1a. Re: Composite integer function
     Posted by: "Yann Guidon" whygee@... yasep16
     Date: Wed Nov 18, 2009 9:02 am ((PST))

Hello Kermit,

it seems that my emails can't reach you
due to some unexplainable blacklist on some router near you.
So I answer on the list :


can you please elaborate ?




Hello Yann.


..........s=1... s=2 ... s=3 ... s=4 ..  s=5 . s=6
m=1  015    021    027    033    039    045
m=2  035    045    055    065    075    085
m=3  063    077    091    105    119    133
m=4  099    117    135    153    171    189
m=5  143    165    187    209    231    253
m=6  195    221    247    273    299    325

The table extends to arbitrarily  large values.

Table entry at row m and column s is (2 * m + 1) * (2 * m + 1 + 2 * s)

For example, 153 at row 4 and column 4 is (2 * 4 + 1) * (2 * 4 + 1 + 2 *
4) = 9 * 17

Adjacent table entries have an additive relationship to each other.

For example 117 in row 4, column 2,
and  165 in row 5, column 2,
are related as follows.

165 = 117 + 48 = 117 + 8 * 4 + 4 * 2 + 8

Table entry in row (m+1) and column s
= table entry in row (m) and column s,
plus 8 times row number m,
plus 4 times column number s,
plus 8.

There is a similar addition rule to calculate entries in the next column
over.

The extended table lists all the non-square odd composite positive integers,
and has both additive and multiplicative rules for determining
table entries.


It is not trivial to find the location of a large number in the table.

The easiest way to find a large integer in the table is to factor the
integer.

However, if some other algorithm for locating a given number in the
table is developed, that algorithm would also be a factoring algorithm.

Kermit.

--- In primenumbers@yahoogroups.com, "Jens Kruse Andersen" <jens.k.a@...> wrote:
>
> Mark wrote:
> > Given any prime expressed as a+b, is there always some  a,b
> > such that 2^a*3^b is one away from a prime?
>
> I expect infinitely many counter examples but there are none below 7500.
> I only computed one prime for each prime sum a+b.
>

I think you're right Jens, infinitely many.

By observation, the average number of solutions for a given prime seems to be
roughly constant, around 10.

For instance the first 13 primes after 2000 yield the following number of hits:

[2003, 8]  [2011, 9]  [2017, 7]  [2027, 12]  [2029, 10]  [2039, 15]  [2053, 7] 
[2063, 15]  [2069, 4]  [2081, 9]  [2083, 6]  [2087, 12]  [2089, 10] average is
9.5

The first 13 primes after 1000 yield the following:

[1009, 13]  [1013, 11]  [1019, 10]  [1021, 10]  [1031, 10]  [1033, 5]  [1039, 7]
[1049, 9]  [1051, 13]  [1061, 8]  [1063, 9]  [1069, 13]  [1087, 10] average is
9.8

The first 13 primes after 10 yield the following:

[11,5] [13, 10]  [17, 10]  [19, 7]  [23, 9]  [29, 6]  [31, 13]  [37, 10]  [41,
13]  [43, 11]  [47, 12]  [53, 11]  [59, 11]  average is 9.8

Alas  I don't have any heuristic accurate enough to confirm or deny that the
average will stay around 10. :)

Nor am I savvy enough in statistics to deduce from the observed deviations a
ballpark estimate as to when a zero would be expected.

It's nice to see however so many primes nestled up against these "three smooth"
numbers.

Mark

Mark wrote:
> Given any prime expressed as a+b, is there always some  a,b
> such that 2^a*3^b is one away from a prime?

I expect infinitely many counter examples but there are none below 7500.
I only computed one prime for each prime sum a+b.

--
Jens Kruse Andersen

Given any prime expressed as a+b, is there always some  a,b such that 2^a*3^b is
one away from a prime?

I doubt it but have yet to find a counterexample.

Below are primes < 3187 which produced five or fewer solutions. [prime, number
of times that 2^a*3^b is one away from a prime]

[2, 2]  [3, 3]  [5, 5]  [11, 5]  [317, 5]  [347, 3]  [677, 4]  [739, 4]  [809,
5]  [857, 5]  [1033, 5]  [1229, 5]  [1291, 5]  [1319, 5]  [1451, 2]  [1471, 5] 
[1663, 3]  [1721, 5]  [2069, 4]  [2477, 5]  [2659, 4]  [2677, 3]

Mark

Hello Kermit,

it seems that my emails can't reach you
due to some unexplainable blacklist on some router near you.
So I answer on the list :

Kermit Rose wrote:
> You might be interested in the following two variable function.
<snip>
> Factoring a positive integer z is equivalent to finding it in the table.
can you please elaborate ?

thanks,
yg
--
http://ygdes.com / http://yasep.org

You might be interested in the following two variable function.

Define F(m,k) recursively as follows.

F(1,1) = 15
F(m+1,k) = F(m,k) + 4*(2*m + k + 2)
F(m,k+1) = F(m,k) + 2*(2*m + 1)

Then for both m and k positive integers,
F(m,k) = (2 * m + 1) * (2 * m + 2 * k + 1)

which makes it evident that every odd non-square positive composite
integer appears in the table, and that no prime appears in the table.

This function could be used to make an efficient prime number sieve.

Note that F(m,0) = (2*m+1)**2


Factoring a positive integer z is equivalent to finding it in the table.

--- In primenumbers@yahoogroups.com, "Mike Oakes" <mikeoakes2@...> wrote:
>
> Inserting the latest records from Jens's page
> http://users.cybercity.dk/~dsl522332/math/aprecords.htm
> gives this table:-
>
> rank k  d      s=(k+4)*log(d)
> ---- -  -      --------------
> 1    8  1057   83.558
> 2    3  137514 82.817
> 3    12 173    82.453
> 4    25 17     82.163
> 5    5  7009   79.695
> 6    23 19     79.499
> 7    24 17     79.330
> 8    11 195    79.095
> 9    7  1290   78.786
> 10   9  425    78.677
> 11   10 265    78.116
> 12   22 19     76.554  MEDIAN
> 13   14 69     76.214
> 14   19 27     75.803
> 15   4  11961  75.115
> 16   21 20     74.893
> 17   18 29     74.081
> 18   13 78     74.064
> 19   6  1606   73.815
> 20   15 48     73.553
> 21   20 21     73.068
> 22   16 38     72.752
> 23   17 29     70.713
>

That was as per 6 Jun.
Just over 5 months on, the revised table is:-

rank k  d      s=(k+4)*log(d)
---- -  -      ------ -------
1    8  1057   83.558
2    3  137514 82.817
3    12 173    82.453
4    25 17     82.163
5    5  7009   79.695
6    24 18     80.930
7    23 19     79.499
8    11 196    79.172
9    7  1290   78.786
10   9  425    78.677
11   10 274    78.584
12   17 42     78.491 was 23rd
13   6  2145   76.709 was 19th
14   22 19     76.554
15   14 69     76.214
16   19 27     75.803
17   15 54     75.791 was 20th
18   4  11961  75.115 was 15th
19   21 20     74.893 was 16th
20   16 42     74.753
21   18 29     74.081 was 17th
22   13 78     74.064 was 18th
23   20 21     73.068

Note that the lowest score was 70.713 and is now 73.068.
The comments are against k values which have changed rank by more than 2.
The first 11 positions are almost unchanged.

The latest table might help in suggesting to people (but perhaps I shouldn't be
giving these clues:-) which records to try for next.

-Mike Oakes

David,

you are absolutely correct, i misstated that conjecture.
the conjecture asks whether

o_p(2) = p-1

for infinitely many primes.



this would correspond to

limsup o_p(2)/p = 1

thanks,


lou

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

> let o_p(2) = order of 2 in F_p.
...
> o_p(2) = p-1 for p prime

Counterexample: o_137(2) = (137 - 1)/2 = 68

o_p(2) is provably a divisor of p-1 for prime p.
By the Artin conjecture, it is a proper divisor
of p-1 for about
62.604418638079771194527194565348358488837 percent
of the primes. Here are the first few:

  forprime(p=3,137,if(znorder(Mod(2,p))<p-1,print1(p" ")))
7 17 23 31 41 43 47 71 73 79 89 97 103 109 113 127 137

David

Greetings all,
let o_p(2) = order of 2 in F_p.
in other words, o_p(2) = card{ 2^0 (mod p), 2^1 (mod p), ... , 2^p (mod
p)  }
we know that it is an open conjecture that o_p(2) = p-1 for p a
prime.but does anyone actually know what is:
limsup o_p(2)/p = ?where p goes over all primes?
thanks,

lou



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

Mike Oakes wrote:
> Here are new AP16 & AP17 records at 42 digits:-
>
> (263013824+18107251*n)*83#+1 is prime for n=0..16

> The aim was to improve the AP16 record.
> However, most remarkably, an AP17 popped up after only
> 4 AP15's had been found, and before any AP16 at all!

Congratulations!
http://users.cybercity.dk/~dsl522332/math/aprecords.htm is updated.
That AP17 may last a while. I'm not trying to retake it.

--
Jens Kruse Andersen

I have just submitted to Chris's database a pair of primes that, at 25055
digits, will come in at rank 16 on his Top-20 Twins page
http://primes.utm.edu/top20/page.php?id=1

I thought it would be instructive to record the processing steps needed for this
quite unremarkable achievement, and thereby maybe encourage others to more
ambitious goals in this direction.

The twins are these two:-

Primality testing 1035928263*2^83200-1 [N-1/N+1, Brillhart-Lehmer-Selfridge]
Calling N+1 BLS with factored part 100.00% and helper 0.01% (300.01% proof)
1035928263*2^83200-1 is prime! (124.2008s+0.0011s)

Primality testing 1035928263*2^83200+1 [N-1/N+1, Brillhart-Lehmer-Selfridge]
Calling N-1 BLS with factored part 100.00% and helper 0.01% (300.02% proof)
1035928263*2^83200+1 is prime! (79.9587s+0.0016s)

The first step was to ask NewPGen to sieve a block of numbers of the form
k*2^n+/-1, with n=83200 and k=1..4*10^9, to a depth of 179 trillion.
This took 28.4 GHz-days, and left a mere 1,1551,319 candidates.

The next step was to feed the output file to PFGW, which is aware of the NewPGen
output format for Twins, and tests the lower one of each pair, and only if that
is a PRP goes on to test the higher one.
PFGW found the twin after 120 GHz-days, when it was about 1/4 of the way through
the complete NewPGen output file.

Finally, one just needs to give PFGW the 2 PRPs with the -tc processing option,
to confirm that they are indeed prime (giving the output quoted above).

-Mike Oakes

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

> an AP17 popped up after only 4 AP15's had been found

That is, indeed, rather pleasing. Best regards, David

Here are new AP16 & AP17 records at 42 digits:-

(263013824+18107251*n)*83#+1 is prime for n=0..16

All confirmed prime with PFGW -tc

Input/output statistics:-
Numbers tested by NewPGen: 2*10^9
NewPGen reduced these (by sieving to 50 billion) to: 3.71*10^8
PRP's found by PFGW: 171,068,264
AP17's found: 1

Run-time statistics:-
NewPGen: 3 GHz-hrs
PFGW: 36 GHz-hrs
Pascal program to find AP's: 174 GHz-days

The aim was to improve the AP16 record.
However, most remarkably, an AP17 popped up after only 4 AP15's had been found,
and before any AP16 at all!
The Pascal program was killed at this point, which was about 1/3 of the way
through its exhaustive search.

-Mike Oakes

That mathematician is good. All I can see is that the pattern in the first is
repeated somewhere in each of the succeeding eight patterns. And, the 1st, 3rd
and 7th are symmetrical. And it's a good exercise in closing brackets. And they
all start with [.[ and end with ].]

But wait! I sense it has something to do with primes....must be my human-alien
hybrid brain starting to kick in...




--- In primenumbers@yahoogroups.com, Kermit Rose <kermit@...> wrote:
>
> Xeno Riddle
>
> An anthropologist and a mathematician strolled through the forest.
> Suddenly they came upon an extratresterial camp. The two
> extratresterials, equally startled, dropped everything,
> jumped into their nearby spacecraft, and disappeared amid a clap
> of thunder.
>
> The anthropologist rushed over to see what tools they had dropped.
> He found two artifacts. One appeared to be merely a metal ball.
>
> The other artifact appeared to be made of extremely durable
> plastic. Otherwise it resembled an ordinary sheet of paper.
> It had the following symbols embedded within it.
>
> [.[].]
> [.[]..[.[].].]
> [.[.[].]..[.[].].]
> [.[]..[.[].[.[].].].]
> [.[.[].]..[.[]..[.[].].].]
> [.[]..[]..[.[].].]
> [.[.[].]..[]..[.[].].]
> [.[.[]..[.[].].]..[.[].].]
> [.[.[.[].]..[.[].].]..[.[].].]
>
>
> The mathematician, studying the symbols,
> exclaimed, "How could they have built such
> an advanced technology using such clumsy arithmetical
> notation?"
>
> What do the symbols on the paper represent?
>

Xeno Riddle

An anthropologist and a mathematician strolled through the forest.
Suddenly they came upon an extratresterial camp. The two
extratresterials, equally startled, dropped everything,
jumped into their nearby spacecraft, and disappeared amid a clap
of thunder.

The anthropologist rushed over to see what tools they had dropped.
He found two artifacts. One appeared to be merely a metal ball.

The other artifact appeared to be made of extremely durable
plastic. Otherwise it resembled an ordinary sheet of paper.
It had the following symbols embedded within it.

[.[].]
[.[]..[.[].].]
[.[.[].]..[.[].].]
[.[]..[.[].[.[].].].]
[.[.[].]..[.[]..[.[].].].]
[.[]..[]..[.[].].]
[.[.[].]..[]..[.[].].]
[.[.[]..[.[].].]..[.[].].]
[.[.[.[].]..[.[].].]..[.[].].]


The mathematician, studying the symbols,
exclaimed, "How could they have built such
an advanced technology using such clumsy arithmetical
notation?"

What do the symbols on the paper represent?

While I'm here, I'd like to invite everyone to participate in an online prime
search contest.  The contest "Tribal Primes" at
http://www.v-sonline.com/index.pl?C4 has been running for about 12 days now and
it's turning out that one of the lower parts has become an exciting search for
an optimal sequence of primes.  The contest description is as follows:

Your task is to find a lists of N primes such that the sums of every combination
of three in the submitted list generates the most unique primes. For example,
the list of 5 primes (5, 37, 47, 89, 97) generates the 9 unique primes (89, 131,
139, 149, 173, 181, 191, 223, 233). You may submit primes lists such that the
largest three prime sum is less than 2 ^ 26. The contest is divided in to 20
parts where N is from 6 to 25, inclusive.

For N = 6, 7 and 8, the optimals have been found, but the N = 9 optimal of 70
unique primes has proven elusive so far.

Enjoy,

James



p.s. The contests are free to participate in.  I'm basically having fun making
challenging puzzles to solve for the public.

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

Oh my, what a disappointment.  I thought for sure they would have a name. 
Perhaps we can name them?

Three possibilities among many occur to me:  "Unidigital Primes", "Monodigital
Primes", or "Solodigital Primes".

I'm making a puzzle contest and I'm looking for a name to call them if one
exists or can be named.

James

   ----- Original Message -----
   From: Jens Kruse Andersen
   To: primenumbers@yahoogroups.com
   Sent: Thursday, November 12, 2009 6:43 PM
   Subject: Re: [PrimeNumbers] Another question



   James J Youlton Jr wrote:
   > If a prime number does not have the same digit occuring
   > within it more than once, what is that number called (and yes
   > I know there are a finite number of them all being < 10**9).

   They are simply called primes with distinct digits in
   http://www.research.att.com/~njas/sequences/A029743
   which says:
   "This sequence has 283086 terms, the last being 987654103"

   As you apparently know, 3 divides any permutation of 0123456789.

   --
   Jens Kruse Andersen




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

James J Youlton Jr wrote:
> If a prime number does not have the same digit occuring
> within it more than once, what is that number called (and yes
> I know there are a finite number of them all being < 10**9).

They are simply called primes with distinct digits in
http://www.research.att.com/~njas/sequences/A029743
which says:
"This sequence has 283086 terms, the last being 987654103"

As you apparently know, 3 divides any permutation of 0123456789.

--
Jens Kruse Andersen

A quick question if I may.  If a prime number does not have the same digit
occuring within it more than once, what is that number called (and yes I know
there are a finite number of them all being < 10**9).  Example, 97 is such a
number but 101 is not.

Thanks in advance,

James

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

I happened to look at
http://primes.utm.edu/bios/top20.php?type=person&by=PrimesRank
and was struck by the fact that Bouk de Water,
without needing to prove any more primes, becomes
ever more distinguished in Chris Caldwell's rankings.

Observe that
> 14 Bouk de Water 101.367 38.3070
has more than 100 primes,
with a "crunching-score" less than 39.
But no-one else in the top-20 by primes
has a "crunching-score" less than 45.

If we "move down list", we see that Ken does well:
> 31 Ken Davis 32 39.4760
and Norman even better
> 36 Norman Luhn 29.5 33.8234

My own performance is miserable:
> 9 David Broadhurst 156.233 45.8131
with a crunching score above 45,
placing me in the ranks of those
who seem to prefer brawn to brains.

Ah well, we can't all be original, all of the time.
But it's a shame to see the verification page
http://primes.utm.edu/primes/status.php?hours=96
looking so uninspired, in the last few days.

David (ruefully)

> theorem: iff F(n)== -1 mod (2^(n-1) +1), then F(n) is prime!

did you mean "if" ?

But it's still wrong for n=16 (and n=36) :
F(16) = -1 (mod 2^15+1)
but F(16) is known to be composite,
as Peter wrote, it is divisible by 825753601.


> so, if (r +1)*b = 2^(2^n)+2, then b must equal 2; so, r +1=

I think here you use that
2^(2^n-1)+1 has no factor of the form 2^h+1,
but this is not true in general...
For example with n=16 you have
2^65535+1 which is divisible by 2^3+1 = 9.

[Sorry, in my previous mail I misread the definition of r vs 2^h, please
ignore...]

Maximilian

whoa, I got a lot of responses!; didn't have time to look at any of them.
I made the correction to the typo earlier, but sent the e-mail to myself;
this proof makes it rock solid with the typo displayed and supplying the
proof... hope no one's mad; do you like it PROFESSOR Caldwell ???  your
website is immense if nothing else... done with maths for a while...


Professor Caldwell, et. al.

I couldn't resist... here's my idea for proof:


theorem: iff F(n)== -1 mod (2^(n-1) +1), then F(n) is prime!


if F(n)== -1 mod (r +1), then r= 2^h >1 and (r+1)*b= F(n) +1;

so, if (r +1)*b = 2^(2^n)+2, then b must equal 2; so, r +1=

2^(2^n-1) +1 implies... 2^h= 2^(2^n -1) implies... h= 2^n -1;

and if 2^n is replaced by n, then 2^n -1 is rel. prime to n-1.

now, (2^(n-1) +1) is mutually exclusive with (2^(2^n -1) +1),

and thus, if both modulators produce the same result, then

iff F(n)== -1 mod (2^(n-1) +1), then F(n) is prime; the other

case of (r -1) would be proved similarly.

*QED

I noticed it in an hour the night before & proved it last night.

Bill


--- On Wed, 11/11/09, leavemsg1 <leavemsg1@...> wrote:

> From: leavemsg1 <leavemsg1@...>
> Subject: Re: Fermat conjecture; please post it
> To: "Bill Bouris" <leavemsg1@...>
> Date: Wednesday, November 11, 2009, 9:16 AM
> small typo... look below.
>
> --- In primenumbers@yahoogroups.com,
> Bill Bouris <leavemsg1@...> wrote:
> >
> > Hello, Professor Caldwell.
> >
> > I like your website.  Could you please post the
> following conjecture:
> >
> > if n>= 2 and F(n)= 2^(2^n)+1, then iff [F(n) mod
> (2^(n-1)+1) == -1]
> >
> > (or) [F(n) mod (2^(n-1)+1) == -1], then F(n) is
> prime.
>
> the second half should read ... F(n) mod (2^(n-1)-1) == -1;
> sorry.
>
> >
> >
> > it requires someone as formidable as Lucas, Lehmer,
> etc. to prove it.
> >
> > examples...
> > n= 2; F(n)= 17;  17 mod 3 == -1; 17 mod 1 == 0; 17 is
> prime!
> >
> > n= 3; F(n)= 257;  257 mod 5 == +2; 257 mod 3 == -1;
> 257 is prime!
> >
> > n= 4; F(n)= 65537; 655377 mod 9 == -1; 65537 mod 7 ==
> +3; 65537 is prime!
> >
> > n= 5; F(n)= 4294967297; F(n) mod 17 == +2; F(n) mod 15
> = +2; composite!
> >
> > I wish someone had the technical expertise to prove
> it; it's valid, and
> > I've studied it... trying to come up with a
> proof.  Share it with a close
> > colleague, if you like.
> >
> > Thanks in advance,
> >
> > Bill Bouris
> >
>
>
>

Hi, All.

It would be great if I could get some feedback on this.  I have completed a
third proof of the Riemann Hypothesis, this one being the only one I consider
worthwhile. Please give feedback or comments as you wish, or forward it to
someone you know who can effectively review it.

Here's the abstract:

Abstract:

A proof of the Riemann Hypothesis is proposed in six lemmas, where five of the
six are proven using elementary arithmetic. It is shown that by applying all the
zeros of the zeta function to a ratio, having an infinite number of numerators
and divisors equal to the same value, the modulus of a variable z used to
calculate the ratio are all equal to the square root of one divided by fourteen
for all the zeros of zeta of s, trivial or non-trivial. Using the common
modulus, it is shown that the value of the ratio for all the non-trivial zeros
is a fixed constant, whereby allowing one to calculate the only possible
positive Real part of s for the non-trivial zeros. Such proof suggests that the
greatest common multiple and lowest common denominator of this ratio for all the
zeros of zeta of s lie in the non-trivial zeros with a fixed Real part one half.

To download the entire file (30 pages, 2MB), you can do this from
www.JeffreyNCook.com or more directly, this link:

http://www.jeffreyncook.com/cook%20riemann%20hypothesis%20proof.htm

Much thanks,

Jeff

--- In primenumbers@yahoogroups.com,
Jack Brennen <jfb@...> wrote:

> A couple of easy counterexamples can be found,
> for n = 16 and n = 36

It also fails for n = 256:

  {if(Mod(2,36986355*2^258+1)^(2^256)+1==0
  &&Mod(2,2^255+1)^(2^256)+2==0,print(fail))}
fail

David

Hi Bill,

> if n>= 2 and F(n)= 2^(2^n)+1, then iff
> [F(n) mod (2^(n-1)+1) == -1] (or) [F(n) mod (2^(n-1)+1) == -1],
> then F(n) is prime.
>
> n= 2; F(n)= 17;  17 mod 3 == -1; 17 mod 1 == 0; 17 is prime!
> n= 3; F(n)= 257;  257 mod 5 == +2; 257 mod 3 == -1; 257 is prime!
> n= 4; F(n)= 65537; 65537 mod 9 == -1; 65537 mod 7 == +3; 65537 is prime!
> n= 5; F(n)= 4294967297; F(n) mod 17 == +2; F(n) mod 15 = +2; composite!

I believe you were trying to say something like this (at least that's what
your sample calculation suggests):

If n >= 2 and F(n) = 2^(2^n) + 1. then, if either of the following two
conditions holds, F(n) is a prime.
Condition 1: F(n) mod (2^(n-1)+1) == -1
Condition 2: F(n) mod (2^(n-1)-1) == -1

> I wish someone had the technical expertise to prove it; it's valid, and
> I've studied it... trying to come up with a proof.

> it requires someone as formidable as Lucas, Lehmer, etc. to prove it.
> examples...

... and it takes a few milliseconds in Pari to prove it... wrong.

Let n = 16. Then, we have
2^(n-1)+1 = 32769,
F(n) = 2^(2^16) + 1,
F(n) mod 32769 = -1.

Yet, F(n) mod 825753601 = 0, thereby proving compositeness of F(n) (in
order to avoid nitpickers, note that F(n) > 825753601).

Myth^H^H^H^HConjecture busted!

Peter

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

A couple of easy counterexamples can be found, for n = 16 and n = 36,
both of which meet your hypothesis, but both of which are known not
to be prime.



Bill Bouris wrote:
> Hello, Professor Caldwell.
>
> I like your website.  Could you please post the following conjecture:
>
> if n>= 2 and F(n)= 2^(2^n)+1, then iff [F(n) mod (2^(n-1)+1) == -1]
>
> (or) [F(n) mod (2^(n-1)+1) == -1], then F(n) is prime.
>
>
> it requires someone as formidable as Lucas, Lehmer, etc. to prove it.
>
> examples...
> n= 2; F(n)= 17;  17 mod 3 == -1; 17 mod 1 == 0; 17 is prime!
>
> n= 3; F(n)= 257;  257 mod 5 == +2; 257 mod 3 == -1; 257 is prime!
>
> n= 4; F(n)= 65537; 655377 mod 9 == -1; 65537 mod 7 == +3; 65537 is prime!
>
> n= 5; F(n)= 4294967297; F(n) mod 17 == +2; F(n) mod 15 = +2; composite!
>
> I wish someone had the technical expertise to prove it; it's valid, and
> I've studied it... trying to come up with a proof.  Share it with a close
> colleague, if you like.
>
> Thanks in advance,
>
> Bill Bouris
>
>
>
>
>
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>
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> The Prime Pages : http://www.primepages.org/
>
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>
>
>
>
>

Hello, Professor Caldwell.

I like your website.  Could you please post the following conjecture:

if n>= 2 and F(n)= 2^(2^n)+1, then iff [F(n) mod (2^(n-1)+1) == -1]

(or) [F(n) mod (2^(n-1)+1) == -1], then F(n) is prime.


it requires someone as formidable as Lucas, Lehmer, etc. to prove it.

examples...
n= 2; F(n)= 17;  17 mod 3 == -1; 17 mod 1 == 0; 17 is prime!

n= 3; F(n)= 257;  257 mod 5 == +2; 257 mod 3 == -1; 257 is prime!

n= 4; F(n)= 65537; 655377 mod 9 == -1; 65537 mod 7 == +3; 65537 is prime!

n= 5; F(n)= 4294967297; F(n) mod 17 == +2; F(n) mod 15 = +2; composite!

I wish someone had the technical expertise to prove it; it's valid, and
I've studied it... trying to come up with a proof.  Share it with a close
colleague, if you like.

Thanks in advance,

Bill Bouris

---- Chris Caldwell <caldwell@...> wrote:
> > I must test numbers in the form k*2^n+1.
> > Is Pfgw faster than prp.exe?
>
> Others will correct me if I am wrong, but I think they use the same
> arithmetic engine now.  Of course prp.exe does not prove primality.  LLR
> might be the fastest depending on k, but the key thing to do is to
> prescreen well--that is where you can save a great deal of time.

For base 2, I recommend LLR, since it will automatically do a Proth test for
k*2^n+1 numbers.  You could also use PFGW, but you would need to force a
primality test by using the -tm option.  The problem with PFGW is that it will
not produce a residue for primality tests.  LLR produces a residue for all
tests.  LLR should be just as fast as PFGW for this base.

I suspect that George is keeping PRP up to date with his changes to gwnum, but I
think he has slowly been moving those functions to Prime95.  As Chris said, PRP
cannot do a primality test.

--Mark

> I must test numbers in the form k*2^n+1.
> Is Pfgw faster than prp.exe?

Others will correct me if I am wrong, but I think they use the same
arithmetic engine now.  Of course prp.exe does not prove primality.  LLR
might be the fastest depending on k, but the key thing to do is to
prescreen well--that is where you can save a great deal of time.