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Two new prime generators

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  • Chris Caldwell
    Eric Rowland notes that if you set a(1) = 7, and for n ™ 2 set a(n) = a(n-1) + gcd(n,a(n-1)); then a(n) - a(n-1) (the first differences ) are all either 1
    Message 1 of 4 , Jul 21, 2008
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      Eric Rowland notes that if you set a(1) = 7, and for n ≥ 2 set

      a(n) = a(n-1) + gcd(n,a(n-1));

      then a(n) - a(n-1) (the "first differences") are all either 1 or prime.

      Not useful, but hey, why should it be? See Shallit's fine page

      http://recursed.blogspot.com/2008/07/rutgers-graduate-student-finds-new.html




      Jeff also notes that Benoit Cloitre proved that if you set b(1) = 1 and
      for n ≥ 2 set

      b(n) = b(n-1) + lcm(n,b(n-1))

      then b(n)/b(n-1)-1 is either 1 or prime for all n ≥ 2.



      Note the key here is these are simple formulas and **proven** These should
      be any person's goals if they are creating prime number formulas.


      CC.
    • Bob Gilson
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      Message 2 of 4 , Aug 11, 2008
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        Does Eric Rowland's formula not also=9Ainfer (prove?)=9Athat the prime numb=
        er sequence =0A3,5,7,11 ... is incidentally an arithmetic progression, even=
        though we cannot pinpoint precisely what the arithmetic progression is?=0A=
        =0A=0A----- Original Message ----=0AFrom: Chris Caldwell <caldwell@...>=
        =0ATo: primenumbers@yahoogroups.com=0ASent: Monday, July 21, 2008 4:14:08 P=
        M=0ASubject: [PrimeNumbers] Two new prime generators=0A=0AEric Rowland note=
        s that if you set a(1) =3D 7, and for n =99 2 set =0A=0A=9A =9A a(n) =3D a(=
        n-1) + gcd(n,a(n-1));=0A=0Athen a(n) - a(n-1) (the "first differences") are=
        all either 1 or prime.=0A=0ANot useful, but hey, why should it be?=9A See =
        Shallit's fine page=0A=0A=9A =9A http://recursed.blogspot.com/2008/07/rutge=
        rs-graduate-student-finds-new.html=0A=0A=0A=0A=0AJeff also notes that Benoi=
        t Cloitre proved that if you set b(1) =3D 1 and =0Afor n =99 2 set=0A=0A=9A=
        =9A b(n) =3D b(n-1) + lcm(n,b(n-1)) =0A=0Athen b(n)/b(n-1)-1 is either 1 o=
        r prime for all n =99 2.=0A=0A=0A=0ANote the key here is these are simple f=
        ormulas and **proven**=9A These should=0Abe any person's goals if they are =
        creating prime number formulas. =0A=0A=0ACC.=0A=0A-------------------------=
        -----------=0A=0AUnsubscribe by an email to: primenumbers-unsubscribe@yahoo=
        groups.com=0AThe Prime Pages : http://www.primepages.org/=0A=0AYahoo! Group=
        =9A =9A http://groups.yahoo.com/group/primenumbers/join=0A=9A =9A (Yahoo! I=
        Content-Type: text/html; charset=koi8-r
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        <html><head><style type=3D"text/css"><!-- DIV {margin:0px;} --></style></he=
        ad><body><div style=3D"font-family:times new roman, new york, times, serif;=
        font-size:12pt"><DIV>Does Eric Rowland's formula not also infer (prove=
        ?) that the prime number sequence </DIV>=0A<DIV>3,5,7,11 ... is incide=
        ntally an arithmetic progression, even though we cannot pinpoint precisely =
        what the arithmetic progression is?</DIV>=0A<DIV style=3D"FONT-SIZE: 12pt; =
        FONT-FAMILY: times new roman, new york, times, serif"><BR>=0A<DIV style=3D"=
        FONT-SIZE: 13px; FONT-FAMILY: arial, helvetica, sans-serif">----- Original =
        Message ----<BR>From: Chris Caldwell <caldwell@...><BR>To: primen=
        umbers@yahoogroups.com<BR>Sent: Monday, July 21, 2008 4:14:08 PM<BR>Subject=
        : [PrimeNumbers] Two new prime generators<BR><BR>Eric Rowland notes that if=
        you set a(1) =3D 7, and for n =99 2 set <BR><BR>    a(n) =3D a(n=
        -1) + gcd(n,a(n-1));<BR><BR>then a(n) - a(n-1) (the "first differences") ar=
        e all either 1 or prime.<BR><BR>Not useful, but hey, why should it be? =
        ; See Shallit's fine page<BR><BR>    <A href=3D"http://recursed.b=
        logspot.com/2008/07/rutgers-graduate-student-finds-new.html" target=3D_blan=
        k>http://recursed.blogspot.com/2008/07/rutgers-graduate-student-finds-new.h=
        tml</A><BR><BR><BR><BR><BR>Jeff also notes that Benoit Cloitre proved that =
        if you set b(1) =3D 1 and <BR>for n =99 2 set<BR><BR>    b(n) =3D=
        b(n-1) + lcm(n,b(n-1)) <BR><BR>then b(n)/b(n-1)-1 is either 1 or
        prime for all n =99 2.<BR><BR><BR><BR>Note the key here is these are simpl=
        e formulas and **proven**  These should<BR>be any person's goals if th=
        ey are creating prime number formulas. <BR><BR><BR>CC.<BR><BR>-------------=
        -----------------------<BR><BR>Unsubscribe by an email to: <A href=3D"mailt=
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        R>The Prime Pages : <A href=3D"http://www.primepages.org/" target=3D_blank>=
        http://www.primepages.org/</A><BR><BR><A href=3D"http://groups.yahoo.com/" =
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      • Billy Hamathi
        I have taken time in to write something worthy of your valuable time.   Nothing really new, just my attempt at understanding the relationship between
        Message 3 of 4 , Aug 11, 2008
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          I have taken time in to write something worthy of your valuable time.
           
          Nothing really new, just my attempt at understanding the relationship between composite numbers and prime numbers.
           
          First of all all natural numbers greater than 1, are eitheir prime or composite, never both. As such to study the other, you use the other. Meaning if you want to see if there is a pattern in prime numbers, it follows that you will use composites. Also if you want to study primes, you would then use composites. The Goldbach Conjecture is an example which uses primes to relate primes to composites.
           
          I have for sometime now, been struggling with expressing all composite numbers based on prime numbers. In short, I have been trying to establish a pattern in primes using only composites.
           
          I have observed the following:
           
          Intervals in composite numbers are determined by the squares of primes. The meaning of this is that, the first composite [4] is a square of the first prime [2]. All the composites that follow next are even numbers [f(x) = 2x], until the square of the next prime [9]. After [9], the next composites will only be multiples of 2 and 3 ONLY [f(x) = 2x,3x] until [25]. After [25], the next composites will be f(x) = 2x, 3x, 5x, until [49] and so on.
           
          The pattern of how composite progress in relation to primes is as shown below:
           
          (p1)^2,
          [f(x) = (p1)*x],
          (p2)^2,
          [f(x) = (p1)*x, (p2)*x],
          (p3)^2,
          [f(x) = (p1)*x, (p2)*x, (p3)*x],
          (p4)^2,
          [f(x) = (p1)*x, (p2)*x, (p3)*x, (p4)*x],
          :
          :
          :
          (pn)^2,
          [f(x) = (p1)*x, (p2)*x, (p3)*x, (p4)*x, ... , (pn)*x],
          :
           
          So far this is what I can share with you for now without wasting anymore of your time. If you dont mine, please share your thoughts.
           
          Thanks.
           
          Billy.


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        • Chris Caldwell
          ... Perhaps I do not understand the question, but the primes are trivially *not* arithmetic. They do however contain arithmetic progressions of all lengths,
          Message 4 of 4 , Aug 11, 2008
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            > Does Eric Rowland's formula not also=9Ainfer (prove?)=9Athat
            > the prime numb= er sequence =0A3,5,7,11 ... is incidentally
            > an arithmetic progression, even= though we cannot pinpoint
            > precisely what the arithmetic progression is?=0A=

            Perhaps I do not understand the question, but the primes are trivially *not* arithmetic. They do however contain arithmetic progressions of all lengths, but tat follows from Green and Tao, not Rowland.
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