- #1

Victor Sorokine

- 70

- 0

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.