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An Exploration of the Arithmetic Alaina Sandhu Research Midterm Report, Summer 2006 Under Supervision: Dr. McCallum, Ben Levitt, Cameron McLeman

1 Introduction

The Arithmetic Derivative is a recently defined function of whose properties directly relate to some of the most well known conjectures within . The arithmetic derivative of an is defined to be the map which sends every prime integer to 1 and that satisfies the Leibnitz rule, for all a, b ∈ Z,(ab)0 = a0b + ab0. From this definition, we can conjecture that for any a ∈ N, there exists a solution to the differential equation n0 = 2a. A proof of Goldbach’s conjecture would validate this statement, as the derivative of the product of 0 two primes (p1 · p2) is the sum. Our goal is to explore the properties of the arithmetic derivative and propose further conjectures with regard to the nature of the function as well as its implications for the Conjecture and Goldbach Conjecture, among others.

2 Definition

The arithmetic derivative n0 of any is defined as follows:

• p0 = 1 for any prime p.

• (ab)0 = a0b + ab0 for any a, b ∈ N (Leibniz rule). • 00 = 0.

To illustrate this,

150 = (3 · 5)0 = 30 · 5 + 3 · 50 = 1 · 5 + 3 · 1 = 8.

We now examine the derived formula and ensure the function is well-defined.

1 Qk ni Theorem 1 For any natural number n, if n = i=1 pi is the prime factorization of n, then

k X ni n0 = n . (1) p i=1 i

Qk 0 Pk 1 First we prove by induction on k that if n = pi, then n = n . i=1 i=1 pi Qk Proof. Consider n ∈ N and express n = i=1 pi, where all pi are not necessarily distinct. 0 Qk When k = 1, it is clear that n = 1. Assume that n = i=1 pi. Then,

0 0 0 (npk+1) = n pk+1 + n(pk+1) k X 1 = n + np k+1 p i=1 i k+1 X 1 = np k+1 p i=1 i

Qk ni Now, if we express n = i=1 pi , our formula becomes n0 = n Pk ni i=1 pi Now, we check that our formula is consistent with the Leibniz rule for the product of two Qk ai Qk bi natural numbers; Let a = i=1 pi and b = i=1 pi . Then,

k k ! k ! X ai + bi X ai X bi ab = a b + a b p p p i=1 i i=1 i i=1 i . This definition maintains some natural properties, such as (nk)0 = knk−1n0. Also, because 10 = (1 · 1)0 = 10 · 1 + 1 · 10 = 2 · 10,, the derivative of 1 is defined as 10 = 0 .

2.1 Bounds for the Arithmetic Derivative Theorem 2 The only positive integer which satisfies n0 = 0 is n = 1.

Proof. This is a direct result of the definition of our function.

Theorem 3 The only solutions to n0 = 1 in natural numbers are prime numbers.

Proof. A composite number can be expressed as the product of prime numbers, of which the derivative of (by the product rule) is the sum of at least two positive , which is greater than 1.

Theorem 4 [1] For any positive integer n

n log n n0 ≤ 2 . (2) 2

2 If n is composite, √ n0 ≥ 2 n. (3) Furthermore, if n is a product of k factors larger than 1, then

0 k−1 n ≥ kn k . (4)

Qk ni (Proof taken directly from [1]) Proof. If n = i=1 pi , then

k k Y ni X n ≥ 2 ⇒ log2 n ≥ ni. i=1 i=1 Using Theorem 1, this inequality translates to

k Pk X ni n ni n log n n0 = n ≤ i=1 ≤ 2 . p 2 2 i=1 i

If n = n1n2n3 ··· nk then, according to the Leibnitz rule,

0 0 0 0 0 n = n1n2n3 ··· nk + n1n2n3 ··· nk + n1n2n3 ··· nk + ... + n1n2n3 ··· nk ≥

n2n3n4 ··· nk + n1n3n4 ··· nk + n1n2n4 ··· nk + ... + n1n2 ··· nk−1 = 1     k 1 1 1 1 1 1 −1 k−1 n + + ... + ≥ n · k · ··· = k · n · n k = k · n k . n1 n2 nk n1 n2 nk Here we have replaced the arithmetic mean by the geometric mean.

3 Properties of the Arithmetic Derivative

While the arithmetic derivative has only recently been defined, there has been some signifi- cant research published on various properties of this function, notably ”How to Differentiate a Number” by Ufnarovski and Ahlander [1]. We will now address the theorems and properties relevant to our research, and discuss the related questions we seek to resolve.

3.1 Solutions to n0 = n Numerical derivations have shown positive integers of the form pp (where p is prime) solve the differential equation n0 = n. X p (pp)0 = pp = pp (5) p This unique property of pp is useful in the following theorems.

Theorem 5 [1] If n = pp · m for some prime p and natural m > 1, then n0 = pp(m + m0) (k) and limk→∞ n = ∞.

3 Proof. The derivative of n = pp · m, n0 = (pp)0 · m + pp · m0 = pp(m + m0) > n and further proof by induction shows that n(k) ≥ n + k.

Theorem 6 [1] Let pk be the highest power of prime p that divides the natural number n. If 0 < k < p, then pk−1 is the highest power of p that divides n0. Furthermore, each derivative n, n0, n00, . . . , n(k) is distinct.

Proof. Let n = pkm. Then n0 = kpk−1m + pkm0 = pk−1(km + pm0), and because k < p, the inside term is not divisible by p. We see that indeed, the only solutions to this equation are integers of the form pp.

Theorem 7 [1] For n ∈ N, n0 = n if and only n = pp, where p is a prime. Furthermore, there is an infinite number of solutions to the equation.

Proof. Assume n0 = n. Then by Theorem 6, if p | n at least pp | n or else it would contradict n0 = n. Conversely, assume n = pp. As seen in (5), n0 = pp = n. These properties lead to the conjecture that n(k) = n has only numbers of the form pp as solutions. As we will discuss later, we do not expect the function n(k) to cycle for any arbitrary n not of the form pp.

3.2 Solutions to n0 = a As previously discussed in Section 2.1, we have specific solutions to the differential equations n0 = 0 and n0 = 1. Now we turn our attention to the existence of solutions to the equation n0 = a, where a > 1.

First, we observe that there can only be a finite number of solutions to n0 = a because a2 all potential solutions are bounded above by 4 .

One direct application of n0 = a concerns the Goldbach Conjecture, which states that every even number larger than 3 is the sum of two distinct prime numbers. In terms of our function, we can restate it as follows:

Conjecture 1 For any a ∈ N, there exists a solution to the equation n0 = 2a.

The Goldbach Conjecture allows us to represent 2a = p1 + p2, so if we take the derivative 0 0 0 of the product (p1 · p2) = (p1) (p2) + p1(p2) = p2 + p1 = 2a. Additionally, we find some solutions for n0 = a where a is odd:

Theorem 8 For any prime, p, and a ∈ N which can be expressed a = p + 2, 2p is a solution to the n0 = a.

Proof. (2p)0 = 20p + 2p0 = p + 2. It is useful to explore the properties of the function I(a) which calculates all positive integers n such that n0 = a and the corresponding function i(n) which denotes the number of such solutions. After implementing the program within Pari, initial explorations supported

4 our above conjecture that the each even integer through 100, had at least one solution to n0 = a. We observed that a substantial amount of prime numbers (through 1000) did not have any anti-, which may prove useful when considering bounds for the function i(n) and I(n).

Theorem 9 [1] The function i(n) is unbounded for n > 1.

(This proof was taken directly from [1]) Proof. Suppose that i(n) < C for all n > 1 for some constant C. Then 2n X i(k) < 2Cn k=2 for any n. But for any two primes p, q the product pq belongs to I(p + q) thus

2n X X 0 π(n)(π(n) + 1) π(n)2 i(k) > 1 = > , 2 2 k=2 p≤q≤n where P0 means that the sum runs over the primes, and π(n) is the number of primes not exceeding n. This leads to the inequality

π(n)2 √ 2Cn > ⇒ π(n) < 2 Cn, 2 n which contradicts the known asymptotic behavior π(n) ≈ ln n . We are going to use this proof as a starting point towards a stronger statement regarding the infinitude of i(n) for n > 1. We would like to yield some asymptotic relationship for the function.

3.3 Dynamics of n(k) as k → ∞ It appears that for any natural number, the trajectory of its n-th derivative can follow any three paths:

Conjecture 2 For any n ∈ N, one of the following could happen: • n(k) = 0 as k → ∞

• n(k) = ∞ as k → ∞

• n = pp for some prime p., and thus n(k) = pp as k → ∞

We would like to focus on the proofs of the former two points, as well as prove that not only are cycles infinite, but they never repeat themselves. Additionally, we are forming bounds with which to describe integers that go to infinity as well as those which go to zero. 1 2 A trivial example would be that at least 4 of all integers go to infinity (factors of 2 ).

The following conjecture is related to positive integers which go to zero:

5 Conjecture 3 There exist infinitely many composite numbers n such that n(k) = 0 for sufficiently large natural k. This conjecture is dependent upon the validity of the Twin Prime Conjecture which states that there exists infinitely many prime pairs of the form (p, p + 2). This idea is applied to our derivative as follows: As seen in Theorem 8, (2p)0 = 2 + p. Subsequently (2 + p)0 = 1 and 10 = 0. Thus, a proof of the Twin Prime Conjecture would validate this proof.

Our current research is mainly concerned with the preceding theorems and conjectures within the boundaries of natural numbers. However it is important to know that the deriva- tive can be extended to integers as well as rational numbers, with which comes many more interesting conjectures and research problems.

4 Codes in Pari

The following are codes for various functions with the arithmetic derivative. The comment line above each code describes its function.

This code will compute the arithmetic derivative of any integer or rational number:

{f(n)=sign(n)*abs(n)*sum(i=1, matsize(factor(abs(n)))[1],}

{factor(abs(n))[i,2]/factor(abs(n))[i,1])}

This code will compute the solutions, a for a0 = n (the I(n) function in the paper):

{I(n) = for(i=1, n^2/4+1, if((f(i))==n, print(i)));} This code computes the number of solutions, a, to solve a0 = n (the i(n) function in the paper):

{ i(n)= count = 0; k=n; for(i=1, k^2/4+1, if((f(i))== k, count = count + 1)); print(count);}

[Note: We also have the code to compute the n−th derivative of a number, but there is one minor change we want to make to it. It will be on the final report with some additional codes]

6 7 5 Table of n to the k−th derivative, for n ≥ 100 and k ≥ 10

n n(1) n(2) n(3) n(4) n(5) n(6) n(7) n(8) n(9) n(10) 1 0 0 0 0 0 0 0 0 0 0 2 1 0 0 0 0 0 0 0 0 0 3 1 0 0 0 0 0 0 0 0 0 4 4 4 4 4 4 4 4 4 4 4 5 1 0 0 0 0 0 0 0 0 0 6 5 1 0 0 0 0 0 0 0 0 7 1 0 0 0 0 0 0 0 0 0 8 12 16 32 80 176 368 752 1520 3424 8592 9 6 5 1 0 0 0 0 0 0 0 10 7 1 0 0 0 0 0 0 0 0 11 1 0 0 0 0 0 0 0 0 0 12 16 32 80 176 368 752 1520 3424 8592 20096 13 1 0 0 0 0 0 0 0 0 0 14 9 6 5 1 0 0 0 0 0 0 15 8 12 16 32 80 176 368 752 1520 3424 16 32 80 176 752 1520 3424 8592 20096 70464 235072 17 1 0 0 0 0 0 0 0 0 0 18 21 10 7 1 0 0 0 0 0 0 19 1 0 0 0 0 0 0 0 0 0 20 24 44 48 112 240 608 1552 3120 8144 16304 21 10 7 1 0 0 0 0 0 0 0 22 13 1 0 0 0 0 0 0 0 0 23 1 0 0 0 0 0 0 0 0 0 24 44 48 112 240 608 1552 3120 8144 16304 32624 25 10 7 1 0 0 0 0 0 0 0 26 15 8 12 16 32 80 176 368 752 1520 27 27 27 27 27 27 27 27 27 27 27 28 32 80 176 368 752 1520 3424 27 27 70464 29 1 0 0 0 0 0 0 0 0 0 30 31 1 0 0 0 0 0 0 0 0 31 1 0 0 0 0 0 0 0 0 0 32 80 176 368 752 1520 3424 8592 20096 70464 235072 33 14 9 6 5 1 0 0 0 0 0 34 19 1 0 0 0 0 0 0 0 0 35 12 16 32 80 176 368 752 1520 3424 8592 36 60 92 96 272 560 1312 3312 8976 22288 47872 37 1 0 0 0 0 0 0 0 0 0 38 21 10 7 1 0 0 0 0 0 0 39 16 32 80 176 368 752 1520 3424 8592 20096 40 68 72 156 220 284 288 912 2176 7744 24640 41 1 0 0 0 0 0 0 0 0 0 42 41 1 0 0 0 0 08 0 0 0 43 1 0 0 0 0 0 0 0 0 0 44 48 112 240 608 1552 3120 8144 16304 32624 65264 45 39 16 32 80 176 368 752 1520 3424 8592 n n(1) n(2) n(3) n(4) n(5) n(6) n(7) n(8) n(9) n(10) 46 25 10 7 1 0 0 0 0 0 0 47 1 0 0 0 0 0 0 0 0 0 48 112 240 608 1552 3120 8144 16304 32624 65264 130544 49 14 9 6 5 1 0 0 0 0 0 50 45 39 16 32 80 176 368 752 1520 3424 51 20 24 44 48 112 240 608 1552 3120 8144 52 56 92 96 272 560 1312 3312 8976 22288 47872 53 1 0 0 0 0 0 0 0 0 0 54 81 108 216 540 1188 2484 5076 10260 23112 57996 55 16 32 80 176 368 752 1520 3424 8592 20096 56 92 96 272 560 1312 3312 8976 22288 47872 198656 57 22 13 1 0 0 0 0 0 0 0 58 31 1 0 0 0 0 0 0 0 0 59 1 0 0 0 0 0 0 0 0 0 60 92 96 272 560 1312 3312 8976 22288 47872 198656 61 1 0 0 0 0 0 0 0 0 0 62 33 14 9 6 5 1 0 0 0 0 63 51 20 24 44 48 112 240 608 1552 3120 64 192 640 2368 7168 36864 245760 1851392 12976128 120127488 1012858880 65 18 21 10 7 1 0 0 0 0 0 66 61 1 0 0 0 0 0 0 0 0 67 1 0 0 0 0 0 0 0 0 0 68 72 156 220 284 288 912 2176 7744 24640 84608 69 26 15 8 12 16 32 80 176 368 752 70 59 1 0 0 0 0 0 0 0 0 71 1 0 0 0 0 0 0 0 0 0 72 156 220 284 288 912 2176 7744 24640 84608 296256 73 1 0 0 0 0 0 0 0 0 0 74 39 16 32 80 176 368 752 1520 3424 8592 75 55 16 32 80 176 368 752 1520 3424 8592 76 80 176 368 752 1520 3424 8592 20096 70464 235072 77 18 21 10 7 1 0 0 0 0 0 78 71 1 0 0 0 0 0 0 0 0 79 1 0 0 0 0 0 0 0 0 0 80 176 368 752 1520 3424 8592 20096 70464 235072 705280 81 108 216 540 1188 2484 5076 10260 23112 57996 135648 82 43 1 0 0 0 0 0 0 0 0 83 1 0 0 0 0 0 0 0 0 0 84 124 128 448 1408 5056 15232 56384 169216 677120 2902784 85 22 13 1 0 0 0 0 0 0 0 86 45 39 16 32 80 176 368 752 1520 3424 87 32 80 176 368 752 1520 3424 8592 20096 70464 88 140 188 192 640 2368 7168 36864 245760 1851392 12976128 89 1 0 0 0 0 0 0 0 0 0 90 123 44 48 ‘ 240 608 1552 3120 8144 16304 91 20 24 44 48 112 2409 608 1552 3120 8144 92 96 272 560 1312 3312 8976 22288 47872 198656 1094656 93 34 19 1 0 0 0 0 0 0 0 94 49 14 9 6 5 1 0 0 0 0 95 24 44 48 112 240 608 1552 3120 8144 16304 n n(1) n(2) n(3) n(4) n(5) n(6) n(7) n(8) n(9) n(10) 96 272 560 1312 3312 8976 22288 47872 198656 1094656 5474304 97 1 0 0 0 0 0 0 0 0 0 98 77 18 21 10 7 1 0 0 0 0 99 75 55 16 32 80 176 368 752 1520 3424 100 140 188 192 640 2368 7168 36864 245760 1851392 12976128

References

[1] Ahlander, Bo and Ufnarovski, Victor. “How to differentiate a Number.” Journal of Integer Sequences. Vol. 6 (2003).

[2] Barbeau, E.J. “Remark on an arithmetic derivative.” Canadian Mathematical Bulletin. Vol. 4 (1961): 117-122.

[3] Buium, A. “Arithmetic analogues of derivations.” Journal of Algebra. Vol. 198 (1997) : 290-299.

[4] Hardy, G.H. and Wright, E.M. An Introduction to the Theory of Numbers 5th ed., Oxford University Press, 1980.

[5] Niven, Ivan et. al. An Introduction to the Theory of Numbers 5th ed., Wiley Textbooks, 1991.

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