I've programmed the Fermat prime number sieve and tested it out on a few
different ranges.


If the occasion ever arose to factor all numbers in an interval,  this
algorithm could be extended to find factors for all the numbers in the
interval almost as quickly as Fermat factoring could factor the most
difficult number in the interval.


  >>> SieveFermat(10**3,10**3+100)

   number of clears is  260
[0, 0, 0, 0, 0, 0, 0, 0, 0, 1009, 0, 0, 0, 1013, 0, 0, 0, 0, 0, 1019, 0,
1021, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1031, 0, 1033, 0, 0, 0, 0, 0, 1039, 0,
0, 0, 0, 0, 0, 0, 0, 0, 1049, 0, 1051, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1061,
0, 1063, 0, 0, 0, 0, 0, 1069, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 1087, 0, 0, 0, 1091, 0, 1093, 0, 0, 0, 1097, 0, 0, 0]



  >>> SieveFermat(10**4,10**4+100)

   number of clears is  2135
[0, 0, 0, 0, 0, 0, 0, 10007, 0, 10009, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10037, 0, 10039, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10061, 0, 0, 0,
0, 0, 10067, 0, 10069, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10079, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 10091, 0, 10093, 0, 0, 0, 0, 0, 10099, 0]



  >>> SieveFermat(10**5,10**5+100)

   number of clears is  21255
[0, 0, 0, 100003, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100019,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
100043, 0, 0, 0, 0, 0, 100049, 0, 0, 0, 0, 0, 0, 0, 100057, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 100069, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]



  >>> SieveFermat(10**6,10**6+100)

   number of clears is  213878
[0, 0, 0, 1000003, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1000033, 0, 0, 0, 1000037, 0, 1000039,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1000081, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1000099, 0]



  >>> SieveFermat(10**5,10**5+1000)

   number of clears is  22491
[0, 0, 0, 100003, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100019,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
100043, 0, 0, 0, 0, 0, 100049, 0, 0, 0, 0, 0, 0, 0, 100057, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 100069, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100103, 0, 0,
0, 0, 0, 100109, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 100129, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 100151, 0, 100153, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
100169, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100183, 0, 0, 0, 0, 0,
100189, 0, 0, 0, 100193, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100207,
0, 0, 0, 0, 0, 100213, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 100237, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100267, 0, 0, 0, 100271, 0,
0, 0, 0, 0, 0, 0, 100279, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100291, 0, 0,
0, 0, 0, 100297, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100313, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100333, 0, 0, 0,
0, 0, 0, 0, 0, 0, 100343, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100357,
0, 0, 0, 100361, 0, 100363, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
100379, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100391, 0, 100393, 0, 0, 0, 0,
0, 0, 0, 0, 0, 100403, 0, 0, 0, 0, 0, 0, 0, 100411, 0, 0, 0, 0, 0,
100417, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 100447, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100459,
0, 0, 0, 0, 0, 0, 0, 0, 0, 100469, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 100483, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100493, 0, 0, 0, 0, 0, 0, 0,
100501, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100511, 0, 0, 0, 0, 0, 100517, 0,
100519, 0, 0, 0, 100523, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100537,
0, 0, 0, 0, 0, 0, 0, 0, 0, 100547, 0, 100549, 0, 0, 0, 0, 0, 0, 0, 0, 0,
100559, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100591, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 100609, 0, 0, 0, 100613, 0, 0, 0, 0, 0, 0, 0, 100621,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 100649, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 100669, 0, 0, 0, 100673, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 100693, 0, 0, 0, 0, 0, 100699, 0, 0, 0, 100703, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 100733, 0, 0, 0, 0, 0, 0, 0, 100741, 0, 0, 0, 0, 0, 100747, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100769, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100787, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 100799, 0, 100801, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100811, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 100823, 0, 0, 0, 0, 0, 100829, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100847, 0, 0, 0, 0, 0, 100853, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 100907, 0, 0, 0, 0, 0, 100913, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 100927, 0, 0, 0, 100931, 0, 0, 0, 0, 0, 100937, 0, 0, 0, 0, 0,
100943, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100957, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100981, 0, 0, 0, 0,
0, 100987, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 100999, 0]
  >>>


Kermit   <   kermit@...  >



   ----------

def SieveFermat(start,end):
     rstart = ksqrt(start-1)
     if start == 0:
         rstart = -1
     rend = ksqrt(end)
     jar = range(start,end+1)

     plex = 0

#####   zero out 1 if it is present

     if start < 2:
         jar[1 - start] = 0

####   zero out all even numbers

     mds = start + start%2
     mde = end + end%2
     while mds < mde + 1:
         jar[mds - start] = 0
         mds = mds + 2


#####  outer loop :  Set larger square for difference of squares

     rs = rstart + 0
     if rs < 2:
         rs = 2
     rsupper = (end + 10)/6
     while rs < rsupper:
         rs = rs + 1
         rs2 = rs * rs
         if rs2 < start:
             continue

#######      set smaller square for difference of squares.  Difference must be
odd number.

         loop2 = 0
         sidelow  = rs2 - end
         if sidelow < 1:
             sidelow = 1
         sidehigh = rs2 - start + 1
         dexlow = ksqrt(sidelow - 1)
         dexhigh = ksqrt(sidehigh - 1)
         if dexhigh > rs -4:
             dexhigh = rs - 4
         dex = dexlow - dexlow%2 -rs%2 -1

#####      dex remains opposite parity from rs

         while  dex < dexhigh:
             dex = dex + 2

             plex = plex + 1

             dex2 = dex * dex
             d1 = rs - dex
             if d1 < 2:
                 break
             d2 =  rs2 - dex2
             if d2 < start:
                 break
             if d2 > end:
                 continue
             pos = d2 - start
             jar[pos] = 0

     print "\n  number of clears is ",plex
     return jar



def ksqrt(start):
     if start < 1:
         return 0
     r = 1
     r2 = r * r
     while r2 < start:
         r = r * 2
         r2 = r2 * 4
     s = 0
     s2 = 0
     while r > 0:
         s = s + r
         s2 = s * s
         if s2 > start:
             s = s - r
         r = r/2
     return s


[Non-text portions of this message have been removed]

Hi, Group, et. al.

I have a rather simple observation that may imply that the number of
Fermat primes is finite.

Conjecture: If a Fermat number of the form F(k,m)= 2^ (2^ ((2^k)*m))
+1 is prime, then [2^ (2^ ((2^k)*m)) +1 == 2^q (mod 2^(m-k) +1)]
where 'k' is /-1 or a whole number/ and 2^k <= 'm' which is a /whole
number/ for some /odd number/ 'q'.

Since 'q' is /odd/, this argument further reduces to...
[((2^k)*m) == (2n+1) mod (m-k)] and is only satisfied when k= -1 and
m= 0, 2; or k= 0 and m= 1, 3; or k= 1 and m= 2.  For all other 'k's &
whole numbers 'm' > 3, the last equation gives the incompatible re-
sult of 2n == some /odd number/.

Just examine the following equations which imply that n= 0 or 1 from
the reduced modulo equation/argument...

k=-1, m=0;
[0 == (2n+1) mod 1]=> [-0 == (2n) mod 1]=> [2n= 0]=> [n= 0]; F0=3,

k=0, m=1;
[1 == (2n+1) mod 1]=> [0 == (2n) mod 1]=> [2n= 0]=> [n= 0]; F1=5,

k=-1, m=2;
[1 == (2n+1) mod 3]=> [0 == (2n) mod 3]=> [2n= 0]=> [n= 0]; F2=17,

k=0, m=3;
[3 == (2n+1) mod 3]=> [2 == (2n) mod 3]=> [2n= 2]=> [n= 1]; F3=257,

k=1, m=2;
[2 == (2n+1) mod 1]=> [1 == (2n) mod 1]=> [2n= 0]=> [n= 0]; F4=65537.

Other values for 'k' and 'm' would make 2n== some /odd number/.
Now, if the observation held for only one or two of the Fermat prime
numbers, then I would've been hesitant to imply that the modulo ar-
gument is, in fact, a reliable method for determining whether F(k, m)
= 2^ (2^ ((2^k)*m)) +1 IS prime or not.

This is just a rough draft... but I believe it's a great indicator.

Could someone work a few other modulo sentences to convince me that
it's not just a hoax?

Respectfully,

Bill Bouris

I make the following statement based only on a limited number of
calculations. I was unable to find any web references. Has anyone come
across anything like this?
For any two consecutive positive even integers A and B, there exists at
least one set of three consecutive primes C<D<E such that A equals (D-C)
and B equals (E-D) OR that A equals (E-D) and B equals (D-C).
For example for the 2 consecutive even integers A=2 and B=4, the 3
consecutive primes are C=5, D=7 and E=11.
For the 2 consecutive even integers A=10 and B=12, the 3 consecutive
primes are C=619, D=631 and E=641.
For the 2 consecutive even integers A=94 and B=96, the 3 consecutive
primes are C=327418141, D=327418237 and E=327418331.
I tested all pairs of consecutive even integers (2, 4) to (98, 100). Each
pair had a matching prime triad. As expected, the larger the pair of
consecutive even integers, the longer it takes to find the triad. I have
only limited access to a computer so I stopped checking.
My reason for posting this is to ask whether there is a neat way of
selecting ranges of primes among which one would be most likely to find a
record triad example, assuming the statement is true. Also I thought this
idea of "twin gaps" might interest someone in the group. Thanks folks for
your time.
Bill Sindelar

--- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...>
wrote:
>
> Hi, Group, et. al.
>
> I have a rather simple observation that may imply that the number
of
> Fermat primes is finite.
>
> Conjecture: If a Fermat number of the form F(k,m)= 2^ (2^ ((2^k)
*m))
> +1 is prime, then [2^ (2^ ((2^k)*m)) +1 == 2^q (mod 2^(m-k) +1)]
> where 'k' is /-1 or a whole number/ and 2^k <= 'm' which is
a /whole
> number/ for some /odd number/ 'q'.
>
> Since 'q' is /odd/, this argument further reduces to...
> [((2^k)*m) == (2n+1) mod (m-k)] and is only satisfied when k= -1
and
> m= 0, 2; or k= 0 and m= 1, 3; or k= 1 and m= 2.  For all other 'k's
&
> whole numbers 'm' > 3, the last equation gives the incompatible re-
> sult of 2n == some /odd number/ or...

>>>
((would imply that 'n' > 1 (now, I think my argument is preserved.))
>>>

>
> Just examine the following equations which imply that n= 0 or 1
from
> the reduced modulo equation/argument...
>
> k=-1, m=0;
> [0 == (2n+1) mod 1]=> [-0 == (2n) mod 1]=> [2n= 0]=> [n= 0]; F0=3,
>
> k=0, m=1;
> [1 == (2n+1) mod 1]=> [0 == (2n) mod 1]=> [2n= 0]=> [n= 0]; F1=5,
>
> k=-1, m=2;
> [1 == (2n+1) mod 3]=> [0 == (2n) mod 3]=> [2n= 0]=> [n= 0]; F2=17,
>
> k=0, m=3;
> [3 == (2n+1) mod 3]=> [2 == (2n) mod 3]=> [2n= 2]=> [n= 1]; F3=257,
>
> k=1, m=2;
> [2 == (2n+1) mod 1]=> [1 == (2n) mod 1]=> [2n= 0]=> [n= 0];
F4=65537.
>
> Other values for 'k' and 'm' would make 2n== some /odd number/ or

>>>
((would imply that 'n' > 1 (now, I think my argument is preserved.))
>>>

> Now, if the observation held for only one or two of the Fermat
prime
> numbers, then I would've been hesitant to imply that the modulo ar-
> gument is, in fact, a reliable method for determining whether F(k,
m)
> = 2^ (2^ ((2^k)*m)) +1 IS prime or not.
>
> This is just a rough draft... but I believe it's a great indicator.
>
> Could someone work a few other modulo sentences to convince me that
> it's not just a hoax?
>
> Respectfully,
>
> Bill Bouris
>
It's very difficult to express one's ideas via e-mails; final draft.

I have to say, I have a lot of trouble following any
argument that uses a modulus of 1.  Could you use more
words, more generalized equations, describe more
specifically where your special cases arise, and maybe
outline the overall idea?  The conjecture itself is
not at all clear to me.  I don't understand the /
notation you're using.

Joseph.


--- leavemsg1 <leavemsg1@...> wrote:

> --- In primenumbers@yahoogroups.com, "leavemsg1"
> <leavemsg1@...>
> wrote:
> >
> > Hi, Group, et. al.
> >
> > I have a rather simple observation that may imply
> that the number
> of
> > Fermat primes is finite.
> >
> > Conjecture: If a Fermat number of the form F(k,m)=
> 2^ (2^ ((2^k)
> *m))
> > +1 is prime, then [2^ (2^ ((2^k)*m)) +1 == 2^q
> (mod 2^(m-k) +1)]
> > where 'k' is /-1 or a whole number/ and 2^k <= 'm'
> which is
> a /whole
> > number/ for some /odd number/ 'q'.
> >
> > Since 'q' is /odd/, this argument further reduces
> to...
> > [((2^k)*m) == (2n+1) mod (m-k)] and is only
> satisfied when k= -1
> and
> > m= 0, 2; or k= 0 and m= 1, 3; or k= 1 and m= 2.
> For all other 'k's
> &
> > whole numbers 'm' > 3, the last equation gives the
> incompatible re-
> > sult of 2n == some /odd number/ or...
>
> >>>
> ((would imply that 'n' > 1 (now, I think my argument
> is preserved.))
> >>>
>
> >
> > Just examine the following equations which imply
> that n= 0 or 1
> from
> > the reduced modulo equation/argument...
> >
> > k=-1, m=0;
> > [0 == (2n+1) mod 1]=> [-0 == (2n) mod 1]=> [2n=
> 0]=> [n= 0]; F0=3,
> >
> > k=0, m=1;
> > [1 == (2n+1) mod 1]=> [0 == (2n) mod 1]=> [2n=
> 0]=> [n= 0]; F1=5,
> >
> > k=-1, m=2;
> > [1 == (2n+1) mod 3]=> [0 == (2n) mod 3]=> [2n=
> 0]=> [n= 0]; F2=17,
> >
> > k=0, m=3;
> > [3 == (2n+1) mod 3]=> [2 == (2n) mod 3]=> [2n=
> 2]=> [n= 1]; F3=257,
> >
> > k=1, m=2;
> > [2 == (2n+1) mod 1]=> [1 == (2n) mod 1]=> [2n=
> 0]=> [n= 0];
> F4=65537.
> >
> > Other values for 'k' and 'm' would make 2n== some
> /odd number/ or
>
> >>>
> ((would imply that 'n' > 1 (now, I think my argument
> is preserved.))
> >>>
>
> > Now, if the observation held for only one or two
> of the Fermat
> prime
> > numbers, then I would've been hesitant to imply
> that the modulo ar-
> > gument is, in fact, a reliable method for
> determining whether F(k,
> m)
> > = 2^ (2^ ((2^k)*m)) +1 IS prime or not.
> >
> > This is just a rough draft... but I believe it's a
> great indicator.
> >
> > Could someone work a few other modulo sentences to
> convince me that
> > it's not just a hoax?
> >
> > Respectfully,
> >
> > Bill Bouris
> >
> It's very difficult to express one's ideas via
> e-mails; final draft.
>
>




________________________________________________________________________________\
____
Need a vacation? Get great deals
to amazing places on Yahoo! Travel.
http://travel.yahoo.com/

I have one rather simple explanation for why I
think that the number of Fermat primes is finite.

Conjecture:

If a Fermat number of the form F(k,m)= 2^(2^((2^k)*m)) +1
is prime, then [2^(2^(2^k)*m) +1 == 2^q (mod 2^(2^k*m) +1)]
where 'k' is /-1 or a whole number/ and 'k' <= 'm' which is
a /whole number/ for some /odd number/ 'q'.

This argument further reduces to...[2^(2^k)*m) == (2n +/-1) mod
(2^k*m -k)] and is only satisfied when a combination of 'k' and 'm'
can be found to confirm that n= 0. The modulo equation is nasty,
but the results are wonderful!


Examine the following modulo equations:

expanded...
k=-1, m=0; [(2^((2^(-1))*0)) == (2n+1) mod ((2^(-1))*0) -(-1)]
=>[2^0 == (2n+1) mod (0+1)] => [1 == (2n+1) mod 1]
=> [0 == (2n) mod 1]=> [2n= 0] => [n= 0]; F0= 3 is prime,


continued...
k= 0, m=1; [2 == (2n-1) mod 1] => [0 == (2n) mod 1]
=> [2n= 0] => [n= 0]; F1= 5 is prime,

k= 1, m=1; [4 == (2n+1) mod 1] => [0 == (2n) mod 1]
=> [2n= 0] => [n= 0]; F2= 17 is prime,

k= 0, m=3; [8 == (2n-1) mod 3] => [0 == (2n) mod 3]
=> [2n= 0] => [n= 0]; F3=257 is prime,
k= 1, m=2; [16 == (2n+1) mod 3] => [0 == (2n) mod 3]
=> [2n= 0] => [n= 0]; F4= 65537 is prime.

but...
k= 0, m=5; [32 == (2n-1) mod 5] => [3 == (2n) mod 5]
=> [2n= 3] => [n= fraction]; and F5 is composite!

k= 0, m=6; [64 == (2n+1) mod 6] => [3 == (2n) mod 6] =>
[2n= 3] => [n= fraction]; and k= 1,m=3; [64 == (2n+1) mod 5]
=> [3 == (2n) mod 5] => [2n= 3] => [n= fraction]; & F6 is composite!

k= 0, m=7; [128 == (2n-1) mod 7]=> [3 == (2n) mod 7]
=> [2n= 3] => [n= fraction]; so F7 is composite!

and...
k= 0, m=10;[256 == (2n+1) mod 10]=> [5 == (2n) mod 10
=> [2n= 5]=> [n= fraction]; and k= 2,m=5; [256 == (2n+1) mod 8]
=> [7 == (2n) mod 8]=> [2n= 7] => [n= fraction]; & F10 is composite!
etc.

Hence, the number of Fermat primes is finite because the pattern
ends abruptly on the 5th number in the sequence, and thereafter.

Thanks, Bill

Hello,

the k-tuplets page is down since several days.

In addition Tony Forbes' e-mail adress @ltkz.demon.co.uk is not working.

Does anyone know an actual e-mail adress of Tony or if his k-tuple page
( old adress: http://www.ltkz.demon.co.uk/ktuplets.htm ) moved
somewhere else?

Regards,
Dirk

Dirk Augustin wrote:
> Does anyone know an actual e-mail adress of Tony or if his k-tuple page
> ( old adress: http://www.ltkz.demon.co.uk/ktuplets.htm ) moved
> somewhere else?

See http://mersenneforum.org/showthread.php?t=8832
The page is archived at
http://web.archive.org/web/20070405115024/www.ltkz.demon.co.uk/ktuplets.htm
Tony can be reached at the address at http://www.m500.org.uk/contacts.htm
I also have a private address but I don't know whether he wants it
published.

--
Jens Kruse Andersen

Let p < q be consecutive prime numbers.
Does sum(1/(q-p*ln q), p=2..inf, converge exactly to 6?

10347747270980*3^n+1, n from 1 to 10 all prime

Do these prime runs have a name?(base 2 this would be a Cunningham Chain)

Robert wrote:
> 10347747270980*3^n+1, n from 1 to 10 all prime
>
> Do these prime runs have a name?(base 2 this would be a Cunningham Chain)

http://primes.utm.edu/glossary/page.php?sort=CunninghamChain says:
"Note that some authors extend the definition of Cunningham Chain to
all sequences of primes p_i the form p_(i+1) = a*p_i+b where a and b
are fixed, relatively prime integers with a > 1."

It has been called a generalized Cunningham chain, for example at
http://ourworld.compuserve.com/homepages/hlifchitz/Henri/us/CunnGenus.htm,
but that term has also been used about other variations of
Cunningham chains.
Your chain corresponds to (a, b) = (3, -2):  p_(i+1) = 3*p_i-2.

--
Jens Kruse Andersen

Does it even converge?

If gap size at p is about ln(p) then (with all equalities
being "approximately equal to") q=p+ln(p) so q-p*ln(q)=p+ln(p)-p*ln(p+ln
(p)) is on the order of p*ln(p) and the improper integral of 1/[x*ln
(x)] diverges.

It should it diverge to -oo about as fast as ln(ln(p)).

Adam

>
> Let p < q be consecutive prime numbers.
> Does sum(1/(q-p*ln q), p=2..inf, converge exactly to 6?
>

That's right. One could also say: q ~ p, for p/q ~ 1. Then 1/(q-p*ln
q) ~ 1/(p-p*ln p) and int(1/(x-x*ln x)) =  -ln (ln x - 1) -> -inf.
Thanks. Can you nevertheless find out p when sum(1/(q-p*ln(q)) < 6
for the first time?

Werner



--- In primenumbers@yahoogroups.com, "Adam" <a_math_guy@...> wrote:
>
> Does it even converge?
>
> If gap size at p is about ln(p) then (with all equalities
> being "approximately equal to") q=p+ln(p) so q-p*ln(q)=p+ln(p)-p*ln
(p+ln
> (p)) is on the order of p*ln(p) and the improper integral of 1/[x*ln
> (x)] diverges.
>
> It should it diverge to -oo about as fast as ln(ln(p)).
>
> Adam
>
> >
> > Let p < q be consecutive prime numbers.
> > Does sum(1/(q-p*ln q), p=2..inf, converge exactly to 6?
> >
>

Hi all,

   I just realixzed that the 40th and largest Carol Prime n=253987, was reported
on the prime database on Carol's birthday, May 7, 2007...

   ---Cletus Emmanuel


---------------------------------
Moody friends. Drama queens. Your life? Nope! - their life, your story.
  Play Sims Stories at Yahoo! Games.

[Non-text portions of this message have been removed]

Suppose you are trying to approximate sum over large primes p of a(p)
where a(p) is some monotonically decreasing to 0 function of p.
Instead of writing sum(a(p),p>=P) ~ intgeral(a(x),x=P to infinity), you
should rather approximate it by sum(a(p),p>=P) ~ intgeral(a(x)/log
(x),x=P to infinity), basically because of a frequency count (gap
size).  Say p and q~p+log(p) are consecutive large primes and a(x) is
(roughly) constant on the interval [p,q].  Then integral(a(x),p<=x<=q)
~ a(p)+a(p+1)+ ... +a(q) ~ a(p)*log(p) and is too big to approximate
the summand a(p).  Instead, you should use a(x)/log(x) for the
integrand.  Another way to interpret this is via the dx differential
which should be weighted: a(p) = a(p)*1 ~ a(x) * dx/log(x).

Powers of Two Numbers are

1
21
421
8421
168421
32168421
6432168421
1286432168421
...

The conjecture is that 421 is the only Power of Two Prime.

The other conjecture is that alternate power of two numbers
are divisible by 3.

Milton L. Brown

> The other conjecture is that alternate power of two numbers
> are divisible by 3.

Proof:
    Sum(i=0..N, 10^something(i)*2^i) mod 3 =
    Sum(i=0..N, 2^i) mod 3 =
    2^(N+1)-1 mod 3 =
    0 for odd N, 1 for even N.
Q.E.D.

Peter

--
[Name] Peter Kosinar   [Quote] 2B | ~2B = exp(i*PI)   [ICQ] 134813278

--- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
<jens.k.a@...> wrote:
>
> Robert wrote:
> > 10347747270980*3^n+1, n from 1 to 10 all prime
> >
> > Do these prime runs have a name?(base 2 this would be a Cunningham
Chain)
>
> http://primes.utm.edu/glossary/page.php?sort=CunninghamChain says:
> "Note that some authors extend the definition of Cunningham Chain to
> all sequences of primes p_i the form p_(i+1) = a*p_i+b where a and b
> are fixed, relatively prime integers with a > 1."
>
> It has been called a generalized Cunningham chain, for example at
>
http://ourworld.compuserve.com/homepages/hlifchitz/Henri/us/CunnGenus.htm,

> but that term has also been used about other variations of
> Cunningham chains.
> Your chain corresponds to (a, b) = (3, -2):  p_(i+1) = 3*p_i-2.
>
> --
> Jens Kruse Andersen
>

I can feel another project coming on:

The longest chains k*b^n+/-1 n from n(1) to n(x) all prime, and b= the
primes 2,3,5,7,11,...

It is relatively easy to find chains longer 8 for smaller primes, for
example:
550326588*5^n+1, n from 1 to 10, all prime
943151976*11^n+1, n from 1 to 9, all prime
678979904460*7^n+1, n from 1 to 9 all prime

Regards

Robert Smith

Hi all,

   I am now ready to publish Carol/Kynea numbers.  How do i go about doing that/ 
What journal should I submit to/  Is there a special format?  Any help and
guidance will be greatly appreciated...


   ---Cletus Emmanuel


---------------------------------
Need a vacation? Get great deals to amazing places on Yahoo! Travel.

[Non-text portions of this message have been removed]

No prime p is (p-1)modulo 4, as primes that are(p-1)modulo 2 are
[(p-1)/2]modulo4, and if the order of pmodulo2 is less than p-1, then
the modulo4 of p is either the same as the modulo 2 value or is 1/2 of
the value.

It should be possible therefore to determine values k in the power
series k*4^n+/1, which generate, for all values of n, no factors
smaller than any given prime value through using a modular sieve
process which is much more efficient that choosing the equivalent
primorial or even payam.

As a result, the potential for long Cunningham Chains base 4 becomes
apparent.

So far, the longest I have found (and found after only 3 minutes of
checking!!) is length 13, namely k= 6703351518. A further 5 hours of
checking produced a fair selection of length 10 chains.

Whilst maybe it will be difficult to find one of length 17, it seems
that the trade off between the modular sieve and working in base 4
instead of base 2 might work in favour of the sieve, although I have
no way of proving this.

--- In primenumbers@yahoogroups.com, "Robert" <robert_smith44@...> wrote:
>
> No prime p is (p-1)modulo 4, as primes that are(p-1)modulo 2 are
> [(p-1)/2]modulo4, and if the order of pmodulo2 is less than p-1, then
> the modulo4 of p is either the same as the modulo 2 value or is 1/2 of
> the value.
>
> It should be possible therefore to determine values k in the power
> series k*4^n+/1, which generate, for all values of n, no factors
> smaller than any given prime value through using a modular sieve
> process which is much more efficient that choosing the equivalent
> primorial or even payam.
>
> As a result, the potential for long Cunningham Chains base 4 becomes
> apparent.
>
> So far, the longest I have found (and found after only 3 minutes of
> checking!!) is length 13, namely k= 6703351518. A further 5 hours of
> checking produced a fair selection of length 10 chains.
>
> Whilst maybe it will be difficult to find one of length 17, it seems
> that the trade off between the modular sieve and working in base 4
> instead of base 2 might work in favour of the sieve, although I have
> no way of proving this.
>

Hmm, must have been drunk when I wrote this. For modulo read modulo
order, or multiplicative order.

No base that is square produces a p-1 multiplicative order, 4 is the
smallest square base, it works just as well with bases 9,16,25...

Hello all:

I have obtained this result:

DivisorSigma[2k,m]/DivisorSigma[k,m] is integer
if and only if
m is a perfect square

DivisorSigma[k,m]=sum[d^k, d divisor of m]

Can anyone prove this result?

Sincerely

Sebastian Martin Ruiz

Hello all:

   I have obtained this result:

   DivisorSigma[2k,m]/DivisorSigma[k,m] is integer
   if and only if
   m is a perfect square

   DivisorSigma[k,m]=sum[d^k, d divisor of m]

   Can anyone prove this result?

   Sincerely

   Sebastian Martin Ruiz


---------------------------------

Sé un Mejor Amante del Cine
¿Quieres saber cómo? ¡Deja que otras personas te ayuden!.


[Non-text portions of this message have been removed]

--- In primenumbers@yahoogroups.com, Sebastian Martin <sebi_sebi@...>
wrote:
>
> Hello all:
>
>   I have obtained this result:
>
>   DivisorSigma[2k,m]/DivisorSigma[k,m] is integer
>   if and only if
>   m is a perfect square
>
>   DivisorSigma[k,m]=sum[d^k, d divisor of m]
>
>   Can anyone prove this result?
>
>   Sincerely
>
>   Sebastian Martin Ruiz
>
>
> ---------------------------------
>
> Sé un Mejor Amante del Cine
> ¿Quieres saber cómo? ¡Deja que otras personas te ayuden!.
>
>
> [Non-text portions of this message have been removed]
>

I obtain a lot of small counterexamples with k=1, some with k=2, and
k=3,m=6050 and k=4,m=7203.  E.g., Maple output:

> ifactor(7203);
                                       4
                                (3) (7)
> q1:=sigma[4](7203);
> q2:=sigma[8](7203);
> q2/q1;
                         q1 := 2726235765168410
                 q2 := 7247255655544865674860411008810
                            2658337825414441


All of my counterexamples "appear" to have square factors tho'.....

7203 is NOT a perfect square

   The result is true only if only m is a perfect square: m=n^2

   m= 1, 4, 9, 16, 25, 36, ......

   Adam <a_math_guy@...> escribió:
           --- In primenumbers@yahoogroups.com, Sebastian Martin <sebi_sebi@...>
wrote:
>
> Hello all:
>
> I have obtained this result:
>
> DivisorSigma[2k,m]/DivisorSigma[k,m] is integer
> if and only if
> m is a perfect square
>
> DivisorSigma[k,m]=sum[d^k, d divisor of m]
>
> Can anyone prove this result?
>
> Sincerely
>
> Sebastian Martin Ruiz
>
>
> ---------------------------------
>
> Sé un Mejor Amante del Cine
> ¿Quieres saber cómo? ¡Deja que otras personas te ayuden!.
>
>
> [Non-text portions of this message have been removed]
>

I obtain a lot of small counterexamples with k=1, some with k=2, and
k=3,m=6050 and k=4,m=7203. E.g., Maple output:

> ifactor(7203);
4
(3) (7)
> q1:=sigma[4](7203);
> q2:=sigma[8](7203);
> q2/q1;
q1 := 2726235765168410
q2 := 7247255655544865674860411008810
2658337825414441

All of my counterexamples "appear" to have square factors tho'.....






---------------------------------

¡Descubre una nueva forma de obtener respuestas a tus preguntas!
Entra en Yahoo! Respuestas.


[Non-text portions of this message have been removed]

7203 is not a perfect square!

Maybe you didn't mean "if and only if."  Part of "if and only if" is:

sigma[2k](m)/sigma[k](m) is integer only if m is a square

I demonstrated by example: there exists a non square for which the
fraction is an integer.

--- In primenumbers@yahoogroups.com, Sebastian Martin <sebi_sebi@...>
wrote:
>
> 7203 is NOT a perfect square
>
>   The result is true only if only m is a perfect square: m=n^2
>
>   m= 1, 4, 9, 16, 25, 36, ......
>
>   Adam <a_math_guy@...> escribió:
>           --- In primenumbers@yahoogroups.com, Sebastian Martin
<sebi_sebi@>
> wrote:
> >
> > Hello all:
> >
> > I have obtained this result:
> >
> > DivisorSigma[2k,m]/DivisorSigma[k,m] is integer
> > if and only if
> > m is a perfect square
> >
> > DivisorSigma[k,m]=sum[d^k, d divisor of m]
> >
> > Can anyone prove this result?
> >
> > Sincerely
> >
> > Sebastian Martin Ruiz
> >
> >
> > ---------------------------------
> >
> > Sé un Mejor Amante del Cine
> > ¿Quieres saber cómo? ¡Deja que otras personas te ayuden!.
> >
> >
> > [Non-text portions of this message have been removed]
> >
>
> I obtain a lot of small counterexamples with k=1, some with k=2,
and
> k=3,m=6050 and k=4,m=7203. E.g., Maple output:
>
> > ifactor(7203);
> 4
> (3) (7)
> > q1:=sigma[4](7203);
> > q2:=sigma[8](7203);
> > q2/q1;
> q1 := 2726235765168410
> q2 := 7247255655544865674860411008810
> 2658337825414441
>
> All of my counterexamples "appear" to have square factors tho'.....
>
>
>
>
>
>
> ---------------------------------
>
> ¡Descubre una nueva forma de obtener respuestas a tus preguntas!
> Entra en Yahoo! Respuestas.
>
>
> [Non-text portions of this message have been removed]
>

> DivisorSigma[2k,m]/DivisorSigma[k,m] is integer
> if and only if
> m is a perfect square
>
> DivisorSigma[k,m]=sum[d^k, d divisor of m]

As Adam already disproved the "only if" part, let's concentrate on the
"if" one. First, there are two simple observations to make:
1) If p is a prime,
     DivisorSigma[k, p^e] = 1 + p^k + ... + p^ek = (p^[k(e+1)] - 1)/(p^k-1)
2) If c and d are coprime,
     DivisorSigma[k, c*d] = DivisorSigma[k, c] * DivisorSigma[k, d]

Thus, if m = p1^e2 * p2^e2 ... * pn^en is the prime factorization, we have
DivisorSigma[k, m] =
     (p1^[k(e1+1)] - 1) / (p1^k - 1) *
     (p2^[k(e2+1)] - 1) / (p2^k - 1) *
     ...
     (pn^[k(en+1)] - 1) / (pn^k - 1).
and DivisorSigma[2k, m] =
     (p1^[2k(e1+1)] - 1) / (p1^(2k) - 1) *
     (p2^[2k(e2+1)] - 1) / (p2^(2k) - 1) *
     ...
     (pn^[2k(en+1)] - 1) / (pn^(2k) - 1).

The ratio of these two is ten equal to
DivisorSigma[2k, m] / DivisorSigma[k, m] =
      (p1^[k(e1+1)] + 1) / (p1^k + 1) *
      (p2^[k(e2+1)] + 1) / (p2^k + 1) *
      ...
      (pn^[k(en+1)] + 1) / (pn^k + 1).

Obviously, if all the exponents are even, each fraction of the product
is an integer (the numerator factors algebraically in such case). This
proves the "if" direction of your statement.

The smallest counterexample to the "only if" part seems to be m=20, k=1:
DivisorSigma[20, 2] = 1+4+16+25+100+400 = 546
DivisorSigma[20, 1] = 1+2+4+5+10+20 = 42
546 / 42 = 13

Peter

PS. Now I only have to hope that Yahoogroups do not mangle this mail
      beyond comprehensibility :-)

--
[Name] Peter Kosinar   [Quote] 2B | ~2B = exp(i*PI)   [ICQ] 134813278

Sebastián Martín Ruiz   Said:


1. Divisor Function
     Posted by: "Sebastián Martín Ruiz" sebi_sebi@... sebi_sebi
     Date: Sun Aug 19, 2007 8:31 pm ((PDT))

Hello all:

I have obtained this result:

DivisorSigma[2k,m]/DivisorSigma[k,m] is integer
if and only if
m is a perfect square

DivisorSigma[k,m]=sum[d^k, d divisor of m]

Can anyone prove this result?

Sincerely

Sebastian Martin Ruiz



Kermit Said:




If m is a perfect square then

m = p1^a1  p2^a2 p3^a3 . . .  pj^aj

where each of the a1, a2, .... aj are even.

If d divides m then

d = p1^b1  p2^b2  p3^b3 . . . pj^bj

where each b1,b2,...bj  are less than or equal to the corresponding
a1,a2,...aj.

The
DivisorSigma[k,m]=sum[d^k, d divisor of m]

is  product of

( 1 + p1^k + p1^(2k) + p1^(3k) + . . . + p1^(a1*k) )
(1 + p2^k + p2^(2k) + p2^(3k) + . . .+ p2^(a2*k) )
. . .
(1 + pj^k + pj^(2k) + pj^(3k) + . . . + pj^(aj * k)  )


=

( p1^( a1 * k)  - 1)  / (p1^k - 1)
(p2^(a2 * k)  - 1)/ ( p2^k - 1)
. . .
(pj^(aj*k) - 1) / (pj^k - 1)

When will this product be divisible by

( p1^( a1 * k/2)  - 1)  / (p1^k - 1)
(p2^(a2 * k/2)  - 1)/ ( p2^k - 1)
. . .
(pj^(aj*k/2) - 1) / (pj^k - 1)

?

To see the answer more readily,

define a1 = 2 c1, a2 = 2 c2, ...aj = 2 cj

Then we are asking when is

[ p1^(2 c1 * k) -1 )  ( p2^(2 c2 * k) - 1 ) . . . ( 2 pj ^ ( 2 cj * k )  - 1)  ]
/ [ p1^(c1*k) - 1)     ( p2^(c2 * k) - 1 ) . . . ( 2 pj ^ ( cj * k )  - 1)  ]


= [ p1 ^ (c1 * k) + 1 ) ( p2 ^ (c2 * k) + 1 ) . . .  (pj * ( cj * k) + 1)

an integer?

Main page ( http://www.opertech.com/primes/k-tuples.html )

The latest updates have been posted, with improvements to more than
2200 widths. Currently there are 29 million plus patterns.
( http://www.opertech.com/primes/updates.html )

Of these 2200 improvements about 20 are new first occurence densities.
( http://www.opertech.com/primes/trophycase.html )

The graph is still getting stepper.
( http://www.opertech.com/primes/trophy.bmp ) and
( http://www.opertech.com/primes/varcount.bmp )

--------------------------------------------

As this graph continues to climb, some questions need to be asked and
discussed.

In the following,
k(w) is the maximum number primes permissible in a width of 'w'
integers.
pi(w) is the standard number of primes from 1 to w.

As you look at the graph, the curve 'appears' to be concave up, as
evidenced by the 'hockey stick' start.  The best fit line on the
graph is for tracking purposes and not to imply linearity.
But this line also shows this concave up tendency. Also, the 'flat'
spots at 19000 and 32000 are mainly due to sequence of programming.
The upper bound of the searching was raised from 5200s, to 12000s, to
19000s, to 32000s, and finally to 41741 - meaning the smaller widths
have seen more searching.

Hensley and Richards showed that the two HL conjectures were
incompatible, and left the impression that the k-tuples conjecture
was the true conjecture.

Montgomery and Vaughn showed that k(w) <= 2 * pi(w) using the 'large
sieve'.

If this graph is truly concave up or linear, then the Montgomery and
Vaughn proof would be contradicted, so one of two things must be true,
A) the graph has an inflection point, or
B) the k-tuples conjecture is false.

Concerning A) this inflection point appears highly unlikely as these
quantities can be described as combinatorial objects.

What is the concensus of math world on the HL conjectures?
1: k-tuples true and 2nd HL false
2: k-tuples false and 2nd HL true
or
3: k-tuples false and 2nd HL false

Looking for all the input I can get.

Thomas J Engelsma

--- In primenumbers@yahoogroups.com, "Tom" <tom@...> wrote:


> What is the concensus of math world on the HL conjectures?
> 1: k-tuples true and 2nd HL false
> 2: k-tuples false and 2nd HL true
> or
> 3: k-tuples false and 2nd HL false

This page mentions the incompatibility of the 2 conjectures here
http://en.wikipedia.org/wiki/Second_Hardy-Littlewood_conjecture
and references your website as well as a 1974 article on the
incompatibility of the 2 conjectures.  Seems to indicate a preference
for your #1 above.

I have heard the "k-tuple" conjecture described in different ways.
Are we talking about the k-tuple conjecture in the sense of the first
HL conjecture?  If not, what is the difference?