ ℝ. Because ℝ and ℂ are linear spaces over ℚ, this means that every additive function L : ℝ → ℝ is a ℚ-linear transformation. Additive functions from ℝ into ℂ with a period of 2π are therefore in a 1-1 correspondence with ℚ-linear transformations L : ℝ → ℂ such that L(2π) = 0.

A linear transformation is uniquely determined with its values on any of the bases of its domain. Let B be a base of the space ℝ such that π ∈ B (its existence is guaranteed by Zorn’s lemma). Thus the ℚ-linear transformations L : ℝ → ℂ such that L(2π) = L(π) = 0 are in a 1-1 correspondence with the set of all functions L : B → ℂ such that L(π) = 0.

The base B is obviously not countable, hence the same also holds for the set of all functions L : B \ {π} → ℂ. By the chain of proven 1-1 correspondences, this also holds for the set of all functions D : 𝔼 → 𝔼 satisfying the Leibnitz rule.

iii) If one writes such numbers in their polar form a+bi = re^{iφ}, then D(re^{iφ}) = D(r)e^{iφ} + rD(e^{iφ}) = D(r)e^{iφ} = D((a^{2}+b^{2})^{1/2})e^{iφ} because D(e^{iφ}) = 0 and r = (a^{2}+b^{2})^{1/2}. The Leibnitz rule is still valid because for any z = re^{iφ} and w = se^{iψ} it is: D(zw) = D((re^{iφ})(se^{iψ})) = D(rs)e^{iφ+ψ} = (D(r)s + r(D(s))e^{iφ+ψ} = D(r)e^{iφ}se^{iψ} + re^{iφ}D(s)e^{iψ} = D(z)w + zD(w). The quotient rule is also still valid because \frac{D(z)w−zD(w)}{w^{2}} = \frac{D(r)e^{iφ}se^{iψ}−re^{iφ}D(s)e^{iψ}}{s^{2}e^{i2φ}} = \frac{(D(r)s−rD(s))e^{iφ}}{s^{2}e^{iψ}} = D(\frac{r}{s})e^{i(φ−ψ)} = D(\frac{r}{s}e^{i(φ−ψ)}) = D(\frac{re^{iφ}}{se^{iψ}}) = D(\frac{z}{w}).

Stay [2] generalized the concept of the arithmetic derivative still further, practically for any number, using advanced techniques such as exponential quantum calculus. In this paper I focus on the arithmetic derivative as a function defined for natural numbers and rational numbers.

1.2 Higher derivatives and the logarithmic derivative

Higher derivatives n^{(k)} are defined inductively: n^{(2)} = D^{2}(n) = D(D(n)) = n′′ = (n′)′, n^{(k+1)} = D^{k+1}(n) = D(D^{k}(n)). Many conjectures on the arithmetic derivative focus on the behavior of sequences (n, n′, n′′, …).

Ufnarovski and Åhlander [3] conjectured that for each n ∈ ℕ exactly one of the following can happen: either n^{(k)} = 0 for sufficiently large k, or lim_{k→∞} = ∞, or n = p^{p} for some prime p (in this case n^{(k)} = n for each k ∈ ℕ). They introduced the function L, called the logarithmic derivative, satisfying the condition

L(x) = \frac{x′}{x} = \frac{D(x)}{x}.

For any x = ∏_{i=1}^{k} p_{i}^{x_{i}}, where p_{i} are different primes and x_{i} ∈ ℚ, L satisfies the condition

L(x) = ∑_{i=1}^{k} \frac{x_{i}}{p_{i}} .

For every prime p and every m, n ∈ ℕ the following formulas hold: L(p) = \frac{1}{p}, L(p^{\frac{m}{n}}) = \frac{m}{np}, L(1) = 0. From the definition of the logarithmic derivative also follow the formulas L(−x) = L(x), L(0) = ∞ and D(x) = L(x) · x [3], hence D^{2}(x) = D(L(x))x + L(x)D(x). I also use the notation L(x) = x^{∗}.

The logarithmic derivative is an additive function: L(xy) = L(x)+L(y) for any x, y ∈ ℚ. Consequently, using a table of values L(p) = \frac{1}{p} (computed to sufficient decimal places!) and the formula D(x) = L(x) · x, it is easy to find D(n) for n ∈ ℕ having all its prime factors in the table. For example, D(5·11^{3}) = L(5·11^{3})·5·11^{3} and because L(5·11^{3}) = L(5)+3L(11) = 0.2000 + 3 · 0.0909 = 0.4727, one can calculate D(5 · 11^{3}) = ⌈0.4727 · 6655⌉ = ⌈3145.8185⌉ = 3146.

\begin{matrix}P_{N} & 2 & 3 & 5 & 7 & 11 & 13 & 17 \\ \frac{1}{P_{N}} & 0.5000 & 0.3333 & 0.2000 & 0.1429 & 0.090 & 0.0769 & 0.0588 \end{matrix}

Table 1: Values of the logarithmic derivative for the first seven primes

2 Brief review of known results and conjectures

Barbeau [1] proved that if n is not a prime or unity then D(n) ≥ 2\sqrt{n} with equality only if n = p^{2}, p ∈ ℙ. He showed that, for integers possessing a proper divisor of the form p^{p}, p ∈ ℙ, lim_{k→∞} D^{k}(n) = ∞.

Ufnarovski and Åhlander [3] translated some famous conjectures in number theory (e.g., the Goldbach conjecture, the Prime twins conjecture) into conjectures about the arithmetic derivative. They formulated many other conjectures, mostly related to (arithmetic) differential equations; for example that the equation x′ = 1 has only primes as positive rational solutions (but it has a negative rational solution x = −\frac{5}{4}).

They also conjectured that the equation n′′ = n has no other solutions than n = p^{p}, p ∈ ℙ in natural numbers and that there are some rational numbers without antiderivatives (or ”integrals”). I study these conjectures in Section 3.

2.1 Arithmetic differential equations and integrals

Most of the known results by Ufnarovski and Åhlander [3] focus on arithmetic differential equations of the first and second order. For example:

The only solutions to the equation n′ = n in natural numbers are n = p^{p}, where p is any prime.

The nonzero solutions to x′ = 0 are the rational numbers of the form: x = ± ∏ p_{i}^{α_{i}p_{i}}, where p_{1}, …, p_{k} are different primes and {α_{1}, …, α_{n}} is a set of integers such that Σ_{i=1}^{k}α_{i} = 0.

The rational solutions to differential equations x′ = xα for all rational numbers α = a/b, where gcd(a, b) = 1 and b > 0, are all of the form x = x_{0}y, where x_{0} is a nonzero particular solution and y is any rational solution of the equation y′ = 0.

Let I(a) denote all the solutions of differential equation n′ = a for n ∈ ℕ and let i(a) denote the number of such solutions, called the integrals of n. Because n′ ≥ 2√n if n is not a prime or unity [1], the solutions satisfy n ≤ \frac{a^{2}}{4} , hence i(a) < ∞ for any a ∈ ℕ. Because gcd(n, n′) = 1 if and only if n is square-free, all integrals of primes are products of different primes: p_{1} ⋯ p_{k}. It will be seen (Corollary 21) that there are primes without integrals (e.g., primes 2, 3, 17). If D(n) were known for each n ∈ ℕ, one would know which natural numbers are primes (because the equation n′ = 1 has only primes as solutions in natural numbers). Ufnarovski and Åhlander [3] gave a list of all a ≤ 1000 having no integral, a list of those numbers a ≤ 100 having more than one integral, and a list of those a ≤ 100 for which i(a) = 1. It will be seen (Corollary 25) that something can also be said about the possible factorization structure of the integrals of a given natural number.

3 New results

Ufnarovski and Åhlander [3] solved the equation x′ = αx in rational numbers for every rational number α. Nonetheless it is interesting to know which natural numbers solve this equation when α = m is a natural number.

Proposition 2. Let m ∈ ℕ. The solutions to the equation x′ = mx in natural numbers x are exactly the numbers of the form x = p_{1}^{p_{1}n_{1}} ⋯ p_{k}^{p_{k}n_{k}}, where n_{1} + ⋯ + n_{k} = m.

Proof. Let x = p_{1}^{e_{1}} ⋯ p_{k}^{e_{k}} be the factorization of x. The condition x′ = L(x)x = mx implies L(x)(p_{1} ⋯ p_{k}) = (\frac{e_{1}}{p_{1}} + ⋯ + \frac{e_{k}}{p_{k}})(p_{1} ⋯ p_{k}) = m(p_{1} ⋯ p_{k}). Because the right side of this equation is divisible by any of the primes p_{1}, …, p_{k}, the left side must also be. This is possible only if for every prime p_{i} the corresponding exponent e_{i} is the multiple of this prime e_{i} = n_{i}p_{i}. The derivative of such a number x = p_{1}^{p_{1}n_{1}} ⋯ p_{k}^{p_{k}n_{k}} is D(x) = D(p_{1}^{p_{1}n_{1}} ⋯ p_{k}^{p_{k}n_{k}}) = (n_{1} + ⋯ + n_{k})x (by the Leibnitz rule and because D(p^{p_{i}n_{i}}) = n_{i}p^{p_{i}n_{i}}) and this is equal to mx if and only if n_{1} + ⋯ + n_{k} = m.

3.1 The homogeneous differential equation of the k′th order

What are the solutions x of the differential equation

a_{k}x^{(k)} + a_{k−1}x^{(k−1)} + ⋯ + a_{2}x^{(2)} + a_{1}x′ + a_{0}x = 0

with rational coefficients a_{i}? In order to answer this question I introduce the concept of a logarithmic class.

Definition 3. Let A be any chosen subring of the ring of complex numbers for which the arithmetic derivative is defined. The logarithmic class N_{r,A} of the number r consists of all numbers x ∈ A with the same logarithmic derivative: N_{r,A} = {x ∈ A; x^{∗} = L(x) = r}.

Remark 4. If it is made perfectly clear which A is being worked with the shorter notation N_{r} can be used. For the purpose of this article, let A be the set of all complex numbers with rational real and imaginary parts: A = ℚ[i] = {a + bi, a, b ∈ ℚ}.

Example 5. Besides the number 0 and the complex numbers z = e^{iφ} ∈ A on the unit circle in the Gaussian plane the class N_{0} also contains rational solutions x = ± ∏ p_{i}^{α_{i}p_{i}}, where p_{1}, …, p_{k} are different primes and {α_{1}, …, α_{n}} is a set of integers such that Σ_{i=1}^{k}α_{i} = 0; for instance, x = \frac{4}{27} [3]. All x ∈ N_{0} solve the equation x′ = 0.

The class N_{1,ℚ} contains all numbers p^{p}, where p is any prime because D(p^{p}) = p^{p}. The class N_{−1,ℚ} contains all numbers p^{−p}, where p is any prime because D(p^{−p}) = −p^{−p}. Do N_{1,ℚ} and N_{−1,ℚ} also contain other rational solutions? Yes:

Proposition 6. If (\frac{a}{b})′ = ±\frac{a}{b} ∈ ℚ and if a and b have no common factors greater than 1, then a = p_{1}^{p_{1}n_{1}} ⋯ p_{k}^{p_{k}n_{k}}, b = q_{1}^{q_{1}s_{1}} ⋯ q_{k}^{q_{k}n_{s}}, and a^{∗}−b^{∗} = ±1, where n_{i}, m_{j}, k, l are nonnegative integers and a^{∗}, b^{∗} logarithmic derivatives. All such numbers are solutions because: (\frac{a}{b})′ = (\frac{a}{b})^{∗}(\frac{a}{b}) = ±(\frac{a}{b}).

Proof. If (\frac{a}{b})′ = ±\frac{a′b−b′a}{b^{2}} = ±\frac{a}{b} then a′b − b′a = ±ab, hence gcd(a, b) = 1 implies a′ = ma and b′ = nb, where m = a^{∗} and n = b^{∗} are natural numbers. Hence (by Proposition 2) a = p_{1}^{p_{1}n_{1}} ⋯ p_{k}^{p_{k}n_{k}}, where n_{1} + ⋯ + n_{k} = m, and b = q_{1}^{q_{1}s_{1}} ⋯ q_{k}^{q_{k}n_{s}}, where s_{1} + ⋯ + s_{k} = n. Because (\frac{a}{b})′ = ±\frac{a}{b} implies (\frac{a}{b})^{∗} = ±1, it must be a^{∗} − b^{∗} = ±1.

Proposition 7. i) The derivative D sends logarithmic classes into logarithmic classes as follows: D(N_{r}) ⊆ N_{r^{∗}+r}. Consequently D(N_{r}) ⊆ N_{r} if and only if r^{∗} = 0 and therefore r′ = 0.

ii) If r^{∗} + r = 0 and r ≠ 0 then D(N_{r}) = N_{0} − {0} ≠ N_{0}. So in this case D(N_{r}) ⊇ N_{r^{∗}+r} is not true.

iii) If r^{∗} + r ≠ 0 then D(N_{r}) = N_{r^{∗}+r}.

Proof. If r = 0 then N_{r} = N_{0} = N_{r^{∗}+r}. So let us now assume that r ≠ 0.

i) D(N_{r}) ⊆ N_{r^{∗}+r} is true because x ∈ N_{r} implies x′ = rx, hence (x′)′ = r′x + rx′ = \frac{r′x′}{r} + rx′ = (\frac{r′}{r} + r)x′ = (r^{∗} + r)x′.

ii) It is possible indeed that r^{∗} + r = 0 while r ≠ 0 (an example is r = \frac{1}{5}, r′ = −\frac{1}{5^{2}} , r^{∗} = −\frac{1}{5}). If r^{∗} + r = 0 then 0 ∈ N_{r^{∗}+r} because 0′ = 0. Now suppose there is an x ∈ N_{r} such that D(x) = 0. Wowever, this implies x ∈ N_{0}, hence r = 0 and there is a contradiction with the assumption r ≠ 0. This means that in this case it is not true D(N_{r}) ⊇ N_{r^{∗}+r} = N_{0} because there is no x ∈ N_{r} such that D(x) = 0.

Now let it be proved that for each nonzero y ∈ N_{r^{∗}+r} there is an x ∈ N_{r} such that D(x) = y. If there is any such x ∈ N_{r}, then it must be y = x′ = rx, hence there is at most one such x and it is defined by the formula x = \frac{y}{r}. Now suppose the derivative of this x is not equal to y. This assumption leads to a contradiction because D(\frac{y}{r}) = \frac{D(y)r−yD(r)}{r^{2}} = \frac{(r^{∗}+r)yr−yr^{∗}r}{r^{2}} = \frac{yr^{2}}{r^{2}} ≠ y implies 1 ≠ 1. Because D(x) = y = rx, this x is indeed a member of N_{r}. Thus it has been proved that D(N_{r}) ⊇ N_{0} − {0}.

Because it is already known that i) is true and because it has been shown that there is no x ∈ N_{r} such that D(x) = 0, the equation D(N_{r}) = N_{0} − {0} holds. Moreover, because 0 ∈ N_{0}, it is also true that N_{0} is a proper subset of N_{0} − {0}.

iii) If r^{∗} + r ≠ 0 then r ≠ 0, and N_{r^{∗}+r} contains only nonzero elements because 0 ∈ N_{0}. Now it is possible to repeat the reasoning as in ii) and for any y ∈ N_{r^{∗}+r} one finds an x ∈ N_{r} such that y = D(x). Hence D(N_{r}) ⊇ N_{r^{∗}+r} and this together with i) implies D(N_{r}) = N_{r^{∗}+r}.

Remark 8. It is already known that there are many rational solutions to the equation r′ = 0; for example, r = 1, −1, 0, hence: D(N_{1}) ⊆ N_{1}, D(N_{−1}) ⊆ N_{−1}, D(N_{0}) ⊆ N_{0}.

Proposition 9. All the derivatives x^{(k)} of any number x can be expressed as functions of the logarithmic derivative L(x) = x^{∗} as follows: x^{(k)} = f_{k}(x^{∗})x where f_{1}(x^{∗}) = x^{∗} and f_{k+1}(x^{∗}) = ((f_{k}(x^{∗}))′ + f_{k}(x^{∗}))x^{∗}. Thus x′ = x^{∗}x, x^{(2)} = ((x^{∗})′ + (x^{∗})^{2})x etc.

Proof. This is true for k = 1. Suppose x^{(k)} = f_{k}(x^{∗})x. Then by the Leibnitz rule: x^{(k+1)} = (f_{k}(x^{∗}))′x + f_{k}(x^{∗})x′ = ((f_{k}(x^{∗}))′ + f_{k}(x^{∗})x^{∗})x.

Proposition 10. Any homogeneous differential equation f(x) ≡ a_{k}x^{(k)} + a_{k−1}x^{(k−1)} + ⋯ + a_{2}x^{(2)}+a_{1}x′+a_{0}x = 0 reduces to an equation g(x^{∗}) = 0. Consequently, if the set of nontrivial solutions to any homogeneous differential equation f(x) = 0 is not empty, then it consists of some classes N_{r}.

Proof. Let r = x^{∗} = \frac{x′}{x} . Because x^{(k)} = f_{k}(x^{∗})x, one can divide the differential equation by x and get an equation of the form g(r) = 0. Thus f(x) = 0 if and only if g(x^{∗}) = 0. Hence whether f(x) = 0 or not depends only on the logarithmic class N_{r} = N_{x^{∗}}. Note also that the degree of the polynomial g is k − 1, one less than the degree of the polynomial f.

Example 11. The nonzero solutions to the equation x′′−x = 0 implying ((x^{∗})′+(x^{∗})^{2})x−x = 0 satisfy the equation ((x^{∗})′ + (x^{∗})^{2} − 1 = 0 or (x^{∗})′ = 1 − (x^{∗})^{2}. Thus the nonzero solutions to x′′ − x = 0 exist if and only if the nonhomogeneous equation r′ = 1 − r^{2} can be solved.

Proposition 12. For every r ∈ ℚ the class N_{r} is not empty because it contains the numbers p^{pr}, where p is any prime.

Proof. L(p^{pr}) = r for any prime p, hence p^{pr} ∈ N_{r}.

Thus in solving the homogenous differential equation f(x) = 0 one can always search for the solutions of the form x = p^{pr}. Of course, other solutions are also possible. Remark 13. Numbers p^{pr} behave in the first derivative just like the exponential function e^{xr} whose derivative is re^{xr}. However, for the higher derivatives the analogy is no longer valid because (p^{p}r)′′ = (r′ + r^{2})p^{pr}, while (e^{xr})′′ = r^{2}e^{xr}. This is one of the reasons why solving an arithmetic differential equation is harder than solving the analogous problem for functions.

Proposition 14. i) In addition to the trivial solutions x ∈ N_{0} the equation x′′ = 0 also has solutions x ∈ N_{1/p}, where p is any prime.

ii) All solutions x ∈ N_{r} of x′′ = 0 satisfy the condition: r^{∗} + r = 0 (hence r ∈ N_{−r} and r′ = r^{∗}r = −r^{2}) and all such x are solutions to x′′ = 0. If r = \frac{a}{b} ∈ ℚ then \frac{a}{b} = b^{∗} −a^{∗} = (\frac{b}{a})^{∗}, hence (\frac{b}{a})′ = 1.

Proof. i) If x′ = \frac{1}{p} then x′′ = −\frac{1}{p^{2}} x + \frac{1}{p} \frac{1}{p}x = 0 = (x′)^{∗}x′, hence (x′)^{∗} = 0.

ii) Any solutions x ∈ N_{r} of x′′ = 0 satisfy the condition D^{2}(N_{r}) ⊆ D(N_{r^{∗}+r}) = D_{0}, hence r^{∗} + r = 0. Conversely, r^{∗} + r = 0 implies D^{2}(N_{r}) ⊆ D(N_{r^{∗}+r}) = D(N_{0}) = {0}, hence x′′ = 0. If r = \frac{a}{b} then r′ = \frac{a′b−ab′}{b^{2}} = \frac{a′b−ab′}{ab} \frac{a}{b}, hence (\frac{a}{b})^{∗} = \frac{a′b−ab′}{ab} = −r = −\frac{a}{b}, therefore \frac{a′}{a} − \frac{b′}{b} = −\frac{a}{b} and \frac{a}{b} = b^{∗} − a^{∗} = (\frac{b}{a})^{∗}. Hence (\frac{b}{a})′ = \frac{b}{a}^{∗} \frac{b}{a} = \frac{a}{b} \frac{b}{a} = 1.

Example 15. Ufnarovski and Åhlander [3] observed that if x = −\frac{5}{4} then x′ = 1 (hence x′′ = 0). Then r = x^{∗} = −\frac{4}{5}, r′ = \frac{−4·5−(−4)·1}{5^{2}} = −\frac{16}{25} = \frac{4}{5} \frac{−4}{5}, hence r^{∗} = \frac{4}{5} and r^{∗} + r = \frac{4}{5} + (−\frac{4}{5}) = 0. Hence all x ∈ N_{− \frac{4}{5}} solve x′′ = 0.

3.2 The graph of derivatives of natural numbers

Of interest is the structure of the infinite directed graph G_{D}, whose vertices correspond to natural numbers n and whose arcs n → D(n) connect the number and its derivative. The corresponding dynamic system: n → D(n) has two obvious attractors: 0 and ∞. There are numbers n with an increasing sequence of derivatives n < n′ < n′′ < ⋯ < n^{(k)} < n^{(k+1)} < ⋯ (e.g., n = p^{pk}, p ∈ ℙ), so there are paths of infinite length in the graph G_{D}.

Ufnarovski and Åhlander [3] conjectured that the equation n^{(k)} = n has only trivial solutions p^{p}, where p ∈ ℙ, satisfying n′ = n. If this is true then the only cycles in G_{D} are the loops in these fixed points. They have shown that if m′ = n and n′ = m then m and n must be square-free numbers: n = ∏_{i=1}^{k} p_{i} and m = ∏_{j=1}^{l} q_{j}, where all p_{i} are distinct from all q_{i}. I present further constraints (Propositions 16, 17, 18, 19) on the structure of possible solutions of the equation n′′ = n in natural numbers.

However, the equation x′′ = x has non-trivial rational solutions of the form x = p^{−p} where p ∈ ℙ and they also satisfy the equation x′ = −x.

Let us first show that any eventual nontrivial solutions m, n (different from m = n = p^{p} where p ∈ ℙ) of the system n′ = m, m′ = n in natural numbers cannot be just a product of two primes; at least one of the numbers m, n ∈ ℕ solving such a system must have at least three different prime factors.

Proposition 16. The system n′ = m, m′ = n has no solutions in natural numbers of the form m = p_{1}p_{2} and n = q_{1}q_{2}, where p_{1}, p_{2}, q_{1}, q_{2} ∈ ℙ.

Proof. Because p_{1}, p_{2}, q_{1}, q_{2} must be different primes, it is not possible for both m and n to have the factor 2. Therefore one can assume that p_{1} and p_{2} are both odd. Then m′ = p_{1} + p_{2} = n = 2q_{2} and n′ = 2 + q_{2} = m = p_{1}p_{2}, hence 2q_{2} = 2p_{1}p_{2} − 4 = p_{1} + p_{2} and p_{1} = \frac{p_{2}+4}{2p_{2}−1} = 1 + \frac{5−p_{2}}{2p_{2}−1}, and this implies p_{2} ≤ 5 because p_{1} must be a natural number. Moreover, it was assumed that p_{2} is odd. However, for p_{2} = 3 one would get p_{1} = 1 + \frac{2}{5} = \frac{7}{5} and for p_{2} = 5 one would get p_{1} = 1.

A computer search showed that there are no natural solutions to x′′ = x such that x < 10000. This result can be improved, at least if the smaller of the numbers m and n is odd.

Proposition 17. Let n′ = m, m′ = n and let n = 2j − 1 < m. Then there are at least nine primes in the factorization of n and n > 3 · 5 · 7 · 11 · 13 · 17 · 19 · 23 · 29.

Proof. It is known that m and n must be products of different primes n = ∏_{i=1}^{k} p_{i} and m = ∏_{j=1}^{l} q_{j}, where all p_{i} are distinct from all q_{i}. Because n < m = n′ = L(n)n we have L(n) = ∑_{i=1}^{k} \frac{1}{p_{i}} > 1. Because n is odd, all p_{i} are greater than 2. The sum of the reciprocals of the first eight odd primes is: \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{11} + \frac{1}{13} + \frac{1}{17} + \frac{1}{19} + \frac{1}{23} = 0.9987 ⋯ < 1. Thus k ≥ 9 and n > 3 · 5 · 7 · 11 · 13 · 17 · 19 · 23 · 29.

If n = 2j < m then the sum \frac{1}{2} + \frac{1}{3} + \frac{1}{5} is already greater than 1 and one does not get any such estimate.

Another possible approach to the system n′ = m, m′ = n in natural numbers is based on the comparison of a square-free number and its derivative; this comparison will imply some inequalities for the smallest primes and biggest primes in the factorizations of n and m.

Proposition 18. Let n = p_{1} ⋯ p_{r}, where p_{1} < ⋯ < p_{r}, and let m = q_{1} ⋯ q_{s} where q_{1} < ⋯ < q_{s} be square-free numbers such that n′ = m, m′ = n. Then: p_{1}q_{s} < rs ≤ (\frac{r+s}{2})^{2}, q_{1}p_{r} < rs ≤ (\frac{r+s}{2})^{2}, p_{1} < r, q_{1} < s. As a consequence n and m must together have at least 34 prime factors p_{i} and q_{j}, thus: r + s ≥ 34. Hence at least one of m and n has at least 17 prime factors and is not smaller than the product of the first 17 primes: 2 · 3 · 5 · 7 · 11 · 13 · 17 · 19 · 23 · 27 · 31 · 37 · 41 · 43 · 47 · 53 · 59. If min{p_{1}, q_{1}} ≥ 3, then r + s ≥ 57, hence at least one of m and n has at least 29 prime factors. If min{p_{1}, q_{1}} ≥ 5, then r + s ≥ 110, hence at least one of m and n has at least 55 prime factors.

Proof. Let n = p_{1} ⋯ p_{r} and m = q_{1} ⋯ q_{s} be square-free numbers such that n′ = m, m′ = n. Then 0 < r\frac{n}{p_{r}} < n′ < r\frac{n}{p_{1}} and 0 < s\frac{m}{q_{s}} < m′ < s\frac{m}{q_{1}}. Hence: p_{1}q_{s} < rs and q_{1}p_{r} < rs. Because r > 1, s > 1 we have p_{r} > r and q_{s} > s, hence p_{1} < r and q_{1} < s. Let r + s = N. Then rs ≤ (\frac{r+s}{2})^{2} = (\frac{N}{2} )^{2}. Thus 2q_{s} ≤ p_{1}q_{s} < (\frac{N}{2} )^{2} and 2p_{r} ≤ q_{1}p_{r} < (\frac{N}{2} )^{2}, hence 2 max\{p_{r}, q_{s}\} < \frac{N^{2}}{4} .

Let P_{i} denote the i-th prime (thus P_{1} = 2, P_{2} = 3, P_{3} = 5, etc.) Because all the primes p_{1}, …, p_{r}, q_{1}, …, q_{s} are distinct, we have max{p_{r}, q_{s}} ≥ P_{r+s} = P_{N}. Thus 2P_{N} = 2P_{r+s} ≤ 2 max\{p_{r}, q_{s}\} < rs ≤ \frac{N^{2}}{4} .

It is possible to directly check that P_{N} ≥ \frac{N^{2}}{8} if N ≤ 33.

\begin{matrix}N & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ P_{N} & 2 & 3 & 5 & 7 & 11 & 13 & 17 & 19 & 23 & 27 & 31 & 37 \\ N^{2} & 1 & 4 & 9 & 16 & 25 & 36 & 49 & 64 & 81 & 100 & 121 & 144 \end{matrix}

\begin{matrix}N & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 \\ P_{N} & 41 & 43 & 47 & 53 & 59 & 61 & 67 & 71 & 73 & 79 & 83 \\ N^{2} & 169 & 196 & 225 & 256 & 289 & 324 & 361 & 400 & 441 & 484 & 529 \end{matrix}

\begin{matrix}N & 24 & 25 & 26 & 27 & 28 & 29 & 30 & 31 & 32 & 33 & 34 \\ P_{N} & 89 & 97 & 101 & 103 & 107 & 109 & 113 & 127 & 131 & 137 & 139 \\ N^{2} & 576 & 625 & 676 & 729 & 784 & 841 & 900 & 961 & 1024 & 1089 & 1156 \end{matrix}

Table 2: The first 34 primes p_{N} and squares N^{2}

Hence it must be that N ≥ 34. If min{p_{1}, q_{1}} ≥ 3, one can get a much better estimate from the condition: p_{N} < \frac{N^{2}}{12} , which is first fulfilled when N ≥ 57. If min\{p_{1}, q_{1}\} ≥ 5, then p_{N} < \frac{N^{2}}{20} , which is first fulfilled when N ≥ 110.

Proposition 19. Suppose n′ = m and n′ = m and n, m are both odd. Let n have r_{1} primes p_{i} ≡ 1 (mod 4) and s_{1} primes q_{j} ≡ −1 (mod 4) and let m have r_{2} primes p_{i} ≡ 1 (mod 4) and s_{2} primes q_{j} ≡ −1 (mod 4). Then (r_{1} − s_{1})(r_{2} − s_{2}) ≡ 1 (mod 4).

Proof. Barbeau [1, pp. 121–122] proved that if n = p_{1} · p_{2} … p_{r} · q_{1} · p_{2} … q_{s}, where p_{i} ≡ 1 (mod 4), q_{j} ≡ −1 (mod 4) are primes, not necessarily distinct, then D(n) ≡ (−1)^{s}(r − s) (mod 4). Hence m = n′ = (−1)^{s_{1}}(r_{1} − s_{1}) and n = m′ = (−1)^{s_{2}}(r_{2} − s_{2}). Because m and n are odd, r_{1} − s_{1} and r_{2} − s_{2} are not even. However, n ≡ (−1)^{s_{1}} (mod 4) and m ≡ (−1)^{s_{2}} (mod 4). Hence mn ≡ (−1)^{s_{1}}(−1)^{s_{2}} ≡ (−1)^{s_{1}}(−1)^{s_{2}}(r_{1} −s_{1})(r_{2} −s_{2}) (mod 4). Consequently (r_{1} − s_{1})(r_{2} − s_{2}) ≡ 1 (mod 4).

3.3 Integrals of natural and rational numbers

Any a such that a′ = b is called an integral of b. The set of all such integrals is denoted as I(b). The same number can have different integrals: 25′ = (5^{2})′ = 2 · 5 = 10 and 21′ = 3 · 7 = 3 + 7 = 10. Because p′ = 1 for any prime, I(1) = ℙ. It is shown that 1 is the only natural number with infinitely many integrals among the natural numbers.

Proposition 20. i) Let b < 2 · 3 · 5 ⋯ P_{n}, where P_{n} is the n-th consecutive prime and let a′ = b, where a ∈ ℕ. Then a = p_{1}^{n_{1}} ⋯ p_{m}^{n_{m}}, where m ≤ n (hence a is divisible by at most n primes p_{i}).

ii) If a′ = b > 1, where a ∈ ℕ, then a ≤ max(2 · 3 · 5 ⋯ P_{n}, b^{bn}). Consequently every b > 1 has at most a finite number of integrals a ∈ ℕ.

Proof. For each natural number b there is a n ∈ ℕ such that b < 2·3·5 ⋯ P_{n}. If a = ∏i=1^{m} p_{i}^{n_{i}} has more than n different prime factors p_{i}, then each summand of a′ = (∑_{i=1}^{m} \frac{n_{i}}{p_{i}}) · b has at least n prime factors, and the smallest of them is not smaller than 2 · 3 · 5 ⋯ P_{n}, therefore in that case a′ > 2 · 3 · 5 ⋯ P_{n} > b, so it cannot be a′ = b. Thus a = p_{1}^{n_{1}} ⋯ p_{m}^{n_{m}} and m ≤ n.

ii) If a contains a factor p^{p}, then b = a′ ≥ a, hence a ≤ b < 2 · 3 · 5 ⋯ P_{n}. The other possibility is that all exponents of a = p_{1}^{n_{1}} ⋯ p_{m}^{n_{m}} are smaller than their primes: n_{i} < p_{i}. It is necessary to consider two cases:

If m = 1 then a = p_{1}^{n_{1}} and a′ = n_{1}p_{1}^{n_{1}−1} = \frac{n_{1}a}{q_{1}} = b. Now b > 1 implies n_{1} ≥ 2 thus p_{1} divides b, hence p_{1} ≤ b and a = \frac{bp_{1}}{n_{1}} ≤ bq_{1} ≤ b^{2}.

If m ≥ 2 then a = p_{1}^{n_{1}} ⋯ p_{m}^{n_{m}} < p_{1}^{p_{1}} ⋯ p_{m}^{p_{m}} < b^{p_{1}+···+p_{m}} < b^{bn} because p_{i} < a′ = b for each p_{i}.

Hence a ≤ max(2 · 3 · 5 ⋯ P_{n}, b^{bn}) and I(b) ⋂ ℕ is finite for any b > 1.

Corollary 21. If a′ = p and p is a prime, then a = p_{1} ⋯ p_{m} and all p_{i} < p. There are some primes without integrals a ∈ ℕ; for example 2,3,17.

Proof. If a = p^{2}c, then a′ = p · (2 + pc′) is not a prime. If a = p_{1} ⋯ p_{m}, then p = a′ > p_{i}. Because 2 < 2 · 3 and 3 < 2 · 3, the only candidates for integrals of 2 and 3 are numbers of the form a = p · q, where p, q ∈ ℙ. However, then (p · q)′ = p + q > 5, hence 2 and 3 can have no integrals a ∈ ℕ. Ufnarovski and Åhlander [3] found their integrals in rational numbers: (−\frac{21}{16})′ = 2, (−\frac{13}{4})′ = 3. Any integral of 17 has at most 3 different prime factors because 17 < 2 · 3 · 5. It cannot have only two different prime factors because if a = pq then a′ = p + q and the sum of any two primes is not 17. However, if a = pqr, then (pqr)′ = pq + pr + qr ≥ 2 · 3 + 2 · 5 + 3 · 5 = 31 > 17. Thus there is no integral of 17.

The numbers s ∈ ℕ not divisible by any square c^{2} where c > 1 are called square-free numbers. They can be either products of different primes or equal to 1. The set of square-free numbers is denoted 𝕊.

Definition 22. Let a = p_{1}^{n_{1}} ⋯ p_{m}^{n_{m}}, where p_{i} ∈ ℙ. The number n_{i} is called the exponent of a prime p_{i} in a. Let us define the following functions of a:

s(a) is the greatest square-free divisor of a, such that gcd(s(a), \frac{a}{s(a)}) = 1,

p(a) = p_{1} ⋯ p_{m} is the product of all prime factors of a,

f(a) = \frac{a}{p(a)} = \frac{a}{p_{1}···p_{m}}, r(a) = a · p(a) = a · (p_{1} ⋯ p_{m}),

h_{max}(a) = max{n_{1}, …, n_{m}}, h_{min}(a) = min{n_{1}, …, n_{m}}. For a = 1 we define s(a) = p(a) = f(a) = r(a) = h_{max}(a) = h_{min}(a) = 1.

From this definition it follows that if a ≠ s(a) then h_{min}(\frac{a}{p(a)}) ≥ 2.

Proposition 23. i) Let a ∈ ℕ. Then a′ = f(a)p(a)a^{∗}, p(a)a^{∗} ∈ ℕ and a = r(f(a))s(a).

ii) If a ∈ ℙ then p(a)a^{∗} = 1 and f(a) = a′. If a ∈ ℕ is not a prime then p(a)a^{∗} > 1 and f(a) < a′.

iii) If a = p_{1}^{n_{1}} ⋯ p_{m}^{n_{m}} and n_{i} is not divisible by p_{i} for all i ∈ {1, 2, …, m} then gcd(f(a), p(a)a^{∗}) = 1 and gcd(a, a′) = f(a).

iv) If h_{min}(a) ≥ 2, then s(a) = 1, hence a = r(f(a)). If h_{max}(a) = 1 then p(a) = a and f(a) = 1, hence r(f(a)) = r(1) = 1.

Proof. i) If a = 1 then a′ = a^{∗} = 0 implies a′ = f(a)p(a)a^{∗}. If a = p_{1}^{n_{1}} ⋯ p_{m}^{n_{m}} then a′ = aa^{∗} = (\frac{a}{p(a)})p(a)a^{∗} and \frac{a}{p(a)} = f(a) = p_{1}^{n_{1}−1} ⋯ p_{m}^{n_{m}−1}. Obviously p(a)a^{∗} = (p_{1} ⋯ p_{m})(\frac{n_{1}}{p_{1}} + ⋯ + \frac{n_{m}}{p_{m}}) ∈ ℕ. Because f(a) is divisible by exactly those primes p_{i} for which p_{i}^{2} divides a, it is r(f(a)) = \frac{a}{s(a)}. Hence a = r(f(a))s(a).

ii) If a ∈ ℙ then p(a) = a, a′ = 1, a^{∗} = \frac{1}{a}, p(a)a^{∗} = a\frac{1}{a} = 1 and f(a) = 1. Now suppose a ∈ ℕ is not a prime. Then either a = p_{1}^{n_{1}}, where n_{1} > 1, or a = p_{1}^{n_{1}} ⋯ p_{m}^{n_{m}}, where m ≥ 2. In the first case p(a) = p_{1}, f(a) = p_{1}^{n_{1}−1}, a′ = n_{1}p_{1}^{n_{1}−1}, a^{∗} = \frac{n_{1}}{p_{1}} , hence p(a)a^{∗} = p_{1} \frac{n_{1}}{p_{1}} = n_{1} > 1. In the second case, p(a)a^{∗} = (p_{1} ⋯ p_{m})(\frac{n_{1}}{p_{1}} + ⋯ + \frac{n_{m}}{p_{m}}) is a sum of at least two natural numbers, hence p(a)a^{∗} > 1. Therefore in both cases f(a) = \frac{a′}{p(a)a^{∗}} < a′.

iii) Now p(a)a^{∗} = (p_{1} ⋯ p_{m})(\frac{n_{1}}{p_{1}} +⋯+ \frac{n_{m}}{p_{m}}) is not divisible by any of the primes p_{i} dividing a because n_{i} is not divisible by p_{i}. Hence p(a)a^{∗} = q_{1}^{u_{1}} ⋯ q_{t}^{u_{t}} and a′ = p_{1}^{n_{1}−1} ⋯ p_{m}^{n_{m}−1}q_{1}^{u_{1}} ⋯ q_{t}^{u_{t}}, where all the primes p_{1}, ⋯, p_{m}, q_{1}, ⋯, q_{t} are distinct. Hence gcd(f(a), p(a)a^{∗}) = 1 and gcd(a, a′) = f(a).

iv) If h_{min}(a) ≥ 2, then p(a) = p(f(a)), hence r(f(a)) = a. If h_{max}(a) = 1 then p(a) = a and f(a) = 1, hence r(f(a)) = 1.

Definition 24. If a = p_{1}^{n_{1}} ⋯ p_{m}^{n_{m}} where all n_{i} < p_{i} then a is in the set 𝕃 whose elements are called ”low” numbers.

Now it is possible to describe the factorization and consequently obtain some bounds of the integrals of ”low” numbers.

Corollary 25. i) If b ∈ 𝕃 then I(b) ⋂ ℕ ⊂ 𝕃. Moreover, every a ∈ I(b) ⋂ ℕ is of the form a = r(c)s(a), where b = cd, c < b, gcd(c, d) = 1, and b = a′ = r′(c)s(a) + r(c)s′(a). Therefore:

i.a) If s(a) = a then r(c) = 1 hence c = 1. Conversely, if c = 1 then s(a) = a.

i.b) If s(a) = 1 then a = r(c).

i.c) If r(c) ≠ 1 then s(a) ≤ \frac{b−r(c)}{r′(c)} ≤ \frac{b}{c}, hence a ≤ r(b) − r(r(c)) ≤ r(b) − 1.

i.d) If r(c) ≠ 1 and s(a) is a prime then s(a) = \frac{b−r(c)}{r′(c)}.

i.e) If r(c) ≠ 1 and s(a) = p_{1}p_{2} then \frac{b−r(c)}{r′(c)+r(c)} ≤ s(a) ≤ \frac{b−5r(c)}{r′(c)}.

ii) For every b = p_{1}^{n_{1}} ⋯ p_{k}^{n_{k}} ∈ 𝕃 there are at most 2^{k} − 1 different divisors c of b such that there is an integral a of b of the form a = r(c)s(a).

iii) If b = p_{1}^{n_{1}} ⋯ p_{k}^{n_{k}} and n_{i} = p_{i} − 1 for all i, then any integral a ∈ I(b) ⋂ ℕ is a square-free number: a = s(a).

Proof. i) If a′ = b and a contains a factor p_{i}^{p_{i}} then a′ = b is also divisible by p_{i}^{p_{i}}, too, hence b is not in 𝕃. Thus, because b ∈ 𝕃, it must be a ∈ 𝕃. Hence f(a) divides a′ = b, f(a) < a′, and r(f(a)) = \frac{a}{s(a)}. Thus one writes c = f(a), one really can get all the integrals of b ∈ 𝕃 in the described form.

i.a) and i.b) are obvious.

i.c) If r(c) ≠ 1 then r′(c) ≠ 0 and s(a) = b − \frac{r(c)s′(a)}{r′(c)} ≤ \frac{b−r(c)}{r′(c)} = \frac{b}{(cp(c))′} = \frac{b}{c′p(c)+cp′(c)} ≤ \frac{b}{c}.

Hence a = r(c)s(a) ≤ c · p(c)\frac{b−r(c)}{c} ≤ p(b)b − p(c)r(c) = r(b) − r(r(c)) ≤ r(b) − 1 because p(c) = p(r(c)).

i.d) If r(c) ≠ 1 and s(a) is a prime, then s′(a) = 1 and s(a) = \frac{b−r(c)}{r′(c)}.

i.e) If r(c) ≠ 1 and s(a) = p_{1}p_{2} then s′(a) = p_{1} + p_{2} ≤ 2 + 3 = 5 hence s(a) = \frac{b−5r(c)}{r′(c)}.

The inequality \frac{b−r(c)}{r′(c)+r(c)} ≤ s(a) follows from the fact that p_{1} + p_{2} ≤ p_{1}p_{2} for any two primes p_{1} and p_{2} implying a′ = (r(c)p_{1}p_{2})′ = r′(c)p_{1}p_{2} + r(c)(p_{1} + p_{2}) ≤ (r′(c) + r(c))p_{1}p_{2}.

ii) Because each factor p_{i}^{n_{i}} of b = cd = p_{1}^{n_{1}} ⋯ p_{k}^{n_{k}} belongs either to c = f(a) or d = \frac{b}{f(a)}, and because c < a′, there are at most 2^{k} − 1 possible factors c dividing b = a′ such that a = r(c)s(a). However, different integrals of b may correspond to the same c, as in the example (3 · 13)′ = (5 · 11)′ = 16, in which both integrals 3 · 13 and 5 · 11 correspond to c = 1.

iii) If p_{i} = n_{i} − 1 for all i, then r(c) = 1 or r(c) contains at least one factor p_{i}^{p_{i}}, but then this is a contradiction because in that case a = b′ would also be divisible by p_{i}^{p_{i}}. Hence r(c) = 1 and a = s(a).

Example 26. Let b = 5^{2}11. There are three possible factorizations b = cd such that gcd(c, d) = 1 and 1 ≤ c < b, corresponding to numbers c_{1} = 1, c_{2} = 5^{2} and c_{3} = 11. Thus the candidates for a ∈ I(b) ⋂ ℕ are:

square-free numbers a = r(1)s(a) = s(a)2 · 5^{2}11, not divisible by 5 or 11,

numbers of the form: a = r(5^{2})s(a) = 5^{3}s(a), where s(a) ≤ \frac{b}{r′(c)} = \frac{5^{2}11}{3·5^{2}} = \frac{11}{3} ≤ 4, hence s(a) ∈ {1, 2, 3}, and

numbers of the form a = r(11)s(a) = 11^{2}s(a), where s(a) ≤ \frac{b}{r′(c)} = \frac{5^{2}11}{2·11} = \frac{5^{2}}{2} , hence s(a) ∈ {1, 2, 3, 5, 6, 7, 10, 11}.

It can easily be seen that s(a) > 1 because s(a) = 1 implies r′(c) = b, but it is 1′ = 0 ≠ b, (5^{3})′ = 3 · 5^{2} ≠ b and (11^{2})′ = 2 · 11 ≠ b. Thus in the case c_{2} = 5^{2} it must be s(a) ∈ {2, 3} and in the case of c_{3} = 11 it is seen that s(a) ∈ {2, 3, 5, 6, 7, 10, 11} is either a prime or a product of two primes.

Hence in the case a = 5^{3}s(a) one can use the formula for the prime s(a) = \frac{b−r(c)}{r′(c)} = \frac{5^{2}11−5^{3}}{3·5^{2}} = \frac{11−5}{3} = 2 and one can verify directly that (2 · 5^{3})′ = 5^{3} + 2 · 3 · 5^{2} = 5^{2}(5 + 6) = b.

Now let a = 11^{2}s(a). If s(a) is a prime one gets s(a) = \frac{b−r(c)}{r′(c)} = \frac{5^{2}11−11^{2}}{2·11)} = \frac{25−11}{2} = 7 and one can check directly that (11^{2} · 7)′ = 2 · 11 · 7 + 11^{2} = 11(14 + 11) = b. It can easily be seen that s(a) cannot be a product of two primes because that would imply a contradiction: 0 ≤ s(a) ≤ \frac{b−5r(c)}{r′(c)} = \frac{5^{2}11−5·11^{2}}{2·11} < 0.

Square-free integrals corresponding to c_{1} = 1 are more difficult to find. It is necessary to check the derivatives of all square-free numbers a ≤ \frac{b^{2}}{4} and compare them with b.

A similar estimate as in Proposition 17 can be made about positive solutions to the equation (\frac{a}{b})′ = \frac{2}{p}, where p ∈ ℙ.

Definition 27. Let P_{i+1} denote the i-th odd prime. For any d ∈ ℝ let O_{d} = 1 · ∏_{i=1}^{m} P_{i+1} denote the product of the first m odd primes such that the sum of their reciprocals, denoted R(d), is not smaller than d.

Thus R(d) = ∑_{i=1}^{m} \frac{1}{P_{i+1}} ≥ d. Because the series ∑_{i=1}^{∞} \frac{1}{P_{i+1}} diverges, O_{d} is well defined for any real d.

Example 28. If d ≤ \frac{1}{3} then O_{d} = P_{2} = 3. If \frac{1}{3} ≤ d ≤ \frac{1}{3} + \frac{1}{5} then O_{d} = P_{2}P_{3} = 3 · 5 = 15. If \frac{1}{3} + \frac{1}{5} ≤ d ≤ \frac{1}{3} + \frac{1}{5} + \frac{1}{7} then O_{d} = P_{2}P_{3}P_{4} = 3 · 5 · 7 = 105.

Proposition 29. Let (\frac{a}{b})′ = \frac{2}{p}, where p ∈ ℙ, p > 2, gcd(a, b) = 1, \frac{a}{b} > 0. Then a = p_{1} ⋯ p_{m} is an odd square-free number with m ≥ 9 prime factors p_{i} ∈ ℙ and b is of the form b = q_{1}^{q_{1}n_{1}} … q_{s}^{q_{s}n_{s}} where q_{i} ∈ ℙ and n_{i} ∈ ℕ. Moreover, L(b) ∈ ℕ, L(a) > L(b) ≥ 1 and a′ > a > O_{1} ≥ 3 · 5 · 7 · 11 · 13 · 17 · 19 · 23 · 29.

Proof. (\frac{a}{b})′ = \frac{2}{p} implies \frac{a′b−ab′}{b^{2}} = \frac{2}{p}, hence b = pc, where c ∈ ℕ. Therefore a′pc − ab′ = 2c^{2}p and because a and b have no common prime factors, we have b′ = kc. Hence a′pc−akc = 2c^{2}p and by dividing this equation by c one gets a′p − ak = 2cp, hence k = pd because p divides b so it cannot divide a. This implies a′p − apd = 2cp, hence a′ − ad = 2c, and b′ = pdc = bd. Thus d = L(b) = b^{∗}. By Proposition 2, b = q_{1}^{q_{1}n_{1}} ⋯ q_{s}^{q_{s}n_{s}}, where n_{1} + ⋯ + n_{s} = d.

Because \frac{a}{b} > 0, it can be assumed a > 0 and b > 0. It must be a > 1 because a = 1 implies a′ = 0 and −d = 2c, implying a contradiction: 0 = 2c + d > 0.

From the equation a′ − ad = 2c follows gcd(a, a′) = 1, because any such common prime factor different from 2 would also divide c and b and this would contradict gcd(a, b) = 1, and if 2 divides a and a′ then it must be a = 4e, hence a′ = 4(e + e′) and 4(e′ + e − 4ed) = 2c would imply that 2 also divides b.

Now it can be seen that a cannot be divisible by 2. In that case 2 would also divide a′ or p. However, this is impossible because gcd(a′, a) = 1 and p > 2.

Because gcd(a, a′) = 1, a must be a square-free number. Because a > 1, it must be: a = p_{1} ⋯ p_{m}, and because a′ − ad = 2c > 0 it is a′ > ad ≥ a, hence L(a) = \frac{a′}{a} = ∑_{i=1}^{m} \frac{1}{p_{i}} > d ≥ 1. Therefore a > O_{d} ≥ O_{1} ≥ 3 · 5 · 7 · 11 · 13 · 17 · 19 · 23 · 29 and m ≥ 9, because the sum of the reciprocals of the first eight odd primes is less than 1: \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{11} + \frac{1}{13} + \frac{1}{17} + \frac{1}{19} + \frac{1}{23} = 0.9987 ⋯ < 1.

Ufnarovski and Åhlander [3] conjectured that some rational numbers have no integrals. The above proposition shows that the integrals of \frac{2}{3} are not easy to find, if they exist at all. The same holds for any \frac{2}{p}, p ∈ ℙ.

In Proposition 7.iii) I showed that if y ∈ N_{y^{∗}} = N_{r+r^{∗}} and if r + r^{∗} ≠ 0 then there is exactly one x ∈ N_{r} such that x′ = y and this is x = \frac{y}{r}. Hence to find a rational integral x of any nonzero rational number y one just has to find a rational number r = x^{∗} such that r + r^{∗} = y^{∗}. In other words, one has to find a logarithmic class N_{r} such that D(N_{r}) = N_{y^{∗}}. The equation r +r^{∗} = y^{∗} translates into r^{2} −y^{∗}r +r′ = 0. Perhaps for some rational numbers y this equation cannot be satisfied by any rational number r, hence such y cannot have a rational integral x. Expressing r as a function of its derivative r′, one gets at most two

different solutions r_{1,2} = ^{\frac{y^{∗}±\sqrt{(y^{∗})^{2}−4r′}}{2}} . Now r ∈ ℚ implies (y^{∗})^{2} − 4r′ ≥ 0 and r′ ≤ \frac{(y^{∗})^{2}}{4} . Because x ∈ ℚ implies r = x^{∗} ∈ ℚ the expression (y^{∗})^{2} − 4r′ = q^{2} must be a square of a rational number q.

Example 30. If y = \frac{2}{3} then y^{∗} = \frac{1}{2} − \frac{1}{3} = \frac{1}{6}. If there is a positive rational number x = \frac{a}{b} ∈ N_{r} such that x′ = y = \frac{2}{3} then r = x^{∗} = (\frac{a}{b})^{∗} = \frac{y}{x} = \frac{2}{3x} > 0 satisfies the equation r + r^{∗} = y^{∗} = \frac{1}{6}. This equation translates into r^{2} − \frac{1}{6}r + r′ = 0. For each r′ one gets at most two different

solutions r_{1,2} = ^{\frac{y^{∗}±\sqrt{(y^{∗})^{2}−4r′}}{2}} = ^{\frac{\frac{1}{6} ±\sqrt{\frac{1}{36} −4r′}}{2}} = \frac{1}{12} ± \sqrt{\frac{1}{144} − r′}. Hence r′ ≤ \frac{(y^{∗})^{2}}{4} = \frac{1}{144}.

It is known from Proposition 29 that r = a^{∗} − b^{∗} = ∑_{i=1}^{m} \frac{1}{p_{i}} − d > 0, where d ∈ ℕ. Thus x = \frac{y}{r} > 0. Now r′ = (\frac{2}{3x})′ = \frac{3x−6}{9x^{2}} = \frac{x−2}{3x^{2}} = \frac{x−2}{3x^{2}} \frac{3x}{2} \frac{2}{3x}, hence r^{∗} = \frac{x−2}{2x}. If x ≤ 2 then r^{∗} ≤ 0 and r = y^{∗} − r^{∗} ≤ y^{∗} = \frac{1}{6} leads to a contradiction: x = \frac{y}{r} ≥ \frac{y}{y^{∗}} = \frac{\frac{2}{3}}{\frac{1}{6}} = 4. Thus it must be

x > 2, hence r^{∗} > 0. Now 0 < r = y^{∗} − r^{∗} < y^{∗} = \frac{1}{6} implies x = \frac{y}{r} > \frac{y}{y^{∗}} = \frac{\frac{2}{3}}{\frac{1}{6}} = 4, hence

a > 4b. Thus there is no positive rational number x such that x ≤ 4 and x′ = \frac{2}{3}.

3.4 Partial derivatives and partial differential equations

Definition 31. Let a = ∏_{i=1}^{k} p_{i}^{x_{i}} be the factorization of a ∈ ℕ into primes. The partial derivative \frac{∂a}{∂pi} = D_{p_{i}}(a) is defined as \frac{∂a}{∂p_{i}} = a\frac{x_{i}}{p_{i}} . If p ∈ ℙ is not in the factorization of a, then \frac{∂a}{∂p_{i}} = 0.

From this definition it immediately follows that D(a) = a ∑_{i=1}^{k} \frac{∂a}{∂p_{i}} . One can also define higher partial derivatives; for example: \frac{∂^{2}a}{∂p_{i}∂p_{j}} = D_{p_{i}p_{j}}^{2} = \frac{∂(∂a/∂p_{j})}{∂p_{i}}, \frac{∂^{2}a}{∂p_{i}∂p_{j}} = D_{p_{i}^{2}}^{2} = \frac{∂(∂a/∂p_{i})}{∂p_{i}}, etc., and study partial differential equations.

Example 32. D_{2}(2^{3} · 5^{4}) = 3 · 2^{2} · 5^{4}, D_{5}(3 · 2^{2}5^{4}) = 3 · 2^{2} · 4 · 5^{3} = 3 · 2^{4} · 5^{3} D_{5}(2^{3} · 5^{4}) = 2^{3} · 4 · 5^{3} = 2^{5} · 5^{3} and D_{2}(2^{5} · 5^{3}) = 5 · 2^{4} · 5^{3}. Thus the order of applying D_{p} and D_{q} is important: D_{p}D_{q} is not always equal to D_{q}D_{p}.

Proposition 33. If a = p^{e}q^{f} and gcd(e, q) = 1 and gcd(f, p) = 1, then D_{p}D_{q}(a) = D_{q}D_{p}(a).

Proof. In that case it is D_{p}D_{q}(p^{e}q^{f}) = D_{p}(p^{e}fq^{f−1}) = ep^{e−1}fq^{f−1} and D_{q}D_{p}(p^{e}q^{f}) = D_{q}(ep^{e−1}q^{f}) = ep^{e−1}fq^{f−1}.

Proposition 34. Let n = p^{e}q^{f}c, where p, q are primes not dividing c. Then D_{p}(p^{e}q^{f}c) = D_{q}(p^{e}q^{f}c) if and only if e = kp and f = kq, where k ∈ ℕ.

Proof. From D_{p}(p^{e}q^{f}c) = ep^{e−1}q^{f}c = fp^{e}q^{f−1}c = D_{q}(p^{e}q^{f}c) follows eq = bp, hence e = kp and f = kq. Then D_{p}(p^{kp}q^{kq}c) = k(p^{kp}q^{kq}c) = D_{q}(p^{e}q^{f}c).

Proposition 35. i) The only solutions to the partial differential equation D_{p^{2}}^{2}(n) = n in natural numbers are n = p^{pk}c, where gcd(c, p) = 1.

ii) If n = p^{e}d where gcd(p, d) = 1 and e ≥ p then D_{p^{2}}^{2}(n) ≥ n.

iii) The only solutions to the partial differential equation (D_{p^{2}}^{2} + D_{q^{2}}^{2})(n) = n in natural numbers are n = p^{p}c and n = q^{q}c, where gcd(pq, c) = 1.

Proof. i) Let n = p^{e}c, where e ∈ ℕ ⋃{0} and c ∈ ℕ is not divisible by p.

If e = 0 then D_{p}(n) = 0, hence D_{p^{2}}^{2}(n) = 0. If e = kp and k ∈ ℕ then D_{p}(n) = D_{p}(p^{pk}c) = pkp^{pk−1}c = p^{pk}c = n hence D_{p^{2}}^{2}(n) = n. If e = 1 then D_{p}(n) = c, hence D_{p^{2}}^{2}(n) = D_{p}(c) = 0.

ii) If e = kp and k ∈ ℕ then D_{p^{2}}^{2}(n) = n, as is already known from i). If gcd(e, p) = 1 and e > p then D_{p}(n) = D_{p}(p^{e}c) = ep^{e−1}c, hence D_{p^{2}}^{2}(n) = D_{p}(ep^{e−1}c) = e(e − 1)p^{e−2}c ≥ (p + 1)pp^{e−2}c > p^{e}c = n.

iii). One can write n = p^{e}q^{f}c, where {e, f} ⊆ ℕ ⋃{0} and c ∈ ℕ is not divisible by p or q. So it is necessary to study the equation (D_{p^{2}}^{2} + D_{q^{2}}^{2})(p^{e}q^{f}c) = p^{e}q^{f}c. If f = 1 then D_{q^{2}}^{2}(p^{e}c) = 0 and one gets the equation D_{p}^{2}(n) = n whose only solution is n = p^{p}c, as is known from i). Likewise if e = 1 one gets n = q^{q}c.

In the case of e ≥ 2 and f ≥ 2, one can use the following argument:

Obviously gcd(qc, p) = 1 implies that D_{p^{2}}^{2}(n) is divisible by q^{f}c. Likewise gcd(pc, q) = 1 implies that D_{q^{2}}^{2}(n) is divisible by p^{e}c. So one can write D_{p^{2}}^{2}(n) = aq^{f}c, D_{q^{2}}^{2}(n) = bp^{e}c, where a, b ∈ ℕ ⋃{0}. Suppose a ≠ 0 and b ≠ 0. Hence one gets the equation aq^{f}c + bp^{e}c = p^{e}q^{f}c. This equation can be solved only if a is divisible by p^{e} and if b is divisible by q^{f}. Suppose a ≠ 0 and b ≠ 0. Then D_{p^{2}}^{2}(n) ≥ n and D_{q^{2}}^{2}(n) ≥ n, hence D_{p^{2}}^{2}(n) + D_{q^{2}}^{2}(n) ≥ 2n and the equality is not possible. Hence it must be either a = 0 or b = 0. However, this is not possible if e ≥ 2 and f ≥ 2.

4 Concluding remarks

I have proved some new results about the arithmetic derivative and integral. I have defined arithmetic partial derivatives and solved some arithmetic partial differential equations. I have shown that, for any solution to the system m′ = n, n′ = m in natural numbers, at least one of the numbers m and n is not smaller than the number 2 · 3 · 5 · 7 · 11 · 13 · 17 · 19 · 23 · 31 · 37 · 41 · 43 · 47 · 53 · 59. The arithmetic derivative can be defined on sequences of numbers as follows: D(a_{1}, a_{2}, …, a_{n}, …) = (D(a_{1}), D(a_{2}), …, D(a_{n}), …)). Thus taking any integer sequence (a) = (a_{1}, a_{2}, …, a_{n}, …) one can get an infinite family of derived sequences D(a), D^{2}(a), …, D^{k}(a), …. I believe many other useful applications of the arithmetic derivative will be discovered.

5 Acknowledgements

I thank the referee for many suggestions that helped me to improve this paper. I am also grateful to Marko Petkovšek for his explanation of why there are uncountably many functions on the unit circle in the complex plane satisfying the Leibnitz rule (Proposition 1.ii).

References

[1] E. J. Barbeau, Remarks on arithmetic derivative, Canad. Math. Bull, 4 (1961), 117–122.

[2] M. Stay, Generalized number derivatives, J. Integer Seq., 8 (2005), Article 05.1.4.

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

2010 Mathematics Subject Classification: Primary 11A25; Secondary 11A41, 11N05, 11N56, 11Y55. Keywords: arithmetic derivative, arithmetic antiderivative, arithmetic partial derivative, Leibnitz rule.

(Concerned with sequences \underline{A000040}, \underline{A000290}, \underline{A051674}.)

Received May 19 2011; revised version received March 18 2012. Published in Journal of Integer Sequences, March 25 2012.

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