On Tue, Feb 5, 2013 at 10:42 PM, James J Youlton Jr
<youjaes@...>wrote:

> **
> For lack of better terminology at this point, I’ll define MLP(n)+1 as the
> first n primes multiplied together plus one
>

s/o else already defined this as http://oeis.org/A006862

see the references there for more (terminology & partial answers to all of
your questions).

Maximilian


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

--- On Wed, 2/6/13, Chris Caldwell <caldwell@...> wrote:
> > I found a small mathematical nit to pick in the press release:
> > http://www.mersenne.org/various/57885161.htm
> > > there certainly are larger Mersenne primes
> > The certainty of that proposition remains unproven to
> > the best of my knowledge.
>
> Doesn't it depend on the universe of discourse?  You
> are absolutely correct about "mathematically certainty"
> (e.g., proof).   But if this is "certainty"
> in the sense that if we flip a fair coin a few thousand
> times we will certainly eventually get heads, then I think
> the statement is fine.  Unproven, not even necessarily
> true, but as certain as most things in our lives.
>
> Wouldn't it be grand if there were no more
> Mersennes?   That, and the reason behind it,
> would be a marvelous discovery!   But without
> any such argument, I see another Mersenne as an unproven
> certainty.  <grin>

Absolutely agreed. Because we don't have the mathematical smarts to either prove
the finiteness or infiniteness of the set of Mersenne primes, either would be a
great step forward.

In some ways, I'm sure GIMPS would be equally happy with either proof too. If
it's proven infinite, then they know that they can happily keep crunching with
the same keenness that they demonstrate presently (which is plenty). But if it's
proven that there are no more, then what could be more fulfilling than knowing
that you *did the whole task to completion*? (There is a whole range of
mathematically-interesting discoveries between these two extremes, of course.)

Until then, all we have is heuristics, and I'm quite happy to map an
experimentally-supported heuristic onto the word "certainty". And the huge
experiment is supporting the heuristics very very well.

Phil
--
()  ASCII ribbon campaign      ()    Hopeless ribbon campaign
/\    against HTML mail        /\  against gratuitous bloodshed

[stolen with permission from Daniel B. Cristofani]

Hi, folks.  Just wanted to inform you all of a little thing I have computed:
 
16072758981106442684006718854529251552093=
 
abcdef+(1/a+1/b+1/c+1/d+1/e+1/f)*109#, where
 
a=7*13*37*83, b=17*19*89, c=11*79*101, d=23*29*67*103, e=47*61*107 and
f=41*53*109.
 
Uniquely, it is the last of 30 primes beginning with 7=1+1+1+1+1+1+1 produced by
sequentially multiplying all but one addend by the primes from 2 through 109.
 
The sequential build-up can be obtained from 485191936591420718030369 (just the
smallest of 62 primes that work) by looking at the number's base-7
representation and taking the digits by order of increasing significance as the
addend not to be multiplied by a prime (Digit equal to 0 means the 1st addend is
skipped and the rest multiplied by the prime, etc., with the units digit
corresponding to the prime 2).
 
Jim Merickel
P.S.  Note that with the strict requirement that a term composed of one 1 and
the rest of the addends 2 be prime would make the start of any larger analogous
sequence start with at least 19.  I doubt such a maximal sequence can be found
for so large a start, but I will try to see if I am wrong. 

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

The problem with 19 turned out to be minimal.  There is a unique (up to
permutations) way to do with 19 for the first 56 primes what was done for 7 with
the first 29.  Further, right now I am conjecturing based upon a lot of
empirical data that 19 sets the permanent record, but this is hedged.  If true,
I should be able to prove this (not by hand) over the weekend.  If false, I
might have a surprise on my computer by the time I get home or at least by
Monday.  The solution for 19 involves leaving 3 of the terms as primorials
through the final multiplications, so there is a lot of room for the alternative
problem where this is not acceptable.
JGM

--- On Thu, 2/7/13, James Merickel <moralforce120@...> wrote:


From: James Merickel <moralforce120@...>
Subject: [PrimeNumbers] Sequence of 30 primes (curio submission)
To: primenumbers@yahoogroups.com
Date: Thursday, February 7, 2013, 12:07 PM



 



Hi, folks.  Just wanted to inform you all of a little thing I have computed:
 
16072758981106442684006718854529251552093=
 
abcdef+(1/a+1/b+1/c+1/d+1/e+1/f)*109#, where
 
a=7*13*37*83, b=17*19*89, c=11*79*101, d=23*29*67*103, e=47*61*107 and
f=41*53*109.
 
Uniquely, it is the last of 30 primes beginning with 7=1+1+1+1+1+1+1 produced by
sequentially multiplying all but one addend by the primes from 2 through 109.
 
The sequential build-up can be obtained from 485191936591420718030369 (just the
smallest of 62 primes that work) by looking at the number's base-7
representation and taking the digits by order of increasing significance as the
addend not to be multiplied by a prime (Digit equal to 0 means the 1st addend
is skipped and the rest multiplied by the prime, etc., with the units digit
corresponding to the prime 2).
 
Jim Merickel
P.S.  Note that with the strict requirement that a term composed of one 1 and
the rest of the addends 2 be prime would make the start of any larger analogous
sequence start with at least 19.  I doubt such a maximal sequence can be found
for so large a start, but I will try to see if I am wrong. 

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








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

Dear groupmembers,

The conjecture, which did not make a whole lot of sense anyway and was already
showing itself unsupported empirically before the following result, turns out to
be false.  I had thought that perhaps there would be an argument modulo 19 or
19#.

The next record of 62 primes in succession occurs with 80671 (arrived c. 8AM
today):

With the 19 numbers
a=2*103*157, b=3*173*233, c=5*41*193*239*271, d=7*113*167*257,
e=11*53*107*199*241*263, f=13*29*101, g=17*43*109*181*229, h=19*97*=
281, i=23*59, j=31, k=37*151, l=47*179*277, m=61*73*137, n=67*83*191*211*283,
o=71*79*127*131*163*197*293, p=89, q=139, r=223*269 and s=227*251,

293#*(1/a+1/b+1/c+1/d+1/e+...+1/q+1/r+1/s+80652)=

164835030772135218479491263344471877765987572654819835433403742
893710632923823067093079399895258247669010672441740020208086391

is the largest of a sequence of 62 primes starting with 80671 addends of 1 =
and produced by multiplying each addend save one by the primes in sequence =
from 2 through 293.

A more reasonable conjecture is that requiring the largest prime in such a =
sequence to have no or at most one term in the sum as the primorial of the =
largest prime multiplied has a solution for 19 and for no larger prime.  This I
have not checked (and may not, since it isn't that interesting).

JGM


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

Dear James

Why not apply your mind to some more basic aspects of prime number problems?

For example, if you inspect the natural numbers line, be they odd or even, N
inevitably appears at the centre of two prime numbers.

For example (N) 7 is at the centre of 3 and 11, (N) 19 is at the centre of the
primes 7 and 31, (N) 8 is at the centre of the primes 3 and 13.

If you could tell us why this is so, not only would you explain the gaps between
the primes, you would also prove Goldbach's Conjecture.

Kind regards

Bob


On 9 Feb 2013, at 16:50, James Merickel <moralforce120@...> wrote:

> Dear groupmembers,
>
> The conjecture, which did not make a whole lot of sense anyway and was already
showing itself unsupported empirically before the following result, turns out to
be false.  I had thought that perhaps there would be an argument modulo 19 or
19#.
>
> The next record of 62 primes in succession occurs with 80671 (arrived c. 8AM
today):
>
> With the 19 numbers
> a=2*103*157, b=3*173*233, c=5*41*193*239*271, d=7*113*167*257,
> e=11*53*107*199*241*263, f=13*29*101, g=17*43*109*181*229, h=19*97*=
> 281, i=23*59, j=31, k=37*151, l=47*179*277, m=61*73*137, n=67*83*191*211*283,
o=71*79*127*131*163*197*293, p=89, q=139, r=223*269 and s=227*251,
>
> 293#*(1/a+1/b+1/c+1/d+1/e+...+1/q+1/r+1/s+80652)=
>
> 164835030772135218479491263344471877765987572654819835433403742
> 893710632923823067093079399895258247669010672441740020208086391
>
> is the largest of a sequence of 62 primes starting with 80671 addends of 1 =
> and produced by multiplying each addend save one by the primes in sequence =
> from 2 through 293.
>
> A more reasonable conjecture is that requiring the largest prime in such a =
> sequence to have no or at most one term in the sum as the primorial of the =
> largest prime multiplied has a solution for 19 and for no larger prime.  This
I have not checked (and may not, since it isn't that interesting).
>
> JGM
>
> [Non-text portions of this message have been removed]
>
>


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

Quite an interesting problem in fact, full of features and properties.

I have been studying Goldbach partitions for even numbers of the form 2p where p
is prime, that leads exactly to what Bob points. I have computed thousands of
this partitions forming triplets of primes (pa, p, pb) where pa + pb = 2p, p at
the center. For example, for p = 43 the triplets would be:

(19, 43, 67), (13, 43, 73), (7, 43, 79), (3, 43, 83)

The interesting part comes when the distances d = pb - p = p - pa are studied.
For the primes above these are:

24, 30, 36 and 40

It is possible to demonstrate that all d are multiples of 6 except one at most,
as is the case with 40, which then has to be the difference d between the prime
p and 3, as is the case in the example for 43.

I am preparing a manuscript with these findings that I plan to share in this
group, but I am aware that many can be well known facts and not findings, so I
am trying to verify each one.

For example, I started with triplets, that can be considered arithmetic
sequences of primes. There could be quartets, quintuplets, etc., in each case
also arithmetic sequences. It can be demonstrated that arbitrarily long
sequences of primes may exists but not infinite sequences. I thought this was a
new result but found Green and Tao's article
(http://arxiv.org/abs/math/0404188v6).

Any help would be greatly appreciated.

All the best,

Leonel


________________________________
  From: "bobgillson@..." <bobgillson@...>
To: James Merickel <moralforce120@...>
Cc: "primenumbers@yahoogroups.com" <primenumbers@yahoogroups.com>
Sent: Saturday, February 9, 2013 11:50 AM
Subject: Re: [PrimeNumbers] Conjecture fails with new record at 80671


 

Dear James

Why not apply your mind to some more basic aspects of prime number problems?

For example, if you inspect the natural numbers line, be they odd or even, N
inevitably appears at the centre of two prime numbers.

For example (N) 7 is at the centre of 3 and 11, (N) 19 is at the centre of the
primes 7 and 31, (N) 8 is at the centre of the primes 3 and 13.

If you could tell us why this is so, not only would you explain the gaps between
the primes, you would also prove Goldbach's Conjecture.

Kind regards

Bob

On 9 Feb 2013, at 16:50, James Merickel moralforce120@...> wrote:

> Dear groupmembers,
>
> The conjecture, which did not make a whole lot of sense anyway and was already
showing itself unsupported empirically before the following result, turns out to
be false.  I had thought that perhaps there would be an argument modulo 19 or
19#.
>
> The next record of 62 primes in succession occurs with 80671 (arrived c. 8AM
today):
>
> With the 19 numbers
> a=2*103*157, b=3*173*233, c=5*41*193*239*271, d=7*113*167*257,
> e=11*53*107*199*241*263, f=13*29*101, g=17*43*109*181*229, h=19*97*=
> 281, i=23*59, j=31, k=37*151, l=47*179*277, m=61*73*137, n=67*83*191*211*283,
o=71*79*127*131*163*197*293, p=89, q=139, r=223*269 and s=227*251,
>
> 293#*(1/a+1/b+1/c+1/d+1/e+...+1/q+1/r+1/s+80652)=
>
> 164835030772135218479491263344471877765987572654819835433403742
> 893710632923823067093079399895258247669010672441740020208086391
>
> is the largest of a sequence of 62 primes starting with 80671 addends of 1 =
> and produced by multiplying each addend save one by the primes in sequence =
> from 2 through 293.
>
> A more reasonable conjecture is that requiring the largest prime in such a =
> sequence to have no or at most one term in the sum as the primorial of the =
> largest prime multiplied has a solution for 19 and for no larger prime.  This
I have not checked (and may not, since it isn't that interesting).
>
> JGM
>
> [Non-text portions of this message have been removed]
>
>

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




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

Leonel

I think what you are seeing is a simple property
of all prime numbers, excepting 3.

If we ignore the special prime 3, all odd primes take
the form 6n+1 or 6n-1 (since 6n+-0, 6n+-2, 6n+-3
and 6n+-4 are obviously composite).  Note that the
form 6n+-5 can be written as 6n-+1):

Case 1:
-------
If your central prime p has the form 6p+1, then
2p has the form 12n+2 or 6m+2.
The only combination of prime forms that add up
to 6m+2 are 6a+1 plus 6b+1 for some {a,b}.
Thus your triple is {6a+1, 6n+1, 6b+1}.
Then your distance:
   d = pb-p = p-pa    is
     = 6b+1-6n-1 = 6n+1-6a-1    or
     = 6(b-n) = 6(n-a) always a multiple of 6

Case 2.
-------
If your central prime p has the form 6p-1,
then 2p has the form 12n-2 or 6m-2.
The only combination of prime forms that add
up to 6m-2 are 6a-1 plus 6b-1 for some {a,b}.
Then your triple is {6a-1, 6n-1, 6b-1}.
Then your distance:
   d = pb-p = p-pa    is
     = 6b-1-6n+1 = 6n-1-6a+1    or
     = 6(b-n) = 6(n-a) again always a multiple of 6

Thus you are correct in that all d are multiples of 6,
with the exception of when 3 is involved.

Regards

Alan Powell


From: Leonel Morales
Sent: Sunday, February 10, 2013 1:25 PM
To: primenumbers@yahoogroups.com
Subject: Re: [PrimeNumbers] Conjecture fails with new record at 80671

Quite an interesting problem in fact, full of features and properties.

I have been studying Goldbach partitions for even numbers of the form 2p where p
is prime, that leads exactly to what Bob points. I have computed thousands of
this partitions forming triplets of primes (pa, p, pb) where pa + pb = 2p, p at
the center. For example, for p = 43 the triplets would be:

(19, 43, 67), (13, 43, 73), (7, 43, 79), (3, 43, 83)

The interesting part comes when the distances d = pb - p = p - pa are studied.
For the primes above these are:

24, 30, 36 and 40

It is possible to demonstrate that all d are multiples of 6 except one at most,
as is the case with 40, which then has to be the difference d between the prime
p and 3, as is the case in the example for 43.

I am preparing a manuscript with these findings that I plan to share in this
group, but I am aware that many can be well known facts and not findings, so I
am trying to verify each one.

For example, I started with triplets, that can be considered arithmetic
sequences of primes. There could be quartets, quintuplets, etc., in each case
also arithmetic sequences. It can be demonstrated that arbitrarily long
sequences of primes may exists but not infinite sequences. I thought this was a
new result but found Green and Tao's article
(http://arxiv.org/abs/math/0404188v6).

Any help would be greatly appreciated.

All the best,

Leonel

________________________________
From: "mailto:bobgillson%40yahoo.com" mailto:bobgillson%40yahoo.com>
To: James Merickel mailto:moralforce120%40yahoo.com>
Cc: "mailto:primenumbers%40yahoogroups.com"
mailto:primenumbers%40yahoogroups.com>
Sent: Saturday, February 9, 2013 11:50 AM
Subject: Re: [PrimeNumbers] Conjecture fails with new record at 80671




Dear James

Why not apply your mind to some more basic aspects of prime number problems?

For example, if you inspect the natural numbers line, be they odd or even, N
inevitably appears at the centre of two prime numbers.

For example (N) 7 is at the centre of 3 and 11, (N) 19 is at the centre of the
primes 7 and 31, (N) 8 is at the centre of the primes 3 and 13.

If you could tell us why this is so, not only would you explain the gaps between
the primes, you would also prove Goldbach's Conjecture.

Kind regards

Bob

On 9 Feb 2013, at 16:50, James Merickel mailto:moralforce120%40yahoo.com> wrote:

> Dear groupmembers,
>
> The conjecture, which did not make a whole lot of sense anyway and was already
showing itself unsupported empirically before the following result, turns out to
be false. I had thought that perhaps there would be an argument modulo 19 or
19#.
>
> The next record of 62 primes in succession occurs with 80671 (arrived c. 8AM
today):
>
> With the 19 numbers
> a=2*103*157, b=3*173*233, c=5*41*193*239*271, d=7*113*167*257,
> e=11*53*107*199*241*263, f=13*29*101, g=17*43*109*181*229, h=19*97*=
> 281, i=23*59, j=31, k=37*151, l=47*179*277, m=61*73*137, n=67*83*191*211*283,
o=71*79*127*131*163*197*293, p=89, q=139, r=223*269 and s=227*251,
>
> 293#*(1/a+1/b+1/c+1/d+1/e+...+1/q+1/r+1/s+80652)=
>
> 164835030772135218479491263344471877765987572654819835433403742
> 893710632923823067093079399895258247669010672441740020208086391
>
> is the largest of a sequence of 62 primes starting with 80671 addends of 1 =
> and produced by multiplying each addend save one by the primes in sequence =
> from 2 through 293.
>
> A more reasonable conjecture is that requiring the largest prime in such a =
> sequence to have no or at most one term in the sum as the primorial of the =
> largest prime multiplied has a solution for 19 and for no larger prime. This I
have not checked (and may not, since it isn't that interesting).
>
> JGM
>
> [Non-text portions of this message have been removed]
>
>

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

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





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

Thanks a lot Alan! I really appreciate you explanation.

And then when 3 is involved d = 6k +- 2. I have calculated thousands of those
and haven't noticed that.

I have looked at the sequences of values of d: for a single prime p all the
possible values of d, and for the set of primes, all the minimum values of d
(http://oeis.org/A078611), all the maximum values, and several others that seem
to be not listed in OEIS. All d for triplets involving 3
is http://oeis.org/A206037.


Because d = p - pa = pb -p then pa*pb = p^2 - d^2 which leads to think of primes
in a triplet as a right triangle:

pa*pb + d^2 = p^2

It can be demonstrated that no pair of such triangles is similar.

I am also trying to look at primes as a network where two primes are connected
if they are consecutive in a triplet. I am using Gephi to analyze such network
and it consistently finds two communities in the network. I am still trying to
figure what are the implications.

Thanks a lot for your answer,

Leonel


________________________________
  From: Alan Powell <AlanPowell@...>
To: primenumbers@yahoogroups.com
Sent: Sunday, February 10, 2013 4:49 PM
Subject: Re: [PrimeNumbers] Conjecture fails with new record at 80671


 
Leonel

I think what you are seeing is a simple property
of all prime numbers, excepting 3.

If we ignore the special prime 3, all odd primes take
the form 6n+1 or 6n-1 (since 6n+-0, 6n+-2, 6n+-3
and 6n+-4 are obviously composite).  Note that the
form 6n+-5 can be written as 6n-+1):

Case 1:
-------
If your central prime p has the form 6p+1, then
2p has the form 12n+2 or 6m+2.
The only combination of prime forms that add up
to 6m+2 are 6a+1 plus 6b+1 for some {a,b}.
Thus your triple is {6a+1, 6n+1, 6b+1}.
Then your distance:
d = pb-p = p-pa    is
= 6b+1-6n-1 = 6n+1-6a-1    or
= 6(b-n) = 6(n-a) always a multiple of 6

Case 2.
-------
If your central prime p has the form 6p-1,
then 2p has the form 12n-2 or 6m-2.
The only combination of prime forms that add
up to 6m-2 are 6a-1 plus 6b-1 for some {a,b}.
Then your triple is {6a-1, 6n-1, 6b-1}.
Then your distance:
d = pb-p = p-pa    is
= 6b-1-6n+1 = 6n-1-6a+1    or
= 6(b-n) = 6(n-a) again always a multiple of 6

Thus you are correct in that all d are multiples of 6,
with the exception of when 3 is involved.

Regards

Alan Powell

From: Leonel Morales
Sent: Sunday, February 10, 2013 1:25 PM
To: primenumbers@yahoogroups.com
Subject: Re: [PrimeNumbers] Conjecture fails with new record at 80671

Quite an interesting problem in fact, full of features and properties.

I have been studying Goldbach partitions for even numbers of the form 2p where p
is prime, that leads exactly to what Bob points. I have computed thousands of
this partitions forming triplets of primes (pa, p, pb) where pa + pb = 2p, p at
the center. For example, for p = 43 the triplets would be:

(19, 43, 67), (13, 43, 73), (7, 43, 79), (3, 43, 83)

The interesting part comes when the distances d = pb - p = p - pa are studied.
For the primes above these are:

24, 30, 36 and 40

It is possible to demonstrate that all d are multiples of 6 except one at most,
as is the case with 40, which then has to be the difference d between the prime
p and 3, as is the case in the example for 43.

I am preparing a manuscript with these findings that I plan to share in this
group, but I am aware that many can be well known facts and not findings, so I
am trying to verify each one.

For example, I started with triplets, that can be considered arithmetic
sequences of primes. There could be quartets, quintuplets, etc., in each case
also arithmetic sequences. It can be demonstrated that arbitrarily long
sequences of primes may exists but not infinite sequences. I thought this was a
new result but found Green and Tao's article
(http://arxiv.org/abs/math/0404188v6).

Any help would be greatly appreciated.

All the best,

Leonel

________________________________
From: "mailto:bobgillson%40yahoo.com" mailto:bobgillson%40yahoo.com>
To: James Merickel mailto:moralforce120%40yahoo.com>
Cc: "mailto:primenumbers%40yahoogroups.com"
mailto:primenumbers%40yahoogroups.com>
Sent: Saturday, February 9, 2013 11:50 AM
Subject: Re: [PrimeNumbers] Conjecture fails with new record at 80671

Dear James

Why not apply your mind to some more basic aspects of prime number problems?

For example, if you inspect the natural numbers line, be they odd or even, N
inevitably appears at the centre of two prime numbers.

For example (N) 7 is at the centre of 3 and 11, (N) 19 is at the centre of the
primes 7 and 31, (N) 8 is at the centre of the primes 3 and 13.

If you could tell us why this is so, not only would you explain the gaps between
the primes, you would also prove Goldbach's Conjecture.

Kind regards

Bob

On 9 Feb 2013, at 16:50, James Merickel mailto:moralforce120%40yahoo.com> wrote:

> Dear groupmembers,
>
> The conjecture, which did not make a whole lot of sense anyway and was already
showing itself unsupported empirically before the following result, turns out to
be false. I had thought that perhaps there would be an argument modulo 19 or
19#.
>
> The next record of 62 primes in succession occurs with 80671 (arrived c. 8AM
today):
>
> With the 19 numbers
> a=2*103*157, b=3*173*233, c=5*41*193*239*271, d=7*113*167*257,
> e=11*53*107*199*241*263, f=13*29*101, g=17*43*109*181*229, h=19*97*=
> 281, i=23*59, j=31, k=37*151, l=47*179*277, m=61*73*137, n=67*83*191*211*283,
o=71*79*127*131*163*197*293, p=89, q=139, r=223*269 and s=227*251,
>
> 293#*(1/a+1/b+1/c+1/d+1/e+...+1/q+1/r+1/s+80652)=
>
> 164835030772135218479491263344471877765987572654819835433403742
> 893710632923823067093079399895258247669010672441740020208086391
>
> is the largest of a sequence of 62 primes starting with 80671 addends of 1 =
> and produced by multiplying each addend save one by the primes in sequence =
> from 2 through 293.
>
> A more reasonable conjecture is that requiring the largest prime in such a =
> sequence to have no or at most one term in the sum as the primorial of the =
> largest prime multiplied has a solution for 19 and for no larger prime. This I
have not checked (and may not, since it isn't that interesting).
>
> JGM
>
> [Non-text portions of this message have been removed]
>
>

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

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

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




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

The set {11,13,17,19, 41,43,47, 71,73,79, 101,103,107,109} shows two sets of
quadruplets 90 apart, and the intervening "+30" and "+60" decades have triplets
(in the decadal sense that I use the term, not necessarily ones with a minimal
span of 6).

Does anyone have a list of the next few occurrences of this pattern?

Sorry, I had initially posted as a forward on my own recent post and then erased
more than I had meant to on a bounce caused by using 'yahoo.com' rather than
'yahoogroups.com'.  This refers to my post preceding.  Not sure if 'out of
context' is a really sound way to consider it, though, being so soon after the
relevant post.  Apologies in any case, because it was slightly off.   

Below, edited to add the accidentally omitted prime 149 and change '62' to '63'.
 
Note: I have had to increase the stack and start over to find the next record,
if I might.
 
JGM
 
--- On Sun, 2/10/13, Maximilian Hasler <maximilian.hasler@...> wrote:


From: Maximilian Hasler <maximilian.hasler@...>
Subject: Re: [PrimeNumbers] Conjecture fails with new record at 80671
To: "James Merickel" <moralforce120@...>
Date: Sunday, February 10, 2013, 12:04 PM


I refrain from asking "which conjecture",
but your post is out of context.


Maximilian



On Sat, Feb 9, 2013 at 12:50 PM, James Merickel <moralforce120@...> wrote:


 



Dear groupmembers,

The conjecture, which did not make a whole lot of sense anyway and was already
showing itself unsupported empirically before the following result, turns out to
be false.  I had thought that perhaps there would be an argument modulo 19 or
19#.

The next record of 63 primes in succession occurs with 80671 (arrived c. 8AM
today):

With the 19 numbers
a=2*103*157, b=3*173*233, c=5*41*193*239*271, d=7*113*167*257,
e=11*53*107*199*241*263, f=13*29*101, g=17*43*109*149*181*229, h=19*97*281,
i=23*59, j=31, k=37*151, l=47*179*277, m=61*73*137, n=67*83*191*211*283,
o=71*79*127*131*163*197*293, p=89, q=139, r=223*269 and s=227*251,

293#*(1/a+1/b+1/c+1/d+1/e+...+1/q+1/r+1/s+80652)=

164835030772135218479491263344471877765987572654819835433403742
893710632923823067093079399895258247669010672441740020208086391

is the largest of a sequence of 63 primes starting with 80671 addends of 1 =
and produced by multiplying each addend save one by the primes in sequence =
from 2 through 293.

A more reasonable conjecture is that requiring the largest prime in such a =
sequence to have no or at most one term in the sum as the primorial of the =
largest prime multiplied has a solution for 19 and for no larger prime.  This I
have not checked (and may not, since it isn't that interesting).

JGM

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

woodhodgson wrote:
> The set {11,13,17,19, 41,43,47, 71,73,79, 101,103,107,109}
> shows two sets of quadruplets 90 apart, and the intervening
> "+30" and "+60" decades have triplets (in the decadal sense
> that I use the term, not necessarily ones with a minimal
> span of 6).
>
> Does anyone have a list of the next few occurrences of this pattern?

There are only 2 admissible patterns, the above and its mirror
{11,13,17,19, 41,47,49, 73,77,79, 101,103,107,109}

A search found 10 occurrences in total below 10^17.
2 of them are the mirror pattern.
The first prime and the number of other primes in the interval:

11, 11 other primes
549758002658141, 2 other primes
1444747726722731, 1 other prime
4869691549793501, 2 other primes
7973040075706331 (mirror pattern), 1 other prime
21603285535472981, 0 other primes
21859392938284241, 1 other prime
23490659029317911, 0 other primes
28423532235584111 (mirror pattern), 1 other prime
94859808174585731, 0 other primes


The first case of 4 prime quadruplets as closely together as admissible is in
http://tech.groups.yahoo.com/group/primenumbers/message/18318:
300000224101777931 + n,
for n in {0,2,6,8, 90,92,96,98, 180,182,186,188, 210,212,216,218}

--
Jens Kruse Andersen

Thank you Jens, also to Maximilian for his replies. It certainly confirms a long
gap to the next occurrences. No doubt these should occur infinitely often
according to general conjectures.

--- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"  wrote:
>
> woodhodgson wrote:
> > The set {11,13,17,19, 41,43,47, 71,73,79, 101,103,107,109}
> > shows two sets of quadruplets 90 apart, and the intervening
> > "+30" and "+60" decades have triplets (in the decadal sense
> > that I use the term, not necessarily ones with a minimal
> > span of 6).
> >
> > Does anyone have a list of the next few occurrences of this pattern?
>
> There are only 2 admissible patterns, the above and its mirror
> {11,13,17,19, 41,47,49, 73,77,79, 101,103,107,109}
>
> A search found 10 occurrences in total below 10^17.
> 2 of them are the mirror pattern.
> The first prime and the number of other primes in the interval:
>
> 11, 11 other primes
> 549758002658141, 2 other primes
> 1444747726722731, 1 other prime
> 4869691549793501, 2 other primes
> 7973040075706331 (mirror pattern), 1 other prime
> 21603285535472981, 0 other primes
> 21859392938284241, 1 other prime
> 23490659029317911, 0 other primes
> 28423532235584111 (mirror pattern), 1 other prime
> 94859808174585731, 0 other primes
>
>
> The first case of 4 prime quadruplets as closely together as admissible is in
> http://tech.groups.yahoo.com/group/primenumbers/message/18318:
> 300000224101777931 + n,
> for n in {0,2,6,8, 90,92,96,98, 180,182,186,188, 210,212,216,218}
>
> --
> Jens Kruse Andersen
>

hi

C_twin=pi^2/12*prod(p>=5 odd primes) (1-2/(p*(p-1)))=0.66016..

You will need tens of thousands of terms to get several decimal places, but the
appearance of pi =3.14159... makes you think it's irrational but the rest of the
construct being related to Artin's primitive root conjecture means that GRH is
being appealed to, doesn't it? So this is currently unprovable.

Still, it's fascinating that what's normally considered related to the sequence
of twin primes is connected to the Artin set as well.

Maybe someone could construct a geometrically convergent version of its
logarithm, with the Lucas series and prime zeta function. I tried and failed.
The product above converges too slowly!

(Must point out this isn't my idea, I found it in a paper on Artin conjectures
and modified it slightly - p22 of this http://arxiv.org/pdf/math/0412262v2.pdf)

Oh, and there are twin primes among the Artin primes, and the sum of their
reciprocals

Sorry to repeat this, but in relationship to the question of what conjecture I
was talking about recently the issue might require calarification.  
It--underlined--is wrong (with the example recently given).
JGM 

--- On Fri, 2/8/13, James Merickel <moralforce120@...> wrote:


From: James Merickel <moralforce120@...>
Subject: Fw: [PrimeNumbers] Sequence of 30 primes (curio submission)--57 PRIMES
To: primenumbers@yahoogroups.com
Date: Friday, February 8, 2013, 11:19 AM







The problem with 19 turned out to be minimal.  There is a unique (up to
permutations) way to do with 19 for the first 56 primes what was done for 7 with
the first 29.  Further, right now I am conjecturing based upon a lot of
empirical data that 19 sets the permanent record, but this is hedged.  If true,
I should be able to prove this (not by hand) over the weekend.  If false, I
might have a surprise on my computer by the time I get home or at least by
Monday.  The solution for 19 involves leaving 3 of the terms as primorials
through the final multiplications, so there is a lot of room for the alternative
problem where this is not acceptable.
JGM

--- On Thu, 2/7/13, James Merickel <moralforce120@...> wrote:


From: James Merickel <moralforce120@...>
Subject: [PrimeNumbers] Sequence of 30 primes (curio submission)
To: primenumbers@yahoogroups.com
Date: Thursday, February 7, 2013, 12:07 PM



 



Hi, folks.  Just wanted to inform you all of a little thing I have computed:
 
16072758981106442684006718854529251552093=
 
abcdef+(1/a+1/b+1/c+1/d+1/e+1/f)*109#, where
 
a=7*13*37*83, b=17*19*89, c=11*79*101, d=23*29*67*103, e=47*61*107 and
f=41*53*109.
 
Uniquely, it is the last of 30 primes beginning with 7=1+1+1+1+1+1+1 produced by
sequentially multiplying all but one addend by the primes from 2 through 109.
 
The sequential build-up can be obtained from 485191936591420718030369 (just the
smallest of 62 primes that work) by looking at the number's base-7
representation and taking the digits by order of increasing significance as the
addend not to be multiplied by a prime (Digit equal to 0 means the 1st addend
is skipped and the rest multiplied by the prime, etc., with the units digit
corresponding to the prime 2).
 
Jim Merickel
P.S.  Note that with the strict requirement that a term composed of one 1 and
the rest of the addends 2 be prime would make the start of any larger analogous
sequence start with at least 19.  I doubt such a maximal sequence can be found
for so large a start, but I will try to see if I am wrong. 

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








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

I expected the twin prime constant to be irrational
because I expected that any constant
that requires EVERY prime in order to calculate it,

would necessarily be irrational.

Kermit Rose

Surely the rationality of irrationality depends on the truth of otherwise of the
twin prime conjecture.

On 13 Feb 2013, at 17:15, Kermit Rose <kermit@...> wrote:

> I expected the twin prime constant to be irrational
> because I expected that any constant
> that requires EVERY prime in order to calculate it,
>
> would necessarily be irrational.
>
> Kermit Rose
>
>


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

What is the product over all of the primes p of:

     (p^2+1)/(p^2-1)  ?

That's a constant that requires EVERY prime in order
to calculate it.

It turns out to be 5/2.  Which is not irrational.



On 2/13/2013 9:15 AM, Kermit Rose wrote:
> I expected the twin prime constant to be irrational
> because I expected that any constant
> that requires EVERY prime in order to calculate it,
>
> would necessarily be irrational.
>
> Kermit Rose
>
>
>
>
> ------------------------------------
>
> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://primes.utm.edu/
>
> Yahoo! Groups Links
>
>
>
>
>

--- In primenumbers@yahoogroups.com, Jack Brennen  wrote:
>
> What is the product over all of the primes p of:
>
>     (p^2+1)/(p^2-1)  ?
>
> That's a constant that requires EVERY prime in order
> to calculate it.
>
> It turns out to be 5/2.  Which is not irrational.

Nice point, Jack.

  print(zeta(2)^2/zeta(4));
2.5000000000000000000000000000000000000

David

On 2/14/2013 7:03 PM, primenumbers@yahoogroups.com wrote:
> 1a. Re: Is the twin prime constant irrational?
>      Posted by: "djbroadhurst"d.broadhurst@...  djbroadhurst
>      Date: Wed Feb 13, 2013 5:14 pm ((PST))
>
>
>
> --- Inprimenumbers@yahoogroups.com, Jack Brennen  wrote:
>> >
>> >What is the product over all of the primes p of:
>> >
>> >     (p^2+1)/(p^2-1)  ?
>> >
>> >That's a constant that requires EVERY prime in order
>> >to calculate it.
>> >
>> >It turns out to be 5/2.  Which is not irrational.
> Nice point, Jack.
>
>   print(zeta(2)^2/zeta(4));
> 2.5000000000000000000000000000000000000
>
> David


Thank you David.

I see how (zeta(2))^2/zeta(4) = (p^2+1)/(p^2-1),

but how do we know what zeta(2) is, and how do we know what zeta(4) is?


zeta(2) = sum(k positive integer)(1/k^2)
= product(p positive prime, J non-negative integer)(sum(1/p^(2J))

= product(p positive prime)(1/(1-1/p^2))

= product(p positive prime)(p^2/(p^2-1))



zeta(4) = sum(k positive integer)(1/k^4)
= product(p positive prime, J non-negative integer)(sum(1/p^(4J))

= product(p positive prime)(1/(1-1/p^4))

= product(p positive prime)(p^4/(p^4-1))



(zeta(2))^2/zeta(4)
= product(p positive prime)(((p^2/(p^2-1)))^2/(p^4/(p^4-1)))

= product(p positive prime)((p^4/(p^2-1)^2)/(p^4/(p^4-1)))

= product(p positive prime)((p^4-1)/(p^2-1)^2))

= product(p positive prime)((p^2+1)/(p^2-1))



Kermit








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

Re: Is the twin prime constant irrational?



Twin prime constant
= (3/2)(1/2)(5/4)(3/4)(7/6)(5/6)(11/10)(9/10)...(p/(p-1))((p-2)/(p-1))...


I expected that it would have been easily determined whether or not
the twin prime constant was rational or irrational.

It would not be possible for the twin prime constant to be rational
because the infinite numerator is odd, and the infinite denominator is
divisible by
2 infinitely many times.

Kermit

How about this infinite product here?

    (99/10)*(111/110)*(1111/1110)*(11111/11110)*...

The partial products are:
    9.9
    9.99
    9.999
    9.9999
    and so on...

The product quite obviously converges to an even number (10), but all of
the numerators are odd and all of the denominators are even.  Even and
odd really have no meaning when it comes to infinity and limits.  As
this example shows, a series of partial products, all of which have
odd numerator and even denominator, can converge to not only a rational
number, but an even integer.


On 2/15/2013 8:53 AM, Kermit Rose wrote:
> Re: Is the twin prime constant irrational?
>
>
>
> Twin prime constant
> = (3/2)(1/2)(5/4)(3/4)(7/6)(5/6)(11/10)(9/10)...(p/(p-1))((p-2)/(p-1))...
>
>
> I expected that it would have been easily determined whether or not
> the twin prime constant was rational or irrational.
>
> It would not be possible for the twin prime constant to be rational
> because the infinite numerator is odd, and the infinite denominator is
> divisible by
> 2 infinitely many times.
>
> Kermit
>
>
>
>
>
>
> ------------------------------------
>
> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://primes.utm.edu/
>
> Yahoo! Groups Links
>
>
>
>
>

--- In primenumbers@yahoogroups.com,
Kermit Rose <kermit@...> wrote:

> how do we know what zeta(2) is

http://arxiv.org/pdf/1004.4238.pdf Problem 3

> and how do we know what zeta(4) is?

Section 2.3 of same essay

David

Absolutely fascinating David, including your CV!

Thank you for sharing this paper with us.

Bob

On 15 Feb 2013, at 22:52, "djbroadhurst" <d.broadhurst@...> wrote:

> --- In primenumbers@yahoogroups.com,
> Kermit Rose wrote:
>
> > how do we know what zeta(2) is
>
> http://arxiv.org/pdf/1004.4238.pdf Problem 3
>
> > and how do we know what zeta(4) is?
>
> Section 2.3 of same essay
>
> David
>
>


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

--- In primenumbers@yahoogroups.com, bobgillson@... wrote:

> Absolutely fascinating David

Jack's ratio is the subject of a notable erratum:

http://www.sciencedirect.com/science/article/pii/037026939500269Q

which links to a .pdf file acknowledging a "pitiable" mistake:

> In our original computer program for the evaluation of the
> three-loop QCD correction to the rho parameter a pitiable mistake
> was overlooked. The wrong value 2/5 was substituted for
> zeta(2)^2/zeta(4) = 5/2.

David

Since you asked...
If F(n) is the nth Fermat prime, then

sum(n odd) 1/F(n)
and
sum(n even) 1/F(n)
and
product(n odd) (1-1/F(n))
and
product(n even) (1-1/F(n))
each are irrational... IFF there are an infinite number of Fermat primes
of the kind in that expression.

One must admit, though, that this is a pretty silly result, for
several reasons.  I guess it is best posed as a puzzle, then :)

Here is a new prime 18-tuplet record (and a new 18 Simultaneous Primes record):

2650778861583720495199114537 + d,
d = 0, 4, 10, 12, 16, 22, 24, 30, 36, 40, 42, 46, 52, 54, 60, 64, 66, 70
(28 digits, Feb 2013, Raanan Chermoni & Jaroslaw Wroblewski)

Jarek

Wojciech Izykowski has discovered AP19 with the smallest possible start of 19:

19 + 13234551541698967679 * 17# * n, n=0,...,18
(27 digits)

At the moment it is not known whether the difference of the AP is the
smallest possible, but Wojciech is working to determine that.

Although Wojciech was using the ideas I have developed in the past, he
created brand new implementation of the search program on his own.

Jarek

--- In primenumbers@yahoogroups.com, Jaroslaw Wroblewski
<jaroslaw.wroblewski@...> wrote:
>
> Wojciech Izykowski has discovered AP19 with the smallest possible start of 19:
>
> 19 + 13234551541698967679 * 17# * n, n=0,...,18
> (27 digits)
>

An amazing and highly nontrivial discovery - many congratulations!

But don't rush to look for an AP23:
heuristics in my NMBRTHRY post of 2001
https://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind0106&L=NMBRTHRY&F=&S=&P=3812
indicate that a complete search would take many times the age of the Universe.

Mike

Jaroslaw Wroblewski wrote:
> Wojciech Izykowski has discovered AP19 with the smallest possible start of
> 19:
>
> 19 + 13234551541698967679 * 17# * n, n=0,...,18
> (27 digits)
>
> At the moment it is not known whether the difference of the AP is the
> smallest possible, but Wojciech is working to determine that.

Congratulations to Wojciech!

http://users.cybercity.dk/~dsl522332/math/aprecords.htm#minimalstart
is updated with a mixed color entry to show that it has the smallest
possible start but the difference may not be the smallest.

--
Jens Kruse Andersen