Hi,

congratulations to Michael Herder and Prime Grid for finding the useful
bi-megaprime 3*2^7033641+1 which has ‎2117338 digits and is ranked 14th
largest prime to date:

http://primes.utm.edu/bios/page.php?id=2298

Paul

Hi,

congratulations to David Broadhurst for discovering the two AP3s based on
Primegrid's prime numbers, in their search for a twin and Sophie Germain pairs.
The exponent 666,666 has been very fruitful,

http://primes.utm.edu/primes/page.php?id=98974
http://primes.utm.edu/primes/page.php?id=98956

There are currently 1551 prime numbers in the Prime Pages database with the
double beastly exponent,

Paul

Dear primenumbers group members,

If anybody might be interested in pursuing the question of whether there might
be a message (an actual message) where I currently think it's most likely to be
found, the homeprime sequence of the repeating portion of the reciprocal of 147
looks pretty good for coincidences.  This reciprocal is the probability of a
whole number being divisible by 42 but not by 4, 9 or 49.

Obviously, the idea of a message here is apt to be seen as ridiculous by most,
but it's not a theoretically unsound notion.  It would entail some manipulation
of our linguistic history obviously, as these numbers are not manipulable.  If I
were looking for something it would be from where and when, first of all, but
then I don't know, and I would not have a clue right now about picking signal
out of noise.

JGM

Paul wrote:
> congratulations to David Broadhurst for discovering the two AP3s
> based on Primegrid's prime numbers, in their search for a twin and
> Sophie Germain pairs. The exponent 666,666 has been very fruitful,
>
> http://primes.utm.edu/primes/page.php?id=98974
> http://primes.utm.edu/primes/page.php?id=98956

Congratulations on these and a third:
http://primes.utm.edu/primes/page.php?id=98982

http://users.cybercity.dk/~dsl522332/math/aprecords.htm is updated.

--
Jens Kruse Andersen

http://perfectcypher.com/


vjl2 f55u euta haoz g62i hdyo crti as67 nz1y h42o kid3 x1ti xa7y sy7u pie3 biu3
s9e5 lh56
xa11 qii9 ta53 zm64 vj84 ck26 bd52 bz7t via8 uo19 r43e weo7 z62l sv74 cg9y ya93
c4oo
gs16 j82p id64 rba3 oy3a xvii qy27 p99z ic4y gsi3 ie11 l329 pua7 r1ra lav5 wnc5
yto7 t4b2
vr84 ny39 qso7 pbve fa47 aezh oeoi huat p69e f2e6 nv95 sk2u mgj6 gcyj iiw7 bdoj
v2c6
9631 w18u oj8i vnhx urn7 yx62 ngms u1zs b2n4 xubw ipy1 s4uo l834 fy81 v827 j85o
neam

--- In primenumbers@yahoogroups.com, "Jens Kruse Andersen" <jens.k.a@...> wrote:
>
> Paul wrote:
> > congratulations to David Broadhurst for discovering the two AP3s
> > based on Primegrid's prime numbers, in their search for a twin and
> > Sophie Germain pairs. The exponent 666,666 has been very fruitful,
> >
> > http://primes.utm.edu/primes/page.php?id=98974
> > http://primes.utm.edu/primes/page.php?id=98956
>
> Congratulations on these and a third:
> http://primes.utm.edu/primes/page.php?id=98982
>
> http://users.cybercity.dk/~dsl522332/math/aprecords.htm is updated.
>
> --

And now a fourth:

http://primes.utm.edu/primes/page.php?id=98991

Paul

--- In primenumbers@yahoogroups.com,
"paulunderwooduk" <paulunderwood@...> wrote:

> And now a fourth:
> http://primes.utm.edu/primes/page.php?id=98991

Mike Oakes visited me and we attended a conference on
general relativity and cosmology. It seems that this
coincided with the discovery of a local region of high
curvature, in probability space. There are clearly a good
few AP3s that can be found by taking two elements from the
PrimeGrid SG/Twin survey, but finding 4 in the space
of a couple of days was unexpectedly non-Euclidean.

It seems to me that PrimeGrid is already slightly unlucky
not to have found an SG or Twin pair by now?

David

--- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
>
> > And now a fourth:
> > http://primes.utm.edu/primes/page.php?id=98991
>
> Mike Oakes visited me and we attended a conference on
> general relativity and cosmology. It seems that this
> coincided with the discovery of a local region of high
> curvature, in probability space. There are clearly a good
> few AP3s that can be found by taking two elements from the
> PrimeGrid SG/Twin survey, but finding 4 in the space
> of a couple of days was unexpectedly non-Euclidean.
>
> It seems to me that PrimeGrid is already slightly unlucky
> not to have found an SG or Twin pair by now?

No longer able to congratulate you over the cornflakes, but anyway: many
congrats on another hit.

My slight concern is: how long before all the rest of us are knocked off both
tables of this page:
http://primes.utm.edu/top20/page.php?id=14
leaving only who=p199?

Mike

--- In primenumbers@yahoogroups.com,
"mikeoakes2" <mikeoakes2@...> wrote:

> My slight concern is: how long before all the rest of us are
> knocked off both tables of this page:
> http://primes.utm.edu/top20/page.php?id=14

I think that PrimeMogul protects you, by allowing
no more than 5 AP3s as "listable" in those tables.

David

--- In primenumbers@yahoogroups.com, "mikeoakes2" <mikeoakes2@...> wrote:
>
> --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@> wrote:
> >
> > > And now a fourth:
> > > http://primes.utm.edu/primes/page.php?id=98991
> >
> > Mike Oakes visited me and we attended a conference on
> > general relativity and cosmology. It seems that this
> > coincided with the discovery of a local region of high
> > curvature, in probability space. There are clearly a good
> > few AP3s that can be found by taking two elements from the
> > PrimeGrid SG/Twin survey, but finding 4 in the space
> > of a couple of days was unexpectedly non-Euclidean.
> >
> > It seems to me that PrimeGrid is already slightly unlucky
> > not to have found an SG or Twin pair by now?
>
> No longer able to congratulate you over the cornflakes, but anyway: many
congrats on another hit.
>
> My slight concern is: how long before all the rest of us are knocked off both
tables of this page:
> http://primes.utm.edu/top20/page.php?id=14
> leaving only who=p199?
>

Take heart: Ken and I and the e-group are trying to top the second table. I am
about 18 of 35 GHz years of PRP'ing, scheduled to finish by July, with hopefully
a few months extra real time of detection. Ken is concurrently searching for an
titanic AP9 now. In the eventuality, the puny numbers found by us will outdo
David and the mighty PrimeGrid.

The lowest ranked doubly beastly exponent is ranked 4481; PrimeGrid can find
another 519 of these before they start to eat their own tail.

Good luck to PrimeGrid with SGs/twins and also to David finding more AP3s,

Paul

--- In primenumbers@yahoogroups.com,
"paulunderwooduk" <paulunderwood@...> wrote:

> Ken and I and the e-group are trying to top the
> second table. I am about 18 of 35 GHz years of PRP'ing, scheduled
> to finish by July, with hopefully a few months extra real time of
> detection. Ken is concurrently searching for an titanic AP9 now.

I shall issue a retraction by King Lear
when that notable feat is accomplished,
as now seems more or less assured.

Best wishes

David

--- On Wed, 3/2/11, cipher <websitequestions@...> wrote:

With an email address like that, I knew as I approved your membership requst
that you might be the source of less useful posts. Seems I was right.

> http://perfectcypher.com/

"""
# "The system must not require secrecy and can be stolen by the enemy without
causing trouble."

Messages written with this cipher can safely be posted on every billboard in an
enemies country without them knowing where the breaks are.
let alone them ever being able to decipher it even if they knew where the breaks
were.
"""

That's a complete and flagrant misinterpretation of the rule. That rule means
that the security of the system remains uncompromised even if the algorithm for
encryption and decryption is known to the enemy. Or in simpler terms "all the
security is in the key".

Certainty in that assertion is usually accompanied by the /ab initio/ publishing
of the algorithm(s). The challenge in your website fails to so  so, and I would
guess is almost certainly quite the opposite of a "perfect" cipher.

Please do not pretend to teach cryptography until you've actually learnt some.

Phil

Hello all:

Let n and k positive integers k<n.

Let P(i) the ith-prime number

We have:

P(n)-P(n-k)-(n-k)P(k)=0   IF AND ONLY IF    P(n) and P(k) are Twin Primes


Sincerely

Sebastián Martín Ruiz




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

{ for(n=1,9,Pn=prime(n);for(k=1,n-1,Pk=prime(k);Pnk=prime(n-k);
(Pn-Pnk==(n-k)*Pk) == (istwin(Pn)&istwin(Pk)) |
print("Counter-example: (n,k)=",[n,k]",
P(n)-P(n-k)-(n-k)*P(k)=",Pn"-"Pnk"-",n-k,"*"Pk))) }

Counter-example: (n,k)=[4, 2], P(n)-P(n-k)-(n-k)*P(k)=7-3-2*3
Counter-example: (n,k)=[5, 2], P(n)-P(n-k)-(n-k)*P(k)=11-5-3*3
Counter-example: (n,k)=[5, 3], P(n)-P(n-k)-(n-k)*P(k)=11-3-2*5
Counter-example: (n,k)=[5, 4], P(n)-P(n-k)-(n-k)*P(k)=11-2-1*7
Counter-example: (n,k)=[6, 2], P(n)-P(n-k)-(n-k)*P(k)=13-7-4*3
Counter-example: (n,k)=[6, 3], P(n)-P(n-k)-(n-k)*P(k)=13-5-3*5
Counter-example: (n,k)=[6, 4], P(n)-P(n-k)-(n-k)*P(k)=13-3-2*7
Counter-example: (n,k)=[7, 2], P(n)-P(n-k)-(n-k)*P(k)=17-11-5*3
Counter-example: (n,k)=[7, 3], P(n)-P(n-k)-(n-k)*P(k)=17-7-4*5
Counter-example: (n,k)=[7, 4], P(n)-P(n-k)-(n-k)*P(k)=17-5-3*7
Counter-example: (n,k)=[7, 5], P(n)-P(n-k)-(n-k)*P(k)=17-3-2*11
Counter-example: (n,k)=[7, 6], P(n)-P(n-k)-(n-k)*P(k)=17-2-1*13
Counter-example: (n,k)=[8, 2], P(n)-P(n-k)-(n-k)*P(k)=19-13-6*3
Counter-example: (n,k)=[8, 3], P(n)-P(n-k)-(n-k)*P(k)=19-11-5*5
Counter-example: (n,k)=[8, 4], P(n)-P(n-k)-(n-k)*P(k)=19-7-4*7
Counter-example: (n,k)=[8, 5], P(n)-P(n-k)-(n-k)*P(k)=19-5-3*11
Counter-example: (n,k)=[8, 6], P(n)-P(n-k)-(n-k)*P(k)=19-3-2*13



On Sat, Mar 5, 2011 at 6:13 AM, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
> Hello all:
>
> Let n and k positive integers k<n.
>
> Let P(i) the ith-prime number
>
> We have:
>
> P(n)-P(n-k)-(n-k)P(k)=0   IF AND ONLY IF    P(n) and P(k) are Twin Primes
>
>
> Sincerely
>
> Sebastián Martín Ruiz
>
>
>
>
> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>
> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://www.primepages.org/
>
> Yahoo! Groups Links
>
>
>
>

> { for(n=1,9,Pn=prime(n);for(k=1,n-1,Pk=prime(k);Pnk=prime(n-k);
> (Pn-Pnk==(n-k)*Pk) == (istwin(Pn)&istwin(Pk)) |
> print("Counter-example: (n,k)=",[n,k]",
> P(n)-P(n-k)-(n-k)*P(k)=",Pn"-"Pnk"-",n-k,"*"Pk))) }
>
> [lots of counterexamples]

> > P(n)-P(n-k)-(n-k)P(k)=0   IF AND ONLY IF    P(n) and P(k) are Twin
Primes

I believe the Sebastian meant "... IFF P(n) and P(k) form a Twin pair",
i.e. P(n) - P(k) = 2. The "IF" direction is trivial, of course.

Peter

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

On 3/5/2011 8:04 AM, "Sebastian Martin Ruiz" s_m_ruiz@... s_m_ruiz
wrote:
> ________________________________________________________________________
> 1. Equivalence for twin primes
>      Posted by: "Sebastian Martin Ruiz" s_m_ruiz@... s_m_ruiz
>      Date: Sat Mar 5, 2011 2:13 am ((PST))
>
> Hello all:
>
> Let n and k positive integers k<n.
>
> Let P(i) the ith-prime number
>
> We have:
>
> P(n)-P(n-k)-(n-k)P(k)=0   IF AND ONLY IF    P(n) and P(k) are Twin Primes
>
>
> Sincerely
>
> Sebastián Martín Ruiz
>
>
>


It is trivial that

if p(n) , p(n-k), and p(k) are distinct primes,
and n>k,
then
P(n)-p(n-k) -(n-k)P(k)=0

if and only if p(n-k) = 2, and n-k = 1.




In the case where p(n) and p(k) belong to the same set of twin primes,
we would have

example:
p(n) = 7,  p(k) = 5

n = 4, k = 3

p(n-k) = p(1) = 2

P(n)-2 -(n-k)P(k)=0

7 - 2 - 1*5 = 5 - 5 = 0

In general , if p(n) and p(k) belong to the same set of twin primes,
and n > k,

it is trivial that

P(n)-P(n-k)-(n-k)P(k)
= p(n) - p( n - [n-1]) - ([n-(n-1)] )p(n-1)
= p(n) - p(1) - 1 * (p(n-1))
= p(n) - 2 - p(n-1)
= (p(n) - p(n-1)) - 2
= 2 - 2 = 0


If
P(n)-P(n-k)-(n-k)P(k) = 0
p(n) - p(n-k) = (n-k) p(k)

n = 2
what are permitted values of k?
k = 1?
3 - p(1) = (2-1)*p(1) ?
3 - 2 = 1 * 2   ?
1 = 2 ?
no

n = 3
p(3) - p(3-k) - (n-k) p(k) = 0
p(3) - p(3-k) = (n-k) p(k)
5 - p(3-k) = (n-k)

5 = p(3-k) + (3-k)
2 = p(3-k) - k

2 + k = p(3-k)

k is odd



In the case where p(n) and p(k) belong to different
sets of twin primes,

example:

p(n) = 13
p(k) = 7

n = 6
k = 4
n-k = 2

P(n)-P(n-k)-(n-k)P(k)
=13 - 3 - 2*7
= -4
is not zero.

 P(n)-P(n-k)-(n-k)P(k)=0   IIF    P(n) and P(k) are a Twin Primes pair

IF P(n) and P(k) are a Twin Primes pair

Then k=n-1 P(n-k)=P(1)=2  P(n)-2-(n-(n-1))P(n-1)=0

Then P(n)-P(n-1)=2

On the other hand If

P(n)-P(n-k)-(n-k)P(k)=0  then

 P(n) and P(k) are a Twin Primes pair

 is more dificult but i think it is also true.


P(n)-P(n-k)-(n-k)*P(k)=7-3-2*3=/=0
Counter-example: (n,k)=[4, 2], P(n)-P(n-k)-(n-k)*P(k)=7-3-2*3
then your contraexample is not true.





 



________________________________
De: Maximilian Hasler <maximilian.hasler@...>
Para: Sebastian Martin Ruiz <s_m_ruiz@...>
CC: primenumbers@yahoogroups.com
Enviado: sáb,5 marzo, 2011 14:12
Asunto: Re: [PrimeNumbers] Equivalence for twin primes

 
{ for(n=1,9,Pn=prime(n);for(k=1,n-1,Pk=prime(k);Pnk=prime(n-k);
(Pn-Pnk==(n-k)*Pk) == (istwin(Pn)&istwin(Pk)) |
print("Counter-example: (n,k)=",[n,k]",
P(n)-P(n-k)-(n-k)*P(k)=",Pn"-"Pnk"-",n-k,"*"Pk))) }

Counter-example: (n,k)=[4, 2], P(n)-P(n-k)-(n-k)*P(k)=7-3-2*3
Counter-example: (n,k)=[5, 2], P(n)-P(n-k)-(n-k)*P(k)=11-5-3*3
Counter-example: (n,k)=[5, 3], P(n)-P(n-k)-(n-k)*P(k)=11-3-2*5
Counter-example: (n,k)=[5, 4], P(n)-P(n-k)-(n-k)*P(k)=11-2-1*7
Counter-example: (n,k)=[6, 2], P(n)-P(n-k)-(n-k)*P(k)=13-7-4*3
Counter-example: (n,k)=[6, 3], P(n)-P(n-k)-(n-k)*P(k)=13-5-3*5
Counter-example: (n,k)=[6, 4], P(n)-P(n-k)-(n-k)*P(k)=13-3-2*7
Counter-example: (n,k)=[7, 2], P(n)-P(n-k)-(n-k)*P(k)=17-11-5*3
Counter-example: (n,k)=[7, 3], P(n)-P(n-k)-(n-k)*P(k)=17-7-4*5
Counter-example: (n,k)=[7, 4], P(n)-P(n-k)-(n-k)*P(k)=17-5-3*7
Counter-example: (n,k)=[7, 5], P(n)-P(n-k)-(n-k)*P(k)=17-3-2*11
Counter-example: (n,k)=[7, 6], P(n)-P(n-k)-(n-k)*P(k)=17-2-1*13
Counter-example: (n,k)=[8, 2], P(n)-P(n-k)-(n-k)*P(k)=19-13-6*3
Counter-example: (n,k)=[8, 3], P(n)-P(n-k)-(n-k)*P(k)=19-11-5*5
Counter-example: (n,k)=[8, 4], P(n)-P(n-k)-(n-k)*P(k)=19-7-4*7
Counter-example: (n,k)=[8, 5], P(n)-P(n-k)-(n-k)*P(k)=19-5-3*11
Counter-example: (n,k)=[8, 6], P(n)-P(n-k)-(n-k)*P(k)=19-3-2*13

On Sat, Mar 5, 2011 at 6:13 AM, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
> Hello all:
>
> Let n and k positive integers k<n.
>
> Let P(i) the ith-prime number
>
> We have:
>
> P(n)-P(n-k)-(n-k)P(k)=0   IF AND ONLY IF    P(n) and P(k) are Twin Primes
>
>
> Sincerely
>
> Sebastián Martín Ruiz
>
>
>
>
> [Non-text portions of this message have been removed]
>
>
>
> ------------------------------------
>
> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://www.primepages.org/
>
> Yahoo! Groups Links
>
>
>
>






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

I have finished up changes for PFGW 3.4.6. Here is a list of changes:

1)  Use gwnum for POWMOD script function when numbers are larger than 650 bits.
2)  Added -C switch to give better control over output for the console version.
      -C takes a single argument:
         quiet - the least amount of output, only gives status updates
         normal - output status updates and newlines after PRPs and primes
(default)
         GFFactors - output factors when using -g switch
         verbose - output result of all tests and output factors found
      The default behavior of the console version is -Cnormal, which is same as
the
      default behavior of WinPFGW.  Use -Cverbose to get the previous behavior of
the
      console version.
3)  Due to addition of -C switch, pfgw will now output test results for all
lines
     from simple input files with two exceptions.  It will not override -Cverbose
     and it will not output factors.

You can get the latest version from here,
http://sourceforge.net/projects/openpfgw/.

If you run into any problems, let me know.

--Mark

Hello all:

My  ​​equivalence for twin primes has been posted today here:
 
http://www.primepuzzles.net/puzzles/puzz_580.htm
 
Sincerely
 
Sebastian Martin Ruiz




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

I have obtained a curious
aproximation to Apery constant Zeta(3)=
1.2020569031595942...

(199/155)(16/165)^(3/2)Pi^3

Can someone test with a super computer expressions of this type:
 
e=N[Zeta[3]/Pi^3,140]
 
 
F[a_,b_,c_,d_]:=N[e(c/d)(a/b)^1.5,140]
 
Do[x=F[a,b,c,d];If[Abs[x-1]<0.0000001,Print[a," ",b," ",c," ",d,"
",x]],{a,1,260},{b,1,260},{c,1,260},{d,1,260}]
 
For larger values ​​of the parameters of course.
 
Or other irrational expressions multiplied by Pi ^ 3
 
Sincerely
 
Sebastián Martín Ruiz




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

--- In primenumbers@yahoogroups.com, Sebastian Martin Ruiz <s_m_ruiz@...> wrote:
>
> I have obtained a curious
> aproximation to Apery constant Zeta(3)=
> 1.2020569031595942...
>
> (199/155)(16/165)^(3/2)Pi^3

Hi.
To follow up on SMR's post and do some crude numerology, I computed Zeta(3)/Pi^3
to 1000 decimals:
0.038768179602916798941119890318721149806234568039552579223126762123777137012286\
855271851...
and then computed its regular continued fraction expansion

[0; 25, 1, 3, 1, 6, 3, 1, 1, 2, 1, 10, 3, 2, 1, 19, 3, 2, 1, 1, 3, 2, 3, 5, 3,
1, 1, 7, 1, 1, 1, 2, 1, 364, 11, 1, 84, 9, 34, 1, 7, 1, 63, 7, 1, 1, 4, 1, 5, 4,
7, 1, 1, 1, 5, 1, 4, 1, 5, 5, 9, 1, 21, 1, 9, 1, 1, 3, 2, 7, 1, 8, 5, 1, 7, 5,
2, 3, 1, 1, 1, 1, 1, 1, 1, 7, 8, 2, 2, 2, 1, 2, 1, 2, 6, 33, 99, 1, 1, 14, 1, 7,
2, 1, 1, 3, 4, 7, 1, 6, 5, 3, 1, 7, 8, 3, 5, 2, 1, 104, 1, 1, 1, 3, 8, 5, 1, 2,
1, 1, 11, 7, 1, 1, 1, 1, 2, 4, 1, 1, 1, 1, 1, 8, 1, 1, 7, 2, 4, 1, 1, 11, 1, 1,
1, 1, 35, 7, 1, 2, 3, 6, 1, 3, 1, 5, 81, 1, 2, 2, 7, 94, 2, 1, 2, 1, 3, 1, 2, 2,
1, 2, 1, 3, 1, 10, 75, 1, 2, 6, 3, 2, 7, 1, 1, 1, 6, 1, 4, 1, 1, 10, 1, 8, 2, 2,
1, 2, 1, 5, 6, 2, 1, 3, 1, 5, 1, 4, 3, 2, 5, 3, 3, 1, 10, 3, 1, 1, 9, 1, 1, 3,
3, 4, 1, 1, 2, 2, 1, 1, 15, 1, 15, 7, 4, 2, 1, 3, 1, 1, 1, 1, 1, 1, 1, 5, 8, 1,
1, 2, 1, 2, 4, 3, 1, 1, 3, 12, 2, 3, 1, 10, 2, 5, 19, 1, 4, 4, 3, 1, 3, 1, 6, 1,
7, 1, 4, 1, 1, 2, 7, 94, 7, 4, 2, 1, 4, 1, 1, 2, 1, 2, 11, 2, 2, 1, 24, 2, 2, 1,
1, 6, 1, 1, 5, 2, 2, 1, 1, 1, 10, 1, 19, 1, 3, 1, 1, 1, 6, 1, 161, 1, 7, 1, 4,
5, 1, 5, 1, 2, 1, 4, 12, 1, 6, 2, 1, 7, 1, 4, 108, 1, 5, 1, 1, 3, 4, 4, 1, 1, 7,
1, 32, 1, 1, 1, 1, 2, 2, 3, 1, 1, 4, 2, 1, 4, 14, 1, 1, 2, 3, 16, 1, 3, 1, 2, 1,
1, 5, 1, 15, 2, 1, 2, 30, 1, 94, 7, 1, 1, 1, 1, 1, 1, 2, 2, 6, 6, 4, 2, 1, 4, 1,
5, 19, 1, 1, 1, 4, 8, 22, 1, 2, 3, 1, 5, 1, 1, 5, 2, 10, 6, 3, 4, 13, 1, 1, 1,
2, 1, 1, 97, 1, 1, 5, 2, 8, 1, 1, 2, 1, 3, 36, 1, 1, 2, 2, 1, 40, 1, 13, 3, 1,
1, 226, 1, 5, 3, 1, 7, 3, 19, 1, 20, 49, 1, 1, 33, 1, 7, 2, 1, 4, 7, 3, 7, 8, 1,
1, 55, 1, 17, 6, 1, 1, 1, 2, 1, 1, 5, 2, 8, 26, 3, 5, 6, 1, 2, 2, 2, 3, 1, 2, 2,
1, 33, 1, 3, 2, 2, 42, 1, 1, 1, 3, 2, 2, 1, 26, 1, 4, 7, 13, 7, 1, 29, 11, 1, 2,
1, 4, 1, 14, 3, 5, 2, 59, 6, 2, 2, 1, 103, 1, 8, 1, 3, 1, 2, 1, 3, 1, 3, 1, 1,
12, 3, 13, 2, 2, 1, 13, 1, 1, 40, 4, 1, 2, 3, 1, 3, 1, 1, 12, 1, 7, 1, 2, 1, 1,
49, 1, 2, 1, 3, 1, 3, 8, 1, 4, 1, 2, 5, 1, 3, 1, 7, 2, 2, 24, 3, 2, 3, 2, 11, 2,
1, 15, 1, 3, 54, 1, 1, 1, 1, 1, 8, 1, 5, 1, 2, 5, 6, 2, 7, 2, 1, 1, 8, 1, 1, 5,
1, 2, 2, 4, 3, 8, 2, 1, 1, 8, 43, 1, 1, 1, 17, 1, 3, 4, 24, 1, 6, 2, 490, 6, 3,
1, 3, 8, 1, 1, 27, 2, 2, 1, 2, 1, 18, 5, 5, 1, 2, 2, 1, 2, 1, 6, 1, 1, 2, 1, 1,
1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 1, 3, 1, 2, 1, 2, 2, 2, 2, 2, 1, 14, 1, 2, 1,
1, 2, 3, 1, 1, 1, 1, 3, 4, 1, 1, 1, 2, 1, 1, 1, 6, 1, 4, 1, 3, 1, 4, 1, 6, 3, 1,
1, 1, 2, 2, 26, 1, 1, 1, 1, 1, 1, 29, 2, 2, 2, 21, 6, 1, 2, 4, 70, 1, 50, 2, 1,
2, 1, 1, 3, 1, 3, 4, 1, 3, 3, 4, 2, 48, 1, 2, 1, 16, 1, 2, 2, 1, 32, 329, 3, 4,
12, 1, 3, 10, 2, 2, 1, 1, 1, 1, 1, 3, 3, 7, 400, 1, 3, 7, 4, 2, 9, 2, 2, 41, 1,
2, 75, 1, 12, 4, 2, 47, 32, 3, 1, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 4, 2, 1, 1, 2,
2, 2, 1, 23, 1, 112, 12, 1, 2, 1, 11, 5, 5, 3, 7, 1, 1, 1, 8, 1, 12, 1, 8, 1, 5,
4, 9, 1, 3, 1, 1, 3, 1, 5, 8, 1, 1, 4, 1, 2, 2, 1, 2, 38, 1, 2, 1, 2, 2, 38, 59,
1, 8, 4, 2, 1, 4, 2, 3, 1, 29, 19, 2, 369, 1, 1, 7, 1, 5, 9, 2, 3, 1, 1, 3, 2,
2, 34, 1, 1, 1, 11, 1, 3, 1, 8, 1, 10, 1, 1, 1, 32, 1, 3, 1, 1, 8, 4, 1, 1, 1,
2, 1, 3, 1, 1, 2, 1, 8, 2, 3, 1, 15, 1, 2, 2, 1, 10, 5, 1, 2, 2, 4, 29, 8, 5, 1,
1, 2, 5, 1, 2, 1, 2, 6, 1, 1, 1, 11, 2, 5, 1, 2, 115, 6, 14, 4, 121, 1, 4, 2,
15, 34, 24, 6, 1, ...]

Now we ask, based on this, "is Sebastien Martin Ruiz on to something
interesting?"

The regular continued fraction (RCF)
expansion of a number X is useful for detecting
if that number is rational -- which is equivalent to the RCF
terminating, for example 355/113=[3;7,16].

It also is useful for detecting if X is a quadratic irrational,
which is equivalent to the RCF being eventually periodic, for
example
sqrt2=[1;2,2,2,2,...]
and
15^(3/2) = [58; repeat(10, 1, 1, 4, 8, 12, 1, 3, 1, 2, 1, 1, 1, 1, 1, 2, 1, 3,
1, 12, 8, 4, 1, 1, 10, 116)].

Finally, it is useful for detecting if X is "unusual."
Almost all real numbers have random RCF partial quotients
sampled from the Gauss-Kuzmin distribution
   http://en.wikipedia.org/wiki/Gauss%E2%80%93Kuzmin_distribution
Any number which violently fails statistical tests for that
is "unusual."  For example
e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, 1, 1, 12, 1, 1, 14,...]

Here are the counts A of how many times N appeared as a partial quotient for
N=1,2,...,100:

A=count for Zeta(3)/Pi^3 = 0.038768
B=corresponding count for Pi+2^(1/3)+7^(1/2) = 7.047265...
G=Gauss-Kuzmin expected count

N, A, B, G
1, 419, 402, 415.0374993
2, 174, 185, 169.9250014
3, 92, 94, 93.10940439
4, 50, 60, 58.89368905
5, 45, 36, 40.64198453
6, 27, 28, 29.74734345
7, 36, 21, 22.72007650
8, 26, 14, 17.92190798
9, 7, 14, 14.49956969
10, 10, 15, 11.97264165
11, 9, 10, 10.05366460
12, 9, 11, 8.562013557
13, 5, 8, 7.379530341
14, 5, 6, 6.426269095
15, 5, 9, 5.646563141
16, 2, 5, 5.000681040
17, 2, 6, 4.459648259
18, 1, 4, 4.001930553
19, 6, 7, 3.611253552
20, 1, 4, 3.275132038
21, 2, 5, 2.983858436
22, 1, 2, 2.729792802
23, 1, 1, 2.506855594
24, 3, 4, 2.310160687
25, 1, 1, 2.135744286
26, 3, 2, 1.980364109
27, 1, 1, 1.841346819
28, 0, 2, 1.716472527
29, 4, 3, 1.603885687
30, 1, 0, 1.502025127
31, 0, 0, 1.409570255
32, 4, 3, 1.325397401
33, 3, 2, 1.248546193
34, 2, 1, 1.178191153
35, 1, 0, 1.113620255
36, 1, 0, 1.054216386
37, 0, 0, 0.9994424311
38, 2, 1, 0.9488293993
39, 0, 1, 0.9019662943
40, 2, 1, 0.8584915734
41, 1, 2, 0.8180862026
42, 1, 2, 0.7804680132
43, 1, 1, 0.7453862058
44, 0, 0, 0.7126180220
45, 0, 0, 0.6819641182
46, 0, 1, 0.6532465388
47, 1, 0, 0.6263056815
48, 1, 0, 0.6009976942
49, 2, 1, 0.5771934627
50, 1, 0, 0.5547760091
51, 0, 0, 0.5336397688
52, 0, 1, 0.5136888564
53, 0, 0, 0.4948361984
54, 1, 0, 0.4770028103
55, 1, 0, 0.4601164965
56, 0, 1, 0.4441111278
57, 0, 0, 0.4289267856
58, 0, 0, 0.4145080281
59, 2, 0, 0.4008043244
60, 0, 0, 0.3877691866
61, 0, 0, 0.3753597376
62, 0, 0, 0.3635365655
63, 1, 0, 0.3522634356
64, 0, 1, 0.3415067120
65, 0, 0, 0.3312352143
66, 0, 0, 0.3214203598
67, 0, 0, 0.3120352996
68, 0, 0, 0.3030553501
69, 0, 1, 0.2944575600
70, 1, 0, 0.2862205664
71, 0, 0, 0.2783244500
72, 0, 0, 0.2707507354
73, 0, 1, 0.2634818134
74, 0, 0, 0.2565019512
75, 2, 1, 0.2497957052
76, 0, 1, 0.2433490748
77, 0, 1, 0.2371487818
78, 0, 0, 0.2311825580
79, 0, 0, 0.2254387130
80, 0, 0, 0.2199062779
81, 1, 0, 0.2145750055
82, 0, 0, 0.2094353705
83, 0, 0, 0.2044781363
84, 1, 0, 0.1996947880
85, 0, 0, 0.1950773880
86, 0, 1, 0.1906182871
87, 0, 0, 0.1863104142
88, 0, 0, 0.1821469860
89, 0, 0, 0.1781215085
90, 0, 0, 0.1742280648
91, 1, 0, 0.1704607381
92, 0, 0, 0.1668144771
93, 0, 0, 0.1632839428
94, 3, 0, 0.1598642283
95, 0, 0, 0.1565508601
96, 0, 0, 0.1533395088
97, 1, 1, 0.1502259897
98, 0, 0, 0.1472062620
99, 1, 0, 0.1442767180
100, 0, 0, 0.1414337502

"Chi Squared" values for departure of observed counts from Gauss-Kuzmin law:

for A:  560.1
for B:  536.4

I played around a bit more and found ChiSquared for several more "random
numbers" like B, for example
    Pi-2^(1/3)+7^(1/2)-exp(1)-ln(3)+erf(1) = 1.55322959...
has ChiSquared=771.9.
So Ruiz's number in (A) is within the typical range of ChiSquared.

In contrast, e has enormous ChiSquared=107491
or sqrt(7) has ChiSquared=189729, both easily detected in this
way as "unusual."

CONCLUSION:
Sorry, the first 1000 RCF quotients do not seem to indicate that
Zeta(3)/Pi^3 is in any way an unusual real number.

You all might want to try larger statistical analyses than this,
but that's what I'm seeing from the first 1000.  (It might be good to create a
fast "unusual number detector" software tool based on
the above techniques and maybe a Kolmogorov-Smirnov test.)

Here is another prime 17-tuplet (and a new 17 Simultaneous Primes record):

730862462126306661846437737 + d,
d = 0, 4, 10, 12, 16, 22, 24, 30, 36, 40, 42, 46, 52, 54, 60, 64, 66
(27 digits, Mar 2011, Raanan Chermoni & Jaroslaw Wroblewski)

Jarek

I also attempted to use PSLQ to figure out whether
Zeta(3)/Pi^3
was a low-degree low-height algebraic number.
Result:
If it has degree<=10 then its height is at least 10^91.

So, sorry again:
Zeta(3)/Pi^3   again fails to look unusual.

No magic here, just the concept that you can find approximations to
basically anything if you have enough degrees of freedom.

For instance:

    (83/779)(605/689)^(3/2)Pi^3 == 2.71828182845...

gives you the first 12 correct digits of e.

    (121/634)(305/724)^(3/2)Pi^3 == 1.6180339887...

gives you the first 11 correct digits of the golden ratio (1+sqrt(5))/2.


Note that there's an even better approximation to Zeta(3) at:

    (25/186)(86/197)^(3/2)Pi^3


On 3/14/2011 9:59 AM, WarrenS wrote:
>
>
> --- In primenumbers@yahoogroups.com, Sebastian Martin Ruiz<s_m_ruiz@...> 
wrote:
>>
>> I have obtained a curious
>> aproximation to Apery constant Zeta(3)=
>> 1.2020569031595942...
>>
>> (199/155)(16/165)^(3/2)Pi^3
>
> Hi.
> To follow up on SMR's post and do some crude numerology, I computed
Zeta(3)/Pi^3 to 1000 decimals:
>
0.038768179602916798941119890318721149806234568039552579223126762123777137012286\
855271851...
> and then computed its regular continued fraction expansion
>
> [0; 25, 1, 3, 1, 6, 3, 1, 1, 2, 1, 10, 3, 2, 1, 19, 3, 2, 1, 1, 3, 2, 3, 5, 3,
1, 1, 7, 1, 1, 1, 2, 1, 364, 11, 1, 84, 9, 34, 1, 7, 1, 63, 7, 1, 1, 4, 1, 5, 4,
7, 1, 1, 1, 5, 1, 4, 1, 5, 5, 9, 1, 21, 1, 9, 1, 1, 3, 2, 7, 1, 8, 5, 1, 7, 5,
2, 3, 1, 1, 1, 1, 1, 1, 1, 7, 8, 2, 2, 2, 1, 2, 1, 2, 6, 33, 99, 1, 1, 14, 1, 7,
2, 1, 1, 3, 4, 7, 1, 6, 5, 3, 1, 7, 8, 3, 5, 2, 1, 104, 1, 1, 1, 3, 8, 5, 1, 2,
1, 1, 11, 7, 1, 1, 1, 1, 2, 4, 1, 1, 1, 1, 1, 8, 1, 1, 7, 2, 4, 1, 1, 11, 1, 1,
1, 1, 35, 7, 1, 2, 3, 6, 1, 3, 1, 5, 81, 1, 2, 2, 7, 94, 2, 1, 2, 1, 3, 1, 2, 2,
1, 2, 1, 3, 1, 10, 75, 1, 2, 6, 3, 2, 7, 1, 1, 1, 6, 1, 4, 1, 1, 10, 1, 8, 2, 2,
1, 2, 1, 5, 6, 2, 1, 3, 1, 5, 1, 4, 3, 2, 5, 3, 3, 1, 10, 3, 1, 1, 9, 1, 1, 3,
3, 4, 1, 1, 2, 2, 1, 1, 15, 1, 15, 7, 4, 2, 1, 3, 1, 1, 1, 1, 1, 1, 1, 5, 8, 1,
1, 2, 1, 2, 4, 3, 1, 1, 3, 12, 2, 3, 1, 10, 2, 5, 19, 1, 4, 4, 3, 1, 3, 1, 6, 1,
7, 1, 4, 1, 1, 2, 7, 94, 7, 4, 2, 1, 4, 1, 1, 2, 1, 2, 11, 2, 2, 1, 24, 2, 2, 1,
1, 6, 1, 1, 5, 2, 2, 1,
1, 1, 10, 1, 19, 1, 3, 1, 1, 1, 6, 1, 161, 1, 7, 1, 4, 5, 1, 5, 1, 2, 1, 4, 12,
1, 6, 2, 1, 7, 1, 4, 108, 1, 5, 1, 1, 3, 4, 4, 1, 1, 7, 1, 32, 1, 1, 1, 1, 2, 2,
3, 1, 1, 4, 2, 1, 4, 14, 1, 1, 2, 3, 16, 1, 3, 1, 2, 1, 1, 5, 1, 15, 2, 1, 2,
30, 1, 94, 7, 1, 1, 1, 1, 1, 1, 2, 2, 6, 6, 4, 2, 1, 4, 1, 5, 19, 1, 1, 1, 4, 8,
22, 1, 2, 3, 1, 5, 1, 1, 5, 2, 10, 6, 3, 4, 13, 1, 1, 1, 2, 1, 1, 97, 1, 1, 5,
2, 8, 1, 1, 2, 1, 3, 36, 1, 1, 2, 2, 1, 40, 1, 13, 3, 1, 1, 226, 1, 5, 3, 1, 7,
3, 19, 1, 20, 49, 1, 1, 33, 1, 7, 2, 1, 4, 7, 3, 7, 8, 1, 1, 55, 1, 17, 6, 1, 1,
1, 2, 1, 1, 5, 2, 8, 26, 3, 5, 6, 1, 2, 2, 2, 3, 1, 2, 2, 1, 33, 1, 3, 2, 2, 42,
1, 1, 1, 3, 2, 2, 1, 26, 1, 4, 7, 13, 7, 1, 29, 11, 1, 2, 1, 4, 1, 14, 3, 5, 2,
59, 6, 2, 2, 1, 103, 1, 8, 1, 3, 1, 2, 1, 3, 1, 3, 1, 1, 12, 3, 13, 2, 2, 1, 13,
1, 1, 40, 4, 1, 2, 3, 1, 3, 1, 1, 12, 1, 7, 1, 2, 1, 1, 49, 1, 2, 1, 3, 1, 3, 8,
1, 4, 1, 2, 5, 1, 3, 1, 7, 2, 2, 24, 3, 2, 3, 2, 11, 2, 1, 15, 1, 3, 54, 1, 1,
1, 1, 1, 8, 1, 5, 1, 2, 5, 6
, 2, 7, 2, 1, 1, 8, 1, 1, 5, 1, 2, 2, 4, 3, 8, 2, 1, 1, 8, 43, 1, 1, 1
> , 17, 1, 3, 4, 24, 1, 6, 2, 490, 6, 3, 1, 3, 8, 1, 1, 27, 2, 2, 1, 2, 1, 18,
5, 5, 1, 2, 2, 1, 2, 1, 6, 1, 1, 2, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 2, 1, 3,
1, 2, 1, 2, 2, 2, 2, 2, 1, 14, 1, 2, 1, 1, 2, 3, 1, 1, 1, 1, 3, 4, 1, 1, 1, 2,
1, 1, 1, 6, 1, 4, 1, 3, 1, 4, 1, 6, 3, 1, 1, 1, 2, 2, 26, 1, 1, 1, 1, 1, 1, 29,
2, 2, 2, 21, 6, 1, 2, 4, 70, 1, 50, 2, 1, 2, 1, 1, 3, 1, 3, 4, 1, 3, 3, 4, 2,
48, 1, 2, 1, 16, 1, 2, 2, 1, 32, 329, 3, 4, 12, 1, 3, 10, 2, 2, 1, 1, 1, 1, 1,
3, 3, 7, 400, 1, 3, 7, 4, 2, 9, 2, 2, 41, 1, 2, 75, 1, 12, 4, 2, 47, 32, 3, 1,
1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 4, 2, 1, 1, 2, 2, 2, 1, 23, 1, 112, 12, 1, 2, 1,
11, 5, 5, 3, 7, 1, 1, 1, 8, 1, 12, 1, 8, 1, 5, 4, 9, 1, 3, 1, 1, 3, 1, 5, 8, 1,
1, 4, 1, 2, 2, 1, 2, 38, 1, 2, 1, 2, 2, 38, 59, 1, 8, 4, 2, 1, 4, 2, 3, 1, 29,
19, 2, 369, 1, 1, 7, 1, 5, 9, 2, 3, 1, 1, 3, 2, 2, 34, 1, 1, 1, 11, 1, 3, 1, 8,
1, 10, 1, 1, 1, 32, 1, 3, 1, 1, 8, 4, 1, 1, 1, 2, 1, 3, 1, 1, 2, 1, 8, 2, 3, 1,
15, 1, 2, 2, 1, 10, 5, 1, 2, 2, 4, 2
9, 8, 5, 1, 1, 2, 5, 1, 2, 1, 2, 6, 1, 1, 1, 11, 2, 5, 1, 2, 115, 6, 14, 4, 121,
1, 4, 2, 15, 34, 24, 6, 1, ...]
>

--- In primenumbers@yahoogroups.com,
"WarrenS" <warren.wds@...> wrote:

> I also attempted to use PSLQ to figure out whether Zeta(3)/Pi^3
> was a low-degree low-height algebraic number.

We should always try to do what most folk think can never be done.

Had such an attempt succeeded, there would be several deeply
unhappy Fields Medalists, who believe (as I do) in the
Drinfeld-Deligne conjecture that the Grottendieck-Teichmueller
algebra has precisely one generator in each odd degree.

Yet that is, emphatically, not a reason for avoiding serious
experiments to investigate a contrary hypothesis. To the best of
my recollection, the possibility that zeta(3)/Pi^3 might be an
algebraic number has been investigated to many tens of thousands
of decimal digits of numerical precision. However, I am unable
to provide a reference. Might someone else do so, please?

David

> > Warren D Smith:
> > I also attempted to use PSLQ to figure out whether Zeta(3)/Pi^3
> > was a low-degree low-height algebraic number.

> DJ Broadhurst:
> Had such an attempt succeeded, there would be several deeply
> unhappy Fields Medalists, who believe (as I do) in the
> Drinfeld-Deligne conjecture that the Grottendieck-Teichmueller
> algebra has precisely one generator in each odd degree.
>
> Yet that is, emphatically, not a reason for avoiding serious
> experiments to investigate a contrary hypothesis. To the best of
> my recollection, the possibility that zeta(3)/Pi^3 might be an
> algebraic number has been investigated to many tens of thousands
> of decimal digits of numerical precision. However, I am unable
> to provide a reference.

--well, you (DJB) probably have better PSLQ and better hardware
than I do (I bet) and could do so yourself...

I see Broadhurst wrote a quantum field theory paper "where do the tedious
products of zetas come from?" (actual title!) which mentions the "Drinfeld
Deligne conjecture"
and "Grothendieck–Teichm:uller algebra" (whatever they are) as well as using
PSLQ.
Unfortunately non-experts (i.e. me) will have a difficult time understanding
this paper.
Even after (trying to) read it, I still have almost no clue what DDC and GTA
are,
and it gives zero cites to D,D,G and T's work.

Anyhow I agree Zeta(3)/Pi^3 being an "unusual number" was a pretty low-chance
proposition, but SM Ruiz's post stimulated me to look, and sure enough, negative
result.
About the idea that you can produce miracles on demand if you have enough
fitting freedom... well, yes, but the approach I used attempts to quantify the
miraculousness,
and the TRUE miracles yield far better approximations than the amount of fitting
freedom would suggest.  Don't accept lame miracles, demand real miracles.

--- In primenumbers@yahoogroups.com,
"WarrenS" <warren.wds@...> wrote:

> Broadhurst wrote a quantum field theory paper
> "where do the tedious products of zetas come from?"

An editor solicited this article and I agreed on the
strict condition that "tedious" be retained in the title.
It happened that I needed to consult it yesterday
and fortunately it was easy to google:
> Ungefähr 84 Ergebnisse (0,18 Sekunden)

David

> About the idea that you can produce miracles on demand if you have enough
fitting freedom... well, yes, but the approach I used attempts to quantify the
miraculousness,
> and the TRUE miracles yield far better approximations than the amount of
fitting freedom would suggest.  Don't accept lame miracles, demand real
miracles.


--for example (this is not due to me; perhaps it was first noticed by
Ramanujan?)
    exp(Pi * sqrt(n))
is within 10^(-12) of being an integer if n=163.

This is beyond what any kind of fitting contrivance would be expected to
produce.
Also, no other n<25000 gets you even within a 100,000 times further away
from being an integer.

PSLQ has in fact discovered truths valid not to 12, but to
an infinite number of decimals :)

--- In primenumbers@yahoogroups.com, "WarrenS" <warren.wds@...> wrote:
>
> --for example (this is not due to me; perhaps it was first noticed by
Ramanujan?)
>    exp(Pi * sqrt(n))
> is within 10^(-12) of being an integer if n=163.
>
> This is beyond what any kind of fitting contrivance would be expected to
produce.
> Also, no other n<25000 gets you even within a 100,000 times further away
> from being an integer.

Even better:
(exp(Pi*sqrt(163))-744)^(1/3) = 640319.999999999999999999999999390...

See Cohen, CCANT, p.383, which explains where astonishing results such as these
come from.

Mike

I suggest that in addition to testing simplicity of

zeta(3)/ pi**3,

that you test

zeta(3)/pi**2

and

zeta(3)/pi**4.


Kermit Rose