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Re: Prime Number Progresions
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 Mark,
The progression of 29 primes I found starts with 31.
31 add 12 = 43
43 add 24 = 67
67 add 36 = 103
103 add 48 = 151
151 add 60 = 211 
This continues for 29 primes. The last prime is 4903.
Thank you for doing the research. I'd like to know if it is the same one.
Virginia
[Nontext portions of this message have been removed]  Hi Virginia
Yours is of the form 6x^2 + 6x + 31, generating primes from x=0 to
x=28.
Interestingly, Gary's (2x^2 88x + 997) generates 29 distinct primes
as well but is a different equation than yours.
I just figured that Gary's equation can be reduced to 2x^2 + 29 to
generate his 29 distinct primes from x=0 to x=28 !
Mark
 In primenumbers@yahoogroups.com, ginnyw@a... wrote:
> Mark,
>
> The progression of 29 primes I found starts with 31.
> 31 add 12 = 43
> 43 add 24 = 67
> 67 add 36 = 103
> 103 add 48 = 151
> 151 add 60 = 211 
>
> This continues for 29 primes. The last prime is 4903.
>
> Thank you for doing the research. I'd like to know if it is the
same one.
>
> Virginia
>
>
>
> [Nontext portions of this message have been removed]  Its nice to know that i'm not the only one looking at
these sequences.
I think and i will need to check I have an improvement
on the sequence below. Generating 60 primes (but like
the sequence below it doubles up and only has 30
distinct primes). I will look this up at home then
post it.
I have been looking at various polynomials and have
notices something quite unusual.
For quadratic polynomials runs of 20+ distinct primes
are quite common. But for cubic polynomials I have yet
to find a run>20 distinct primes. I am aware that the
cubic increases faster therefore each value is less
likely to be prime but I'm wondering if there is
anything else affecting the polynomial.
Eulers record of x^2+x+41 was beaten by Fung and Ruby
with 36x^2810x+2753
See:
http://mathworld.wolfram.com/PrimeGeneratingPolynomial.html
I am certain that a polynomial of the form ax^2+bx+c
can be found that beats this record but it means
searching a hell of a lot of polnomials!!!
Gary
 Mark Underwood <mark.underwood@...>
wrote: > Hi Virginia>
________________________________________________________________________
> Your 29 consecutive prime sequence is very good. I
> wonder what
> equation it is expressed by?
>
> Just in the last week Gary Chaffey reported that the
> equation
> 2*x^288*x+997 generates 51 primes from x=0 to x=
> 50.
>
> Then Dr. Michael Hartley noticed that the same
> equation can be
> slightly modified to produce 57 primes from x=0 to
> x=56 :
>
> 2*t^2112*t+1597
>
> Amazing! I'm not sure if there are any which are
> longer. I had
> thought that the one Euler discovered was the
> longest and was proven
> to be the longest of it's kind, but I guess not?
> Perhaps with Gary's
> sequence the primes are not all distinct, I'm not
> sure.
>
> Mark
>
>
>
>  In primenumbers@yahoogroups.com, ginnyw@a...
> wrote:
> > Mark,
> >
> > Thank you for your email. The progressions are
> the ordinary
> arithmetic type.
> > One of the progressions of 11 primes is:
> > 17 add 44 = 61
> > 61 add 88 = 149
> > 149 add 132 = 281
> > 281 add 176 = 457 and continuing for 6 more terms
> until a nonprime
> 2921 is
> > reached.
> >
> > My program uses a constant to search for the
> progressions. I
> started with
> > pencil and paper and then wrote the program.
> Except for the
> progression which
> > generates 40 primes, the largest progression I
> have to date
> generates 29
> > primes. Perhaps that might be a record. Most of
> the progressions
> are with small
> > numbers because of the limitations of my computer.
> >
> > Virginia W.
> >
> >
> >
> > [Nontext portions of this message have been
> removed]
>
>
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Messenger http://uk.messenger.yahoo.com/   ginnyw@... wrote: > Mark,
>
No it isn't the same sequence this one can be
> The progression of 29 primes I found starts with 31.
> 31 add 12 = 43
> 43 add 24 = 67
> 67 add 36 = 103
> 103 add 48 = 151
> 151 add 60 = 211 
>
> This continues for 29 primes. The last prime is
> 4903.
>
> Thank you for doing the research. I'd like to know
> if it is the same one.
expressed as 6x^2+6x+31 for x in 0..28
but certainly of equal merit.
Gary
________________________________________________________________________
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Messenger http://uk.messenger.yahoo.com/  Mark, Gary,
The equations are very helpful. Thank you. It is Interesting that the two
progressions start at 29 and 31 and generate 29 primes each. This seems to
happen often. Another progression starting with 11 generates 10 primes; another
starting with 17 generates 16 primes, etc. I am interested in hearing more
about your work and plan to change my program based on our discussion.
Virginia
[Nontext portions of this message have been removed] > I just figured that Gary's equation can be reduced
I have just spotted this too.. make y=x22...(is this
> to 2x^2 + 29 to
> generate his 29 distinct primes from x=0 to x=28 !
>
> Mark
the transformation you have spotted Mark???)
Gary
________________________________________________________________________
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Messenger http://uk.messenger.yahoo.com/ Right Gary, I guess that would be the transformation. Actually I just
checked out the progression in the sequence starting with 29, 31,
37 ... and it was easy to see it was of this form. I now see from
http://mathworld.wolfram.com/PrimeGeneratingPolynomial.html
that Legendre is the first reported to have seen this one. Perhaps
2x^2 + 29 generates the longest sequence of consecutive primes of any
two term equation.
And I see that Virginia's sequence of primes is reported in the
Encyclopedia on Integer Sequences as sequence A060834.
The Mathworld link above has alot of informative things to say about
the matter. (Virginia would like to read this!) I agree with you Gary
that there are other polynomials out there that can generate even
longer sequences. How to cleverly find them, that is the question.
But for expressions of the form x^2 + x + p there is no need to look
for a longer one since it has been shown that p = 41 generates the
longest one.
Mark
 In primenumbers@yahoogroups.com, Gary Chaffey <garychaffey@y...>
wrote:> > I just figured that Gary's equation can be reduced
______________________________________________________________________
> > to 2x^2 + 29 to
> > generate his 29 distinct primes from x=0 to x=28 !
> >
> > Mark
> I have just spotted this too.. make y=x22...(is this
> the transformation you have spotted Mark???)
> Gary
>
>
__> Want to chat instantly with your online friends? Get the FREE
Yahoo!
> Messenger http://uk.messenger.yahoo.com/
  Mark Underwood wrote:
> Perhaps 2x^2 + 29 generates the longest sequence of consecutive
If you believe the first HardyLittlewood Conjecture (also known
> primes of any two term equation.
as the ktuple Conjecture), there exist arbitrarily long sequences
of primes from two term equations.
> But for expressions of the form x^2 + x + p there is no need to
Again, the same conjecture implies that arbitrarily long sequences
> look for a longer one since it has been shown that p = 41
> generates the longest one.
of primes exist of the form x^2+x+p.
What has been shown is that p=41 is the largest prime such that
x^2+x+p is prime for all x, 0 <= x <= p2.
It has not been shown that x^2+x+p is never prime for 0 <= x <= 40. > > Perhaps 2x^2 + 29 generates the longest sequence
If we restrict to two term equations then it can't be
> of consecutive
> > primes of any two term equation.
>
> If you believe the first HardyLittlewood Conjecture
> (also known
> as the ktuple Conjecture), there exist arbitrarily
> long sequences
> of primes from two term equations.
of the form 2x^2+p since of a proof (referred to on
Wolfram site) which shows that this type of sequence
can only yield 29 primes.
This implies we must look at ax^2+p for a>2. What
would be interesting is to rule out some more values
for a. I might look at this if I get some spare time.
Gary
________________________________________________________________________
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Messenger http://uk.messenger.yahoo.com/ Thanks for the correction Jack, I would not have discerned that
difference from what I read unless it was pointed out.
I remember the HardyLittlewood k tuple Conjecture but I never did
connect it to polynominals.
Whether the k tuple conjucture is true, I guess I believe it with a
condition, which is that the tuple does not occur if it does not
occur early on, or something like that...
Mark
 In primenumbers@yahoogroups.com, "jbrennen" <jack@b...> wrote:
>  Mark Underwood wrote:
>
> > Perhaps 2x^2 + 29 generates the longest sequence of consecutive
> > primes of any two term equation.
>
> If you believe the first HardyLittlewood Conjecture (also known
> as the ktuple Conjecture), there exist arbitrarily long sequences
> of primes from two term equations.
>
> > But for expressions of the form x^2 + x + p there is no need to
> > look for a longer one since it has been shown that p = 41
> > generates the longest one.
>
> Again, the same conjecture implies that arbitrarily long sequences
> of primes exist of the form x^2+x+p.
>
> What has been shown is that p=41 is the largest prime such that
> x^2+x+p is prime for all x, 0 <= x <= p2.
>
> It has not been shown that x^2+x+p is never prime for 0 <= x <= 40. > Whether the k tuple conjucture is true, I guess I believe it with
a
> condition, which is that the tuple does not occur if it does not
Why would it have to occur early?
> occur early on, or something like that...
>
> Mark
>
>
 Hi Shane
Well I figure that if the tuple doesn't occur early, it is for some
reason. It would fail at higher numbers for the same reason, or a
generalization of that reason. If there is a counterexample I will
gladly eat my hat! I'll make sure I'm wearing chocolate hats from
here on in...
Mark
 In primenumbers@yahoogroups.com, "Shane" <TTcreation@a...> wrote:
> > Whether the k tuple conjucture is true, I guess I believe it
with
> a
> > condition, which is that the tuple does not occur if it does not
> > occur early on, or something like that...
> >
> > Mark
> >
> >
>
>
> Why would it have to occur early? > >
sequences
> > > But for expressions of the form x^2 + x + p there is no need to
> > > look for a longer one since it has been shown that p = 41
> > > generates the longest one.
> >
> > Again, the same conjecture implies that arbitrarily long
> > of primes exist of the form x^2+x+p.
40.
> >
> > What has been shown is that p=41 is the largest prime such that
> > x^2+x+p is prime for all x, 0 <= x <= p2.
> >
> > It has not been shown that x^2+x+p is never prime for 0 <= x <=
Surely all one has to do is find a c in the equation x^2+x+c for
which the following conditions apply:
2+c not divisible by 2, 3, 5
and, 2+c meets all of:
1,5,6 mod 7
3,6,8,9,10 mod 11
1,3,4,5,6,7,10 mod 13
1,3,4,5,8,9,10,12 mod 17
4,5,8,11,12,13,14,16,18 mod 19
1,3,9,10,11,12,14,16,17,20,21 mod 23
3,5,6,7,9,10,12,13,14,16,21,22,26,27 mod 29
4,7,11,12,14,15,17,18,19,20,24,26,28,29,30 mod 31
1,6,7,8,10,11,12,13,15,16,17,22,24,25,28,32,35,36 mod 37
3,4,5,6,7,9,11,14,16,18,19,20,21,22,26,27,30,36,39,40 mod 41
1,5,6,8,10,11,14,17,19,22,23,24,26,27,28,29,30,34,36,37,38 mod 43
c=41 is the first number to reach all of the conditions except the
last, being 1mod7, 10mod11, 4mod13....but 0mod43
Regards
Robert Smith
PS I may have gotten some of the register above incorrect, but
someone will spot an error if I have made one. Thats what I like
about you Primenumbers group.
PPS 2+c = x^2+x+c with x=1
PPPS this is the same logic as used in the determination of Payam
numbers  In primenumbers@yahoogroups.com, "Robert" <100620.2351@c...>
wrote:>
to
> > >
> > > > But for expressions of the form x^2 + x + p there is no need
> > > > look for a longer one since it has been shown that p = 41
<=
> > > > generates the longest one.
> > >
> > > Again, the same conjecture implies that arbitrarily long
> sequences
> > > of primes exist of the form x^2+x+p.
> > >
> > > What has been shown is that p=41 is the largest prime such that
> > > x^2+x+p is prime for all x, 0 <= x <= p2.
> > >
> > > It has not been shown that x^2+x+p is never prime for 0 <= x
> 40.
#
>
> Surely all one has to do is find a c in the equation x^2+x+c for
> which the following conditions apply:
>
> 2+c not divisible by 2, 3, 5
> and, 2+c meets all of:
>
> 1,5,6 mod 7
> 3,6,8,9,10 mod 11
> 1,3,4,5,6,7,10 mod 13
> 1,3,4,5,8,9,10,12 mod 17
> 4,5,8,11,12,13,14,16,18 mod 19
> 1,3,9,10,11,12,14,16,17,20,21 mod 23
> 3,5,6,7,9,10,12,13,14,16,21,22,26,27 mod 29
> 4,7,11,12,14,15,17,18,19,20,24,26,28,29,30 mod 31
> 1,6,7,8,10,11,12,13,15,16,17,22,24,25,28,32,35,36 mod 37
> 3,4,5,6,7,9,11,14,16,18,19,20,21,22,26,27,30,36,39,40 mod 41
> 1,5,6,8,10,11,14,17,19,22,23,24,26,27,28,29,30,34,36,37,38 mod 43
>
> c=41 is the first number to reach all of the conditions except the
> last, being 1mod7, 10mod11, 4mod13....but 0mod43
>
> Regards
>
> Robert Smith
>
> PS I may have gotten some of the register above incorrect, but
> someone will spot an error if I have made one. Thats what I like
> about you Primenumbers group.
>
> PPS 2+c = x^2+x+c with x=1
>
> PPPS this is the same logic as used in the determination of Payam
> numbers
Oops, I forgot to mention that the value of c, which contributes to
a run of >41 primes, must also clear similar mod n hurdles up to
sqrt c   In primenumbers@yahoogroups.com, "Robert" <100620.2351@c...>
wrote:>  In primenumbers@yahoogroups.com, "Robert" <100620.2351@c...>
need
> wrote:
> >
> > > >
> > > > > But for expressions of the form x^2 + x + p there is no
> to
that
> > > > > look for a longer one since it has been shown that p = 41
> > > > > generates the longest one.
> > > >
> > > > Again, the same conjecture implies that arbitrarily long
> > sequences
> > > > of primes exist of the form x^2+x+p.
> > > >
> > > > What has been shown is that p=41 is the largest prime such
> > > > x^2+x+p is prime for all x, 0 <= x <= p2.
the
> > > >
> > > > It has not been shown that x^2+x+p is never prime for 0 <= x
> <=
> > 40.
> >
> > Surely all one has to do is find a c in the equation x^2+x+c for
> > which the following conditions apply:
> >
> > 2+c not divisible by 2, 3, 5
> > and, 2+c meets all of:
> >
> > 1,5,6 mod 7
> > 3,6,8,9,10 mod 11
> > 1,3,4,5,6,7,10 mod 13
> > 1,3,4,5,8,9,10,12 mod 17
> > 4,5,8,11,12,13,14,16,18 mod 19
> > 1,3,9,10,11,12,14,16,17,20,21 mod 23
> > 3,5,6,7,9,10,12,13,14,16,21,22,26,27 mod 29
> > 4,7,11,12,14,15,17,18,19,20,24,26,28,29,30 mod 31
> > 1,6,7,8,10,11,12,13,15,16,17,22,24,25,28,32,35,36 mod 37
> > 3,4,5,6,7,9,11,14,16,18,19,20,21,22,26,27,30,36,39,40 mod 41
> > 1,5,6,8,10,11,14,17,19,22,23,24,26,27,28,29,30,34,36,37,38 mod 43
> >
> > c=41 is the first number to reach all of the conditions except
> > last, being 1mod7, 10mod11, 4mod13....but 0mod43
Payam
> >
> > Regards
> >
> > Robert Smith
> >
> > PS I may have gotten some of the register above incorrect, but
> > someone will spot an error if I have made one. Thats what I like
> > about you Primenumbers group.
> >
> > PPS 2+c = x^2+x+c with x=1
> >
> > PPPS this is the same logic as used in the determination of
> > numbers
to
> #
> Oops, I forgot to mention that the value of c, which contributes
> a run of >41 primes, must also clear similar mod n hurdles up to
#### Now I have had a glass of wine, (Merlot) I see that what I have
> sqrt c
pointed to above relates to (contributes to the thinking behind) the
statement
> > > > What has been shown is that p=41 is the largest prime such
that
> > > > x^2+x+p is prime for all x, 0 <= x <= p2.
#### OK who proved that one? It seems intuitively nonprovable, or
> > > >
only provable by a lot of computer processing. But what do I know?
I think I prefer the statement:
> > > > Again, the same conjecture implies that arbitrarily long
That is something the list of hurdles might contribute to.
> > sequences
> > > > of primes exist of the form x^2+x+p.
#### but the list of hurdles, etc does not contribute to the last
statement
> > > > It has not been shown that x^2+x+p is never prime for 0 <= x
Sorry for the confusion
> <=
> > 40.
Regards
Robert Smith   Mark Underwood <mark.underwood@...>
wrote: >> Thanks for the correction Jack, I would not have
Ditto. I had not noticed this either.
> discerned that
> difference from what I read unless it was pointed
> out.
I am somewhat sceptical that a polynomial of the form
2x^2+p will be found (soon) that yields primes for all
x in [0..29].
I have looked at p upto 3.10^7 and as of yet nothing
gets anywhere close. (best so far x in [0..9]).
I am however going to look a bit deeper.
Gary
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Messenger http://uk.messenger.yahoo.com/  Gary,
> I am somewhat sceptical that a polynomial of the form
I have just sieved up to 10^14 with nothing being found. The best was
> 2x^2+p will be found (soon) that yields primes for all
> x in [0..29].
> I have looked at p up to 3.10^7 and as of yet nothing
> gets anywhere close. (best so far x in [0..9]).
> I am however going to look a bit deeper.
45077834116589, which generated 18 primes for x in [0..29]. The runtime was
395 seconds on this 450 MHz PIII.
Regards,
Paul.
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