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


by Victor Sorokine
Tags: formula, interval, numbers, prime
Victor Sorokine
Victor Sorokine is offline
#1
Sep21-05, 04:25 AM
P: 60
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.
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matt grime
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#2
Sep21-05, 05:15 AM
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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.
ComputerGeek
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#3
Sep21-05, 09:24 AM
P: 534
you are trying to start another discussion of nonsense mathematics like you did with flt..... great. NOT

Hurkyl
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#4
Sep21-05, 06:17 PM
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A formula of prime numbers for interval (q; (q+1)^2)


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.


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