# A formula of prime numbers for interval (q; (q+1)^2), where q is prime number.

1. Jul 21, 2005

### Victor Sorokine

A formula (for amusement only) 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 (France)

P.S.
Explanation:
If M_s is divided by q_i (where q_i < q_k),
then M_t is not divided by q_i
and therefore p = M_s – M_t is not divided by q_i
(q_i = 2, 3, 5, 7, … q_k).
If any p = M_s – M_t < ((q_k) + 2)^2 and p is divided by q_j
(where q_j > q_k), then p is divided by certain q_i (where q_i < q_k),
but p is not divided by q_i. Therefore p is prime.
To date is all, but…
There is an idea for search the function (or algorithm) q_(k+1) = f(Q_k).

Combinations (s) and exponents can be found with the methods
of Mathematical optimization (or programmation).