Number Theory

1901 Submissions

[13] viXra:1901.0227 [pdf] replaced on 2019-01-17 08:32:12

Collatz Conjecture Proof

Authors: James Edwin Rock
Comments: 5 pages. This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License.

Collatz sequences are formed by applying the Collatz algorithm to any 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. If the Collatz conjecture is true eventually you always get back to one. A connected 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

[12] viXra:1901.0193 [pdf] replaced on 2019-01-13 18:40:10

Riemann Hypothesis Further, Two More Counter Example? 0.4999977+i393939944.2571535368  and 0.50001314+i393939946.4889505702488576920

Authors: Toshiro Takami
Comments: 50 Pages.

I calculated it by looking for a counter example, but I can not determine whether this is a counter example or just a normal zero point and post it here. It is a different value from the previous counter example. I searched for points of higher value, but I could not find it for some reason. The point this time is near the previous point, is the number (the number axis) distorted only in this part? It can not be determined. zeta[0.4999977+i393939944.25715353678841792735]= -3.372108136572006... × 10^-19 + 4.002018173119188... × 10^-13 i and zeta[0.50001314+i393939946.4889505702488576920]= -7.000197154138805... × 10^-19 - 2.848659916217643... × 10^-12 i
Category: Number Theory

[11] viXra:1901.0191 [pdf] replaced on 2019-01-16 15:31:22

Riemann Hypothesis 5 Counter Examples

Authors: Toshiro Takami
Comments: 12 Pages.

It presents counter exsample which is close to 0.5 of 5 Riemann expectations but not 0.5. Regardless of how they are calculated, they are all found in the same area for some reason. I could not find it in other areas. This is considered only because the number axis is distorted in this area. Somewhere on the net there is a memory that reads the mathematician's view that "there are countless zero points in the vicinity of 0.5". The value I put out is hand calculation using WolfranAlpha, It seems that it is necessary to strictly correct by supercomputer. zeta[0.5000866+i393939939.3731193515534038924]= -1.60917723458557... × 10^-18 - 1.428779604546702... × 10^-11 i and zeta[0.4999977+i393939944.25715353678841792735]= -3.372108136572006... × 10^-19 + 4.002018173119188... × 10^-13 i and zeta[0.50001314+i393939946.4889505702488576920]= -7.000197154138805... × 10^-19 - 2.848659916217643... × 10^-12 i and zeta[0.4999944+i393939958.90878694741368323631]= 9.30660314868779... × 10^-19 + 1.342928180878699... × 10^-12 i and zeta[0.4999964+i393939964.659437163857861]= -5.914628349384624... × 10^-16 + 6.504227267123851... × 10^-13 i and
Category: Number Theory

[10] viXra:1901.0188 [pdf] submitted on 2019-01-14 01:42:48

Riemann Hypothesis Further, Three More Counter Example?

Authors: Toshiro Takami
Comments: 60 Pages.

Riemann hypothesis Further, three more counter example?
Category: Number Theory

[9] viXra:1901.0155 [pdf] submitted on 2019-01-11 06:22:54

Elements 5 : Three Trigonometric Identities

Authors: Edgar Valdebenito
Comments: 1 Page.

This note presents three trigonometric identities.
Category: Number Theory

[8] viXra:1901.0116 [pdf] submitted on 2019-01-10 02:36:15

A Parametric Equation of the Equation A^5 + B^5 = 2c^2

Authors: Quang Nguyen Van
Comments: 2 Pages.

The equation a^5 + b^5 = c^2 has no solution in integer. However, related to Fermat- Catalan conjecture, the equation a^5 + b^5 = 2c^2 has a solution in integer. In this article, we give a parametric equation of the equation a^5 + b^5 = 2c^2.
Category: Number Theory

[7] viXra:1901.0108 [pdf] submitted on 2019-01-08 11:13:12

Assuming ABC Conjecture is True Implies Beal Conjecture is True

Authors: Abdelmajid Ben Hadj Salem
Comments: 5 Pages. A Proof of ABC conjecture is submitted to the Journal of Number Theory (2019). Comments Welcome.

In this paper, we assume that the ABC conjecture is true, then we give a proof that Beal conjecture is true. We consider that Beal conjecture is false then we arrive to a contradiction. We deduce that the Beal conjecture is true.
Category: Number Theory

[6] viXra:1901.0104 [pdf] submitted on 2019-01-08 18:01:42

The Collapse of the Liemann Empire

Authors: Toshiro Takami
Comments: 5 Pages.

I tried to prove that(Riemann hypothesis), but I realized that I can not prove how I did it. When we calculate by the sum method of (1) we found that the nontrivial zero point will never converge to zero. Calculating ζ(2), ζ(3), ζ(4), ζ(5) etc. by the method of the sum of (1) gives the correct calculation result. This can be considered because convergence is extremely slow in the case of complex numbers, but there is no tendency to converge at all.
Rather, it tends to diffuse. In other words, it is inevitable to conclude that Riemann's hypothesis is a mistake. We will fundamentally completely erroneous ones, For 150 years, We were trying to prove it.   
Category: Number Theory

[5] viXra:1901.0101 [pdf] submitted on 2019-01-09 00:16:39

A Resolution Of The Brocard-Ramanujan Problem

Authors: Johnny E. Magee
Comments: 7 Pages.

We identify equivalent restatements of the Brocard-Ramanujan diophantine equation, $(n! + 1) = m^2$; and employing the properties and implications of these equivalencies, prove that for all $n > 7$, there are no values of $n$ for which $(n! + 1)$ can be a perfect square.
Category: Number Theory

[4] viXra:1901.0058 [pdf] replaced on 2019-01-08 17:53:16

Riemann Hypothesis (Do They Really Converge to 0?)

Authors: Toshiro Takami
Comments: 17 Pages.

2^s/(2^-1)*3^s/(3^-1)*5^s/(5^s-1)*7^s/(7^s-1)……… Whether the above equation converges to 0 was verified. Convergence is extremely slow, and divergence tendency was rather rather abundant when the prime number was 1000 or more. It was thought that the above equation could possibly be an expression that can be composed only of real numbers.
Category: Number Theory

[3] viXra:1901.0046 [pdf] submitted on 2019-01-04 11:35:20

Relation Between the Euler Totient, the Counting Prime Formula and the Prime Generating Functions

Authors: Nazihkhelifa
Comments: 4 Pages.

Relation between The Euler Totient, the counting prime formula and the prime generating Functions The theory of numbers is an area of mathematis hiih eals ith the propertes of hole an ratonal numbers... In this paper I ill intro uie relaton bet een Euler phi funiton an prime iountnn an neneratnn formula, as ell as a ioniept of the possible operatons e ian use ith them. There are four propositons hiih are mentone in this paper an I have use the efnitons of these arithmetial funitons an some Lemmas hiih refeit their propertes, in or er to prove them
Category: Number Theory

[2] viXra:1901.0030 [pdf] submitted on 2019-01-03 17:17:16

Comparison of the Theoretical and Empirical Results for the Benford's Law Summation Test Performed on Data that Conforms to a Log Normal Distribution

Authors: Robert C. Hall
Comments: 24 Pages.

The Benford's Law Summation test consists of adding all numbers that begin with a particular first or first two digits and determining its distribution with respect to these first or first two digits numbers. Most people familiar with this test believe that the distribution is a uniform distribution for any distribution that conforms to Benford's law i.e. the distribution of the mantissas of the logarithm of the data set is uniform U[0,1). The summation test that results in a uniform distribution is true for an exponential function (geometric progression) but not true for a data set that conforms to a Log Normal distribution even when the Log Normal distribution itself closely approximates a Benford's Law distribution.
Category: Number Theory

[1] viXra:1901.0007 [pdf] submitted on 2019-01-01 16:19:56

Characterization of the Integers of the Form (Z^n-Y^n)/(z-y) that Are Divisible by Some Perfect Nth Powers.

Authors: Rachid Marsli
Comments: 11 Pages.

In this work, we show a sufficient and necessary condition for an integer of the form (z^n-y^n)/(z-y)to be divisible by some perfect nth power p^n, where p is an odd prime. We also show how to construct such integers. A link between the main result and Fermat’s last theorem is discussed. Other related ideas, examples and applications are provided.
Category: Number Theory