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A new Perfect number formula

  1. May 26, 2007 #1
    Hi there:)

    I found a quick formula for calculate perfect numbers.

    T(v)={{[1+sqrt(1+8*T(v-1))]*[1+sqrt(1+8*T(k))]}^2}/128 - {[1+sqrt(1+8*T(v-1))]*[1+sqrt(1+8*T(k))]}/16

    T(v-1) is the previous perfect number and T(k) is ONE of the previous, that is to say that k=(1 or 2 or ....or v)

    This formula works fine for first 15 perfect numbers but i dont have the knowledge in computer's to test it in biggers perfects.

    If the formula is right, then to find the 41 perfect number we have to check only 40 numbers!!!

    If anyone has the knowledge in programing to test the formula that will be pretty good!!!
     
    Last edited: May 26, 2007
  2. jcsd
  3. May 26, 2007 #2
    This matter is about 2^(n-1)(2^n-1). In this case 2^n-1 is a prime, to be exact, called a Mersenne prime, of which much research has been done.

    To find the sum of the factors (1+2+4+++2^(n-1)(1+2^n-1) =(2^n)(2^2), which is twice the original number, (since the number itself is counted as a divisor.)

    So that the matter is all about determining if 2^n-1 is a prime or not. That seems to be as easy a way to proceed as any.
     
  4. May 26, 2007 #3
    Looking at what you are doing in more detail, I take it that if we go through all the math, we are really saying that if 2^K-1 is prime and 2^m-1 is also a Mersenne prime, then by adding exponents and subtracting 1 we find that 2^(p+k-1)-1 is also a Mersenne prime.

    Looking at Mersenne primes for N=2,3,5,7,13,17, 19, 31, 67, 127, 257...Of course disregard N=2, but proceeding to 496 and 8128, for N=5 and N=7, we would expect N=11 to be the next case, but N=13 is the next case. However for N=7 and N=13, the case of N=19, works out. Just as it does for 28 and 496, where N=3+5-1 =7, giving us the case of 8128. (The easiest and most likely case to test.)

    You will observe that the list above of M. primes gets thinner and thinner so that no two of them can add up to 257. Your idea would allow an arithmatic sequence of primes. Of course, no such thing is know.
     
    Last edited: May 26, 2007
  5. May 27, 2007 #4

    matt grime

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    Tao and Green: existence of arbitrary length arithmetic sequences of primes.
     
  6. May 27, 2007 #5
    Your idea would allow an arithmatic sequence of primes. Of course, no such thing is know.

    Oops! I guess it is known. But it could not fit this case using any of the perfect numbers, as indicated by harisma.
     
    Last edited: May 27, 2007
  7. May 28, 2007 #6
    Hello again!

    Its not exactly an arithmetic sequence of primes . Its not a sequence because formula gives and others numbers, that they I don’t have some interest. Is a better way to calculate perfect numbers (and Mersenne’s prime’s of course).
    But to be sure for this, must someone test it in bigger perfects!
     
  8. May 28, 2007 #7
    The formula is wrong :/ it's not working in bigger perfects:(

    Thank you!
     
  9. Jun 8, 2007 #8
    harisma: The formula is wrong :/ it's not working in bigger perfects.

    I thought Mr. Harisma that I had explained that: Of course disregard N=2, but proceeding to 496 and 8128, for N=5 and N=7, we would expect N=11 to be the next case, but N=13 is the next case.

    However, I did make the mistake of generalizing this from the standpoint of an arithmetical progression. But there is no way in which any of your cases could allow us to progress to p=257, because the sum of 67 + 127 -1= 193, which is not a Mersenne prime, that is: 2^(193)-1 is not prime. (You will observe that the list of M. primes gets thinner and thinner so that no two of them can add up to 257.)
     
    Last edited: Jun 8, 2007
  10. Jun 27, 2007 #9
    [tex]T_v=\frac{{(1+\sqrt{1+8T_{v-1}})*(1+\sqrt{1+8T_{k}})}^2}{128} - \frac{(1+\sqrt{1+8T_{v-1}})*(1+\sqrt{1+8T_{k}})}{16}[/tex]

    For the sake of making it easier to read, I think that's what you were trying to say^^
     
    Last edited: Jun 27, 2007
  11. Jun 27, 2007 #10
    Can you give us a formula to calculate odd primes? :)
     
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