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Prime Factorial Conjecture

  1. Jun 28, 2004 #1
    Here is a tentative conjecture that needs to be tested.

    [P! + P]/P^2 = INTEGER

    if and only if P is a prime number

    P! is P factorial, e.g. 3*2*1 , 5*4*3*2*1 , 7*6*5*4*3*2*1, etc...
     
  2. jcsd
  3. Jun 28, 2004 #2
    This is true. P!+P = P((P-1)!+1). By Wilson's Theorem, which has been discussed, "Proof of Wilson's Theorem," under Number Theory, we have (P-1)! ==(-1) Modulo p if an only if p is prime. Thus the conjecture is correct.
     
  4. Jun 30, 2004 #3
    Multiply the prime numbers in their correct sequence:

    3*2 = 6 , 6 = 6

    5*3*2 = 30 , 3+0 = 3

    7*5*3*2 = 210 , 2+1+0 = 3

    11*7*5*3*2 = 2310 , 2+3+1+0 = 6

    13*11*7*5*3*2 = 30030 , 3+3 = 6

    17*13*11*7*5*3*2 = 510510 , 5+1+0+5+1+0 = 12 , 1+2 = 3

    etc...

    etc...


    23 ---> 6

    29 ---> 3

    31 ---> 3

    37 ---> 3

    41 ---> 6


    Multiply the sequence of prime numbers then sum the digits[omit 1] of
    the product until it is a single integer. It appears to be 3 or 6.
     
  5. Jul 4, 2004 #4
    [P!+P]/P^2


    [P! +P^2 +P]/P^2 = integers >=1 when P is 1 or prime?



    [P!+P^3+P^2+P]/P^2 = integers >=1 when P is 1 or prime?



    [P! + P^4 + P^3 + P^2 +P]/P^2 = integers >=1 when P is 1 or prime?



    [P! + P^n + P^(n-1) + ...+ P]/P^2 = integers >=1 when P is 1 or prime?
     
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