A formula of prime numbers for interval (q; (q+1)^2)

In summary, there is a formula of prime numbers for the interval (q; (q+1)^2) where q is prime number, given by the expression Q = M_s - M_t, where Q_k is the multitude of the first k prime numbers and M_s and M_t are products of s and t elements respectively. This formula generates only prime numbers in the specified interval and has been previously discussed in this forum.
  • #1
Victor Sorokine
70
0
A formula of prime numbers for interval (q; (q+1)^2),
where q is prime number.

Let:
Q_k – the multitude of first k prime numbers to some extent:
Q_k = (q_0 = 1^0, q_1 = 2^n1, q_2 = 3^n2, q_3 = 5^n3, q_4 = 7^n4, … q_k = u^nk)
(here the expression «_i» signifies lower index, and «^ni» signifies exponent);
M_s – the product of s elements to his extent;
M_t – the product of the rest t = k – s elements.
And now
ALL numbers q = M_s – M_t ( q is function of the combination s and of the exponents n0, n1, … nk) in the interval (q_k ; (q_k)^2) [and in the interval (q_k ; (q_k+1)^2)] are PRIME
(let Q – the multitude of the q, where q_k < q < (q_k+1)^2).

Example:
Q_4 :
q_0 = 1^0, q_1 = 2^n1, q_2 = 3^n2, q_3 = 5^n3, q_4 = 7^n4.
Interval:
7 < q < 9^2 = 81 [< 121].

Q :
11 = 3 x 7 – 2 x 5,
13 = 2^2 x 7 – 3 x 5,
17 = 5 x 7 – 2 x 3^2,
19 = 7^2 – 2 x 3 x 5,
23 = 2 x 3 x 5 – 7,
29 = 5 x 7 – 2 x 3,
31 = 3^2 x 5 – 2 x 7,
37 = 2 x 3 x 7 – 5,
41 = 3 x 5 x 7 – 2^6,
43 = 2 x 5 x 7 – 3^3,
47 = 3 x 5^2 – 2^2 x 7,
53 = 3^2 x 7 – 2 x 5,
59 = 2^4 x 5 – 3 x 7,
61 = 3 x 5^2 – 2 x 7.
67 = 2^4 x 7– 3^2 x 5
71 = 2^3 x 3 x 5 – 7^2,
73 = 3 x 5 x 7 – 2^5,
79 = 2^2 x 3 x 7 – 5,
[and also:
83 = 5^3 – 2 x 3 x 7,
89 = 3 x 5 x 7 – 2^4,
97 = 3 x 5 x 7 – 2^3,
101 = 3 x 5 x 7 – 2^2,
103 = 3 x 5 x 7 – 2,
107 = 3^3 x 5 – 2^2 x 7,
109 = 3^3 x 7 – 2^4 x 5,
113 = 2^2 x 5 x 7 – 3^3,
And only further the formula makes a transient error:
2 x 3^2 x 7 – 5= 121 = 11 х 11.]
Here min(q) = 11.

But now we can write out the multitude
Q_5 :
q_0 = 1^0, q_1 = 2^n1, q_2 = 3^n2, q_3 = 5^n3, q_4 = 7^n4, q_5 = 11^n5
and calculate the prime number in interval
11 < q < 13^2 = 144.
Etc…

In the interval (q_k ; (q_k+1)^2) the formula don't give the composite numbers.

Victor Sorokine

P.S. The fonction q_k+1 = F(q_k) will be done after the recognition of the proof FLT.
PP.S. Bewaring of aggressiveness some professional,
author does not take part in the discussion.
 
Physics news on Phys.org
  • #2
I'm a professional mathematician and, beyond this post, I have no intention of taking part in any discussion. I will merely say I have no idea if that is good or bad in your opinon since I cannot understand what your PP.S. means. This is however in keeping with the rest of your post which also makes no sense at all.
 
  • #3
you are trying to start another discussion of nonsense mathematics like you did with flt... great. NOT
 
  • #4
Victor: your text is impenetrable. I am unable to read most of it.

As for your observation, it isn't new, and it is fairly trivial to prove that in the specified intervals, such formulae can only give prime numbers.

We've had someone come through this very forum presenting this observation (in a readable way), and with an interesting follow-up question.
 

1. What is a formula of prime numbers for interval (q; (q+1)^2)?

A formula of prime numbers for interval (q; (q+1)^2) is a mathematical expression that can be used to generate a list of prime numbers within a specific range of values. It is based on the interval (q; (q+1)^2), where q is a positive integer.

2. How does the formula work?

The formula works by using the concept of a Sieve of Eratosthenes, which is a method for finding prime numbers. It involves creating a list of numbers and crossing out multiples of each prime number until only the prime numbers are left. In the case of the interval (q; (q+1)^2), the formula uses a modified version of this method to generate a list of prime numbers within the specified range.

3. Can the formula be used for any range of numbers?

Yes, the formula can be used for any range of numbers as long as the range is defined by the interval (q; (q+1)^2). This means that the starting number (q) must be a positive integer, and the ending number ((q+1)^2) must be the square of the next consecutive integer.

4. How accurate is the formula?

The formula is highly accurate, as it is based on a proven method for finding prime numbers. However, it is not foolproof and may not generate all prime numbers within the specified interval. It is always recommended to double-check the results using other methods.

5. What are the practical applications of this formula?

This formula can be useful in various mathematical and scientific fields, such as cryptography, number theory, and data encryption. It can also be used for generating large prime numbers for use in computer algorithms and simulations.

Similar threads

  • General Math
Replies
24
Views
2K
  • Linear and Abstract Algebra
Replies
2
Views
759
Replies
1
Views
1K
Replies
3
Views
379
  • Calculus and Beyond Homework Help
Replies
3
Views
503
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Precalculus Mathematics Homework Help
Replies
3
Views
763
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
32
Views
3K
  • Precalculus Mathematics Homework Help
Replies
1
Views
1K
Back
Top