A set of formulas for prime numbers

Received: 12-Nov-2023, Manuscript No. puljpam-24-6921; Editor assigned: 15-Nov-2023, Pre QC No. puljpam-24-6921 (PQ); Accepted Date: Nov 27, 2023; Reviewed: 18-Nov-2023 QC No. puljpam-24-6921 (Q); Revised: 21-Jan-2024, Manuscript No. puljpam-24-6921 (R); Published: 30-Nov-2023

Citation: Rédoane D. A set of formulas for prime numbers. J Pure Appl Math. 2023; 7(6):01-2.

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Abstract

Here I present several formulas and conjectures on prime numbers. I’m interested in studying prime numbers, Euler’s totient function and sum of the divisors of natural numbers.

Key Words

Prime numbers; Sequence; Formula; Divisor function; Euler’s totient

Introduction

Let σ(n) denotes the divisor function which sums the divisors of n, an integer ≥ 1. We introduce the function f such that:

Formula 1

When f(n) = 2n + 1, is 2n + 1 always prime?

And for example for n = 6 we have the prime number 13 that is of the form 2(6) + 1.

The first examples are given by the following sequence:

Carl Schildkraut proved this property [1].

Let {x} = x − [x], let m = n!, and let t = σ(m). Then

the first sum is (1 + m²)(t − 1), and so

Let u = gcd(1 + m², t). The t values {0, 1 + m², 2(1 + m²), . . . , (t − 1)(1 + m²)} modulo t consist of u copies of each multiple of u in (0, t), and so

This means

(In particular, if 1 + (n!)² and σ(n!) are coprime, f(n) = 1).

With this knowledge about f(n), we can tackle the problem at hand. If f(n) = 2n + 1, then, in particular, 2n + 1 divides 1 + (n!)². So, 2n + 1 is relatively prime to n!. This means that 2n + 1 cannot have any factors in the set {2, . . . , n}. However, every number in {n + 1, . . . , 2n} is too large to be a factor of 2n + 1. So, 2n + 1 cannot have any factors strictly between 1 and 2n + 1, and must be prime.

Formula 2

Let x denotes an integer such that x > 1. We define the function f such that:

We have:

(a, b, c, d are integers ≥ 1) We have:

Formula 3

Let k be a positive integer.

Let n be an integer such that n = 6k – 1

Let r be the remainder of the division of (n − 1)! − n by (n + 2)

Property: if 6k + 1 is prime r = 3k + 2

We define the prime 6k+1 such that 6k+1 = r(n)+r(n−1) where r(n) is the sequence of the successive remainders with r(1) = 5 and n ≥ 2. We suppose r(n)≠ 2 and r(n−1)≠ 2.

For example the first 25 values of r are:

5, 8, 11, 2, 17, 20, 23, 2, 2, 32, 35, 38, 41, 2, 2, 50, 53, 56, 2, 2, 65, 2, 71, 2, 77

And we have:

8 + 5 = 13 = 6(2) + 1

11 + 8 = 19 = 6(3) + 1

20 + 17 = 37 = 6(6) + 1

23 + 20 = 43 = 6(7) + 1

35 + 32 = 67 = 6(11) + 1

38 + 35 = 73 = 6(12) + 1

41 + 38 = 79 = 6(13) + 1

53 + 50 = 103 = 6(17) + 1

56 + 53 = 109 = 6(18) + 1

Formula 4

Let a to be a natural number (a ≥ 1), n = 4 * m where m is a natural number ≥ 1) and ϕ is the Euler’s totient function such as:

Prove that if ϕ(aⁿ − 2) + 1 ≡ n − 1 (mod n) then ϕ(aⁿ − 2) + 1 is always aprime number.

Max Alekseyev studied this conjecture but no proof has been found [2].

Formula 5

(a, b) is a couple of twin primes such that b = a + 2 and a > 29. Let N = 4^b and q the quotient which results from the division of N by a and r is the remainder. We calculate P = (q mod b)a + r − 1 Below we prove that P = 3(10b + 1) using Fermat’s little theorem. N = 16 · 4^a ≡ 64 (mod a)then r = 64 (a > 64) 4^b = (b−2)q +64 then 4 ≡ 64−2q (mod b) and q ≡ 30 (mod b) Finally we have P = 30a + 63 = 3(10b + 1)

Formula 6

n is a natural number > 1, φ(n) denotes the Euler’s totient function, Pⁿ is the nth prime number and σ(n) is the sum of the divisors of n. Consider the expression:

F(n) = φ(|P_n+2 − σ(n)|) + 1

Conjecture: when F(n) ≡ 3 (mod 20) then this number is a prime or not. When the number is not a prime it can be a power of prime by calculating |P_n+2−σ(n)| = p^k (p prime, k a natural number > 1).

Examples:

We have n = 680:

F(680) = φ(|P₆₈₂ − σ(680)|) + 1 = φ(5101 − 1620) + 1 = 3423

which is not prime but we have P_n+2 − σ(n) = p², more precisely it is the square of 59.

Interestingly for n ≤ 526388126 (calculations with PARI/GP) all counterexamples are the power of prime.

Another example is found for k = 6, this is n = 526388126. In this case, we have:

F(n) = 10549870323

which is not prime and |P_n+2 − σ(n)| = 47⁶ (here k = 6).

The question is: ”Are there only these two solutions? 1. A power of prime if the result is not a prime 2. Or the result is prime.

References

1. Carl Schildkraut (https://math.stackexchange.com/users/253966/carl-schildkraut), Primes of the form 2n+1, URL (version: 2022-06-23 https://math.stackexchange.com/q/4480961
2. Max Alekseyev (https://math.stackexchange.com/users/147470/max-alekseyev), Eulerâ??stotient function and primes, URL (version: 2022-06-23): https://math.stackexchange.com/q/4478910