[33] **viXra:1812.0497 [pdf]**
*submitted on 2018-12-31 12:45:35*

**Authors:** Pedro Hugo García Peláez

**Comments:** 5 Pages.

Series De Números Cuyos Factores Son La Lista De Los Números Primos Desde El Principio y Enumerando Todos Sin excepción

**Category:** Number Theory

[32] **viXra:1812.0496 [pdf]**
*submitted on 2018-12-31 13:08:20*

**Authors:** Pedro Hugo García Peláez

**Comments:** 5 Pages.

Series of numbers whose factors are the list of prime numbers from the beginning and listing all without exception

**Category:** Number Theory

[31] **viXra:1812.0495 [pdf]**
*replaced on 2019-01-30 16:38:19*

**Authors:** James Edwin Rock

**Comments:** 5 Pages. This replacement paper adds a ratio TPA / 2.06 column to Table 2 and Table 1 contains a comparison of the twin prime pairs formula for [743,743^2] and [19993,19993^2]. It gives a better explanation for the twin prime pairs formula.

Let P_n be the n_th prime. For twin primes P_n – P_n-1 = 2. We exhibit formula for calculating the number of twin primes in the closed interval [P_n, P_n^2]. We prove that as P_n increases the number of twin primes in the interval [P_n, Pn^2] also increases.

**Category:** Number Theory

[30] **viXra:1812.0494 [pdf]**
*replaced on 2019-01-18 17:04:43*

**Authors:** Simon Plouffe

**Comments:** 10 Pages.

A set of new formulas for primes are given. These formulas have a growth rate much smaller than the ones of Mills and Wright.

**Category:** Number Theory

[29] **viXra:1812.0488 [pdf]**
*submitted on 2018-12-30 10:59:58*

**Authors:** Zeolla Gabriel Martín

**Comments:** 8 Pages.

This paper develops the analysis of Simple composite numbers by golden patterns. Examine how the Simple composite numbers are distributed in different combinations of multiples.

**Category:** Number Theory

[28] **viXra:1812.0439 [pdf]**
*submitted on 2018-12-27 18:34:37*

**Authors:** Robert C. Hall

**Comments:** 5 Pages.

While it is fairly easy to prove that the Log Normal distribution becomes a Benford distribution as the standard deviation approaches infinity (see appendix A), it is a bit more difficult to prove that as the standard deviation approaches zero that the distribution becomes a Normal distribution with a mean of e^u where u is the mean of the natural logarithms of the data set values.

**Category:** Number Theory

[27] **viXra:1812.0422 [pdf]**
*replaced on 2019-01-15 03:19:29*

**Authors:** A. A. Frempong

**Comments:** 11 Pages. Copyright © by A. A. Frempong

The author proves the original Beal conjecture that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor. In the numerical equations, two approaches have been used to change the sum, A^x + B^y, of two powers to a single power, C^z. In one approach, the application of factorization is the main principle, while in the other approach, a formula derived from A^x + B^y was applied. The two approaches changed the sum A^x + B^y to a single power, C^z, perfectly. The derived formula confirmed the validity of the assumption that it is necessary that the sum A^x + B^y has a common prime factor before C^z can be derived. It was concluded that if A^x + B^y = C^z, where A, B, C, x, y, z are positive integers and x, y, z > 2, then A, B and C have a common prime factor.

**Category:** Number Theory

[26] **viXra:1812.0418 [pdf]**
*submitted on 2018-12-24 06:01:30*

**Authors:** Nicolò Rigamonti

**Comments:** 15 Pages.

This paper shows the importance of two properties, which are at the base of the Riemann hypothesis. The key point of all the reasoning about the validity of the Riemann hypothesis is in the fact that only if the Riemann hypothesis is true, these two properties, which are satisfied by the non-trivial zeros, are both true. In fact, only if these two properties are both true , all non-trivial zeros lie on the critical line. Leave a comment, a reflection or an opinion about the paper. You shouldn’t keep your doubts and questions in yourself: if you think that it’s right or can’t attack it,share it, otherwise “ Why don’t you start a discussion?.........”.

**Category:** Number Theory

[25] **viXra:1812.0400 [pdf]**
*replaced on 2018-12-25 07:45:30*

**Authors:** Jesús Sánchez

**Comments:** 21 Pages.

b=0;
p=input('Input a number : ');
m=fix((p+1)/2);
for k=2:m+1;
fun=@(w) exp(p.*1j.*w).*exp(-2.*k.*1j.*w).*(1-exp(-m.*k.*1j.*w))./(1-exp(-k.*1j.*w));
a=integral(fun,-pi,pi);
b=b+a;
end;
b=b/(2*pi);
disp(b);
Above simple MATLAB/Octave program, can detect if a number is prime or not. If the result is zero (considering zero being less than 1e-5), the number introduced is a prime.
If the result is an integer, this result will tell us how many permutations of two divisors, the input number has. As you can check, no recurrent division by odd or prime numbers is done. Just this strange integral:
1/2π ∫_(-π)^π▒〖e^pjω (∑_(k=2)^(k=m+1)▒〖e^(-2kjω) ((1-e^(-mkjω))/(1-e^(-kjω) )) 〗)dω 〗
Being p the number that we want to check if it is a prime or not. And being m whatever integer number higher than (p+1)/2(the lowest, the most efficient the operation). As k and ω are independent variables, the sum can be taken outside the integral (as it is in the above program).
To get to this point, we will do the following. First, we will create a domain with all the composite numbers. This is easy, as you can just multiply one by one all the integers (greater or equal than 2) in that domain. So, you will get all the composite numbers (not getting any prime) in that domain.
Then, we will use the Fourier transform to change from this original domain (called discrete time domain in this regards) to the frequency domain. There, we can “ask”, using Parseval’s theorem, if a certain number is there or not. The use of Parseval’s theorem leads to the above integral. If the number p that we want to check is not in the domain, the result of the integral is zero and the number is a prime. If instead, the result is an integer, this integer will tell us how many permutations of two divisors the number p has. And, in consequence information how many factors, the number p has.
So, for any number p lower than 2m-1, you can check if it is prime or not, just making the numerical definite integration (but even this integral, if no further developments are done, the numerical integration is inefficient computing-wise compared with brute-force checking for example). To be added, is the question regarding the level of accuracy needed (number of decimals and number of steps in the numerical integration) to have a reliable result for large numbers. This will be commented on the paper, but a separate study will be needed to have detailed conclusions.
Of course, the best would be that in the future, an analytical result (or at least an approximation) for the summation or for the integration is achieved.

**Category:** Number Theory

[24] **viXra:1812.0362 [pdf]**
*submitted on 2018-12-20 21:07:15*

**Authors:** Ho Soo Shin

**Comments:** 2 Pages. comments welcome!

As is well known, the Collatz sequence, which is also named as the hailstone sequence, follows the rule of Collatz conjecture. The rule requires us to divide any positive even integer by 2. We must multiply every positive odd number by 3 and then add 1 according to the rule. By investigating residues modulo 3, I will prove any integer multiple of 3 cannot appear more than one time in a Collatz sequence, which implies any multiple of 3 cannot be included in a possible cycle of the Collatz sequence.

**Category:** Number Theory

[23] **viXra:1812.0340 [pdf]**
*replaced on 2018-12-21 15:57:08*

**Authors:** Wu ShengPing

**Comments:** 4 Pages.

The main idea of this article is simply calculating integer
functions in module. The algebraic in the integer modules is studied in
completely new style. By a careful construction the result that
two finite numbers is with unequal logarithms in a corresponding module is proven, which result is applied to solving
a kind of high degree diophantine equation.

**Category:** Number Theory

[22] **viXra:1812.0322 [pdf]**
*submitted on 2018-12-18 09:16:24*

**Authors:** James Edwin Rock

**Comments:** 2 Pages. Copyright 2018 James Edwin Rock Create Commons Attribution-ShareAlike 4.0 International License2

Let Pn be the n_th prime. For twin primes Pn – Pn-1 = 2. We give a heuristic argument that in the interval (Pn, Pn2) as Pn gets larger, there is an increasing number of twin primes.

**Category:** Number Theory

[21] **viXra:1812.0312 [pdf]**
*replaced on 2018-12-19 06:26:14*

**Authors:** Julian Beauchamp

**Comments:** 3 Pages.

In 2002 Preda Mihailescu used the theory of cyclotomic fields and Galois modules to prove Catalan's Conjecture. In this short paper, we give a very simple proof. We first prove that no solutions exist for a^x-b^y=1 for a,b>0 and x,y>2. Then we prove that when x=2 the only solution for a is a=3 and the only solution for y is y=3.

**Category:** Number Theory

[20] **viXra:1812.0305 [pdf]**
*replaced on 2018-12-25 04:29:58*

**Authors:** Juan Moreno Borrallo

**Comments:** 24 Pages.

In this paper it is proposed and proved an exact formula for the prime-counting function, finding an expression of Legendre's formula. As corollaries, they are proved some important conjectures regarding prime numbers distribution.

**Category:** Number Theory

[19] **viXra:1812.0300 [pdf]**
*submitted on 2018-12-17 23:40:30*

**Authors:** Aaron chau

**Comments:** 1 Page.

无论是在历史的任何时间，如不谈黎曼猜想则已；如谈，我们大家当然先要彻底弄明白：
黎曼临界线上从小到大的一组零点及其（零点空格），它们在数论上究竟是什么意思呢？

**Category:** Number Theory

[18] **viXra:1812.0299 [pdf]**
*submitted on 2018-12-17 23:47:57*

**Authors:** Aaron chau

**Comments:** 2 Pages.

本文强调：质数与孪生质数分别都是无限的依据是算术中的（加减乘除）。
比如在古希腊，欧几里德证明质数无限，他所应用的是（乘除法）来表述反证法。
又比如，现时在伦敦来证明孪生质数无限，即无限存在二个质数的距离＝2；本文应用的是（加减法）来表述：就在（单数空格）里，单数与奇合数的个数分别在每一数段里的多与少。

**Category:** Number Theory

[17] **viXra:1812.0296 [pdf]**
*submitted on 2018-12-18 01:33:20*

**Authors:** Joseph Olloh

**Comments:** 13 Pages. The content of the paper provides a fundamentally new way of looking at numbers.

We differentiate even and odd numbers into various groups and subgroups. We provide the properties of the forms of numbers which fall into each groups and subgroups. We expound on the relationship of a special group of even numbers and the collatz conjecture, we also derive an accurate formula to calculate the steps involved when an even number of the group is the initial value of the collatz operation.
For each group and subgroup of odd and even numbers, we discuss the observed pattern of their sequences and also derive accurate formulas for each sequence. Throughout, b, d, k, N, n, x, m, and z all denote positive integers, with d, and N denoting odd numbers, x and z denoting even numbers, and b denoting special even-even numbers
The order of priority of the properties of each group is key in the differentiation of the numbers into their various groups and subgroups.

**Category:** Number Theory

[16] **viXra:1812.0287 [pdf]**
*submitted on 2018-12-16 09:20:54*

**Authors:** Zhang Tianshu

**Comments:** 14 Pages.

If regard positive integers which have a common prime factor as a kind, then the positive half line of the number axis consists of infinite many recurring segments which have same permutations of c kinds of integer’s points, where c≥1. In this article, the author shall prove Grimm’s conjecture by the method that changes orderly symbols of each kind of composite number’s points at the original number axis, so as to form consecutive composite number’s points within limits that proven Bertrand's postulate restricts.

**Category:** Number Theory

[15] **viXra:1812.0242 [pdf]**
*replaced on 2018-12-14 12:37:47*

**Authors:** Jean Pierre Morvan

**Comments:** 6 Pages.

L'hypothèse de RIEMANN est fausse parce que la fonction zéta n'est pas nulle avec s = 1/2 + ib

**Category:** Number Theory

[14] **viXra:1812.0208 [pdf]**
*replaced on 2019-01-23 08:21:08*

**Authors:** Kenneth A. Watanabe

**Comments:** 14 Pages. Acknowledgement section was added and proof of delta pi was added

A twin prime is defined as a pair of prime numbers (p1,p2) such that p1 + 2 = p2. The Twin Prime Conjecture states that there are an infinite number of twin primes. A more general conjecture by de Polignac states that for every natural number k, there are infinitely many primes p such that p + 2k is also prime. The case where k = 1 is the Twin Prime Conjecture. In this document, a function is derived that corresponds to the number of twin primes less than n for large values of n. Then by proof by induction, it is shown that as n increases indefinitely, the function also increases indefinitely thus proving the Twin Prime Conjecture. Using this same methodology, the de Polignac Conjecture is also shown to be true.

**Category:** Number Theory

[13] **viXra:1812.0182 [pdf]**
*replaced on 2018-12-30 15:25:04*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 5 Pages. Submitted to the journal Research In Number Theory. Comments welcome.

In this paper, from a,b,c positive integers relatively prime with c=a+b, we consider a bounded of c depending of a,b. Then we do a choice of K(\epsilon) and finally we obtain that the ABC conjecture is true. Four numerical examples confirm our proof.

**Category:** Number Theory

[12] **viXra:1812.0154 [pdf]**
*replaced on 2018-12-13 06:58:30*

**Authors:** M. Sghiar

**Comments:** 7 Pages. Accepted & french version © Copyright 2018 by M. Sghiar. All rights reserved. Respond to the author by email at: msghiar21@gmail.com

I show here that if $ x \in \mathbb{N}^*$ then $1 \in \mathcal{O}_S (x)= \{ S^n(x), n \in \mathbb{N}^* \} $ where $ \mathcal{O}_S (x)$ is the orbit of the function S defined on $\mathbb{R}^+$ by $S(x)= \frac{x}{2} + (x+\frac{1}{2}) sin^2(x\frac{\pi}{2})$, and I deduce the proof of the Syracuse conjecture.

**Category:** Number Theory

[11] **viXra:1812.0150 [pdf]**
*replaced on 2019-04-15 21:48:03*

**Authors:** Phil Aaron Bloom

**Comments:** 3 Pages.

An open problem is proving FLT \emph{simply} (using Fermat's toolbox) for each $n\in\mathbb{N}, n>2$. Our \emph{direct proof} (not BWOC) of FLT is based on our algebraic identity $((r+2q^n)^\frac{1}{n})^n-(2^\frac{2}{n}q)^n=((r-2q^n)^\frac{1}{n})^n$ with arbitrary values of $n\in\mathbb{N}$, and with $r\in\mathbb{R},q\in\mathbb{Q},r>2q^n,n,q,r>0$. For convenience, we \emph{denote} $(r+2q^n)^\frac{1}{n}$ by $s$; we \emph{denote} $2^\frac{2}{n}q$ by $t$; and, we \emph{denote} $(r-2q^n)^\frac{1}{n}$ by $u$. For any given $n>2$ : Since the term $t$ or $2^\frac{2}{n}q$ with $q\in\mathbb{Q}$ is not rational, this identity allows us to relate null set $\{(s,t,u)|s,t,u\in\mathbb{Q},s,t,u>0,s^n-t^n=u^n\}$ with $\{z,y,x|z,y,x\in\mathbb{Q},z,y,x>0,z^n-y^n=x^n\}$ that we subsequently prove null : We show, for $n>0$, that $\{t|s,t,u\in\mathbb{Q},s,t,u>0,s^n-t^n=u^n\}=\{y|z,y,x\in\mathbb{Q},z,y,x>0,z^n-y^n=x^n\}$. Hence, for any given $n\in\mathbb{N},n>2$, it is a true statement that $\{(x,y,z)|x,y,z\in\mathbb{N},x,y,z>0,x^n+y^n=z^n\}=\varnothing$.

**Category:** Number Theory

[10] **viXra:1812.0133 [pdf]**
*submitted on 2018-12-07 13:32:39*

**Authors:** James Edwin Rock

**Comments:** 5 Pages. Copyright 2018 James Edwin Rock Create Commons Attribution-ShareAlike 4.0 International License

Collatz sequences are formed by applying the Collatz algorithm to a positive integer. If it is even repeatedly divide by two until it is odd, then multiply by three and add one to get an even number and vice versa. Eventually you get back to one. The Collatz Structure is created, which contains all positive integers exactly once. The terms of the Collatz Structure are joined together via the Collatz algorithm. Thus, every positive integer forms a Collatz sequence with unique terms terminating in the number one.

**Category:** Number Theory

[9] **viXra:1812.0130 [pdf]**
*submitted on 2018-12-07 21:04:48*

**Authors:** Zhang Tianshu

**Comments:** 12 Pages.

Since there are infinitely many consecutive satisfactory values of ε to enable A+B=C satisfying C>(rad(A, B, C))1+ε, thus the author uses a representative equality, namely 1+2N(2N-2)=(2N-1)2 satisfying (2N-1)2>[rad(1, 2N(2N-2), (2N-1)2)]1+ε, and that first let ε equal a value near the greater end of the infinitely many consecutive satisfactory values to prove the ABC conjecture; again let ε equal a value near the smaller end to negate the ABC conjecture. This shows that the ABC conjecture is in the ambiguity in which case of ε>0.

**Category:** Number Theory

[8] **viXra:1812.0107 [pdf]**
*submitted on 2018-12-06 14:23:59*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 8 Pages. Comments welcome.

In this paper, we give the elliptic curve (E) given by the equation:
y^2=x^3+px+q
with $p,q \in Z$ not null simultaneous. We study a part of the conditions verified by $(p,q)$ so that it exists (x,y) \in Z^2 the coordinates of a point of the elliptic curve (E) given by the equation above.

**Category:** Number Theory

[7] **viXra:1812.0074 [pdf]**
*replaced on 2018-12-10 15:13:28*

**Authors:** Stephen Marshall

**Comments:** 8 Pages.

The Riemann hypothesis is a conjecture that the Riemann zeta function has its zero’s only at the negative even integers and complex numbers with real part 1/2
The Riemann hypothesis implies results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics (Bombieri 2000).
It was proposed by Bernhard Riemann (1859), after whom it is named. The name is also used for some closely related analogues, such as the Riemann hypothesis for curves over finite fields.
The Riemann hypothesis implies significant results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics (Bombieri 2000). The Riemann zeta function is defined for complex s with real part greater than 1 by the absolutely convergent infinite series:
ζ(s) = 1 + 1/2s + 1/3s + 1/4s + ...
The Riemann hypothesis asserts that all interesting solutions of the equation:
ζ(s) = 0
lie on a certain vertical straight line.
In mathematics, the n-th harmonic number is the sum of the reciprocals of the first n natural numbers:
Hn = 1 + 1/2+1/3+1/4+⋯+ 1/n = ∑_(n=1)^n▒1/n
Harmonic numbers have been studied since early times and are important in various branches of number theory. They are sometimes loosely termed harmonic series, are closely related to the Riemann zeta function.
The harmonic numbers roughly approximate the natural logarithm function and thus the associated harmonic series grows without limit, albeit slowly. In 1737, Leonhard Euler used the divergence of the harmonic series to provide a new proof of the infinity of prime numbers. His work was extended into the complex plane by Bernhard Riemann in 1859, leading directly to the celebrated Riemann hypothesis about the distribution of prime numbers.

**Category:** Number Theory

[6] **viXra:1812.0071 [pdf]**
*submitted on 2018-12-04 22:22:02*

**Authors:** Colin James III

**Comments:** 2 Pages. © Copyright 2018 by Colin James III All rights reserved. Respond to the author by email at: info@ersatz-systems dot com.

Using the standard wiki definition of the Collatz conjecture, we map a positive number to imply that a divisor of two implies either an even numbered result (unchanged) or an odd numbered result (changed to the number multiplied by three plus one) to imply the final result of one. This is the shortest known confirmation of the conjecture, and in mathematical logic.

**Category:** Number Theory

[5] **viXra:1812.0022 [pdf]**
*replaced on 2018-12-25 01:58:45*

**Authors:** Pankaj Mani

**Comments:** 12 Pages.

Riemann Hypothesis is TRUE if we look at the Functional Equation satisfied by the Riemann Zeta function upon analytical continuation in Game Perspective way as visualized by David Hilbert. It uses technical game theoretical concepts e.g. Nash Equilibrium to assert that Riemann Hypothesis has to be True. Needs to be looked at the Foundational Principles underlying Mathematics. In other words,it’s the game of arranging Zeros in the complex plane using the functional equation.

**Category:** Number Theory

[4] **viXra:1812.0020 [pdf]**
*submitted on 2018-12-01 15:11:29*

**Authors:** Abdelmajid Ben Hadj Salem

**Comments:** 6 Pages. Comments welcome.

In this paper, we assume that Beal conjecture is true, we give a complete proof of the ABC conjecture. We consider that Beal conjecture is false $\Longrightarrow$ we arrive that the ABC conjecture is false. Then taking the negation of the last statement, we obtain: ABC conjecture is true $\Longrightarrow$ Beal conjecture is true. But, if the Beal conjecture is true, then we deduce that the ABC conjecture is true

**Category:** Number Theory

[3] **viXra:1812.0019 [pdf]**
*replaced on 2018-12-13 09:03:08*

**Authors:** Kamal Barghout

**Comments:** 16 Pages. The material in this article is copyrighted. Please obtain authorization from the author before the use of any part of the manuscript

A probabilistic proof of the Collatz conjecture is described via identifying a sequential permutation of even natural numbers by divisions by 2 that follows a recurrent pattern of the form x,1,x,1…, where x represents divisions by 2 more than once. The sequence presents a probability of 50:50 of division by 2 more than once as opposed to divisions by 2 once over the even natural numbers. The sequence also gives the same 50:50 probability of consecutive Collatz even elements when counted for division by 2 more than once as opposed to division by 2 once and a ratio of 3:1 of high counts. Considering Collatz function producing random numbers and over sufficient iterations, this probability distribution produces numbers in descending order that lead to the convergence of the Collatz function to 1, assuming the only cycle of the function is 1-4-2-1.

**Category:** Number Theory

[2] **viXra:1812.0018 [pdf]**
*replaced on 2019-01-08 21:11:56*

**Authors:** Khalid Ibrahim

**Comments:** 96 Pages.

In this paper, not only did we disprove the Riemann Hypothesis (RH) but we also showed that zeros of the Riemann zeta function $\zeta (s)$ can be found arbitrary close to the line $\Re (s) =1$. Our method to reach this conclusion is based on analyzing the fine behavior of the partial sum of the Dirichlet series with the Mobius function $M (s) = \sum_n \mu (n) /n^s$ defined over $p_r$ rough numbers (i.e. numbers that have only prime factors greater than or equal to $p_r$). Two methods to analyze the partial sum fine behavior are presented and compared. The first one is based on establishing a connection between the Dirichlet series with the Mobius function $M (s) $ and a functional representation of the zeta function $\zeta (s)$ in terms of its partial Euler product. Complex analysis methods (specifically, Fourier and Laplace transforms) were then used to analyze the fine behavior of partial sum of the Dirichlet series. The second method to estimate the fine behavior of partial sum was based on integration methods to add the different co-prime partial sum terms with prime numbers greater than or equal to $p_r$. Comparing the results of these two methods leads to a contradiction when we assume that $\zeta (s)$ has no zeros for $\Re (s) > c$ and $c <1$.

**Category:** Number Theory

[1] **viXra:1812.0002 [pdf]**
*submitted on 2018-12-01 05:01:55*

**Authors:** Zhang Tianshu

**Comments:** 21 Pages.

In this article, the author first classify A, B and C according to their respective odevity, and thereby get rid of two kinds which belong not to AX+BY=CZ. Then, affirm the existence of AX+BY=CZ in which case A, B and C have at least a common prime factor by several concrete equalities. After that, prove AX+BY≠CZ in which case A, B and C have not any common prime factor by the mathematical induction with the aid of the distinct odd-even relation on the premise whereby even number 2W-1HZ as symmetric center of positive odd numbers concerned after divide the inequality in four. Finally, reach a conclusion that the Beal’s conjecture holds water via the comparison between AX+BY=CZ and AX+BY≠CZ under the given requirements.

**Category:** Number Theory