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Old 2016-05-14, 19:25   #2
Batalov's Avatar
Mar 2008

27·7·11 Posts


Conjecture On Prime Number Series

This conjecture and very simple yet very special time in practice is spread over a very simple algorithm:

Nseries (p) * (p)

Nseries and based on a small or large number of predefined numbers ranging from 1 to infinity for example 111 and a series 123456 a other set and where (p) is a prime number. Now the product of the series 111 (p) * (p) = 1117 * 7 = 7819 (where (p) = 7) results in a number not first, and in fact has two unique factors as two prime numbers this case 1117 and 7, the series 111 taken as an example can produce hundreds or thousands of results having as result of the product, a number that has only two factors First.
It remains a guess cause it is not demonstrable that there can not be a series where nothing at all is a set whose product does not comply with the rules and did not in fact only and only two prime factors which respect the algorithm.
The problem divided into two parts.

1. As a first step it must be shown that:

Given an arbitrary number  c belonging to  N written in the formula:

\sum_{i=1}^n c_i \cdot 10^{i-1}

where  n and  c_ {j} belong to  N and  c_ {1} \neq0 , and took a number first arbitrary  p , written in the form

\sum_{j=1}^q c_i \cdot 10^{i-1}

where  q and  d_ {j} belong to  N and  d_ {1} \neq0 , then took the ' infinite set of numbers  B expressible in the form

b = \sum_{i=1}^n c_i \cdot 10^{q+i-1} + \sum_{j=1}^q c_i \cdot 10^{i-1}

necessarily exists at least one number  b belonging to  B , that  b belongs to  P .

To prove it, and 'sufficient to prove that

B \cap P \neq \phi

2. Managed to get this demonstration the second step is to show that:

if there exists a number  b_ {1} of the type shown in eq. (3) that belongs to  B and  P , then necessarily exists at least one other number  b_ {2}> b_ {1} that belongs at the intersection of  P and  B .

The union of the two demonstrations would imply that the set  B \cap P and an infinite set (this would be a demonstration for recursion).

Last fiddled with by Batalov on 2016-05-14 at 20:28
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