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

  1. Jul 21, 2005 #1
    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).
     
  2. jcsd
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