For those not familiar with Wilson's Theory: P is a prime number if,
and only if, (P-1)! + 1 is divisible by P.

Now take the number 1, (1-1)! + 1 = 0! + 1 = 2.
2 is divisible by 1, so why isn't 1 prime?

So I was looking at the Mersenne numbers yesterday and
found a curious property:  3^(2^(p-1)) = -3 (mod 2^p -
1) if (2^p - 1) divides S(p-1).

For the sake of reference, I'm using S(1) = 4, S(n+1)
= S(n)^2 - 2.  Also, by 'Mersenne number' I mean "let
p be prime, then 2^p - 1 is a Mersenne number, namely
M_p".

I got this result by taking the proof of the
sufficiency of the Lucas-Lehmer test (see
http://www.utm.edu/research/primes/notes/proofs/LucasLehmer.html)
and doing some squaring and modding.  I haven't been
able to show (yet?) whether the property I found is a
sufficient condition for primality, but so far it
hasn't found any false positives or missed any actual
primes (tested up through 521 at the time of this
writing).  It does seem to be (at least) a necessary
condition.

Of course... now that I look more closely at my code,
this test appears to be one cycle longer than the
Lucas-Lehmer test, so maybe it isn't really worth
pursuing.  Hard numbers: it was 6 seconds less time to
check M_11213 (94 seconds for Lucas-Lehmer, 88 for my
method, both using algorithms written in Lisp, run
under SBCL 0.8.19 on a 566MHz Celeron), and a little
less memory too...

Joseph.



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I've started looking at at n^a + (n+1)^b, where a,b <= n+1 .

It generates a reasonable number of primes. I'll be examining for any
patterns in prime generation and will let you know of any earth
shattering findings.

But one thing right off that bat which I find somewhat intriguing is
this special case:

n^n + (n+1)^(n+1)

Examples:

1^1 + 2^2 = 5 (prime)
2^2 + 3^3 = 31 (prime)
3^3 + 4^4 = 283 (prime)

Seems like it could produce a good amount of primes at this rate. But
in what appears to be an odds defying feat, no more primes turn up
for any n up to 500.   Perhaps there are no more, for *some* reason.


Mark

In a message dated 01/03/2005 18:12:37 GMT Standard Time,
mark.underwood@... writes:

But one  thing right off that bat which I find somewhat intriguing is
this special  case:

n^n + (n+1)^(n+1)

Examples:

1^1 + 2^2 = 5  (prime)
2^2 + 3^3 = 31 (prime)
3^3 + 4^4 = 283 (prime)

Seems like  it could produce a good amount of primes at this rate. But
in what appears  to be an odds defying feat, no more primes turn up
for any n up to  500.   Perhaps there are no more, for *some*  reason.




Mark,

There is no special "reason" preventing more primes from turning up - it's
just the way the cookie crumbles sometimes.

In Sep 2003 I searched both the following forms up to n=10146 with
PrimeFormGW:-
(n+1)^(n+1) + n^n
and
(n+1)^(n+1) - n^n.

The primes I found were:-
2^2+1^1
3^3+2^2
4^4+3^3
and

2^2-1^1
3^3-2^2
4^4-3^3
7^7-6^6

The PRPs were:-

11^11-10^10
17^17-16^16
106^106-105^105
120^120-119^119
7918^7918-7917^7917

The last of these had already been found earlier (in 2001) by Henri
Lifchitz, who has also recently made a bit of a speciality of the form x^x + y^y
-
see the entries in his Top 5000 PRPs database
_http://ourworld.compuserve.com/homepages/hlifchitz/_
(http://ourworld.compuserve.com/homepages/hlifchitz/)

So yes, you're right, the "+" version is not very productive :-(

-Mike Oakes



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

Here is the definition of Wilson's Theorem from the Prime Pages:
http://www.utm.edu/research/primes/notes/proofs/Wilsons.html

	 Wilson's Theorem.
	 Let p be an integer greater than one. p is prime if and only if (p-1)! = -1
(mod p).

You seem to have left out the first part.

Tom Hadley

-----Original Message-----
From: paxtontravis [mailto:paxtontravis@...]
Sent: Tuesday, March 01, 2005 11:06 AM
To: primenumbers@yahoogroups.com
Subject: [PrimeNumbers] Wilson's Theory and the number 1




For those not familiar with Wilson's Theory: P is a prime number if,
and only if, (P-1)! + 1 is divisible by P.

Now take the number 1, (1-1)! + 1 = 0! + 1 = 2.
2 is divisible by 1, so why isn't 1 prime?






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Yahoo! Groups Links

--- In primenumbers@yahoogroups.com, Chris Caldwell <caldwell@u...>
wrote:
> At 01:43 PM 2/27/2005, Décio Luiz Gazzoni Filho wrote:
> >> 2. Why do people choose to broadcast their solutions to this
problem
> >> on the internet?
> >
> >Beats me. But it makes no difference, because anyone who had a
chance in hell
>
> Depends what you mean by broadcast.  Virtually all mathematics
that will eventually be in reputable journals is first circulated by
preprint; often on the Internet.  Mathematicians often say if you
are waiting to read journals then you are two-years behind the
current research.
> Example: "Primes in P" was circulated on the net, gained
acceptance,
> was improved on repeatedly, all well before its first publication.
>
> You circulate in any media to share what you have.  Often you get
corrections
> and suggestions.  Sometimes references that you missed.  (Amateurs
> often miss the point of references--you must show you know how your
> works fits in with the current literature.)
>
> But if instead you mean why do they have "Goldbach Proof" websites.
> That is because no one would publish their trash.  Audience is the
> key--who are you writing to?

I do hate to being called a crackpot by people who DON'T think what
Chris is saying here is a good way to gain acceptance. I just wish
there where more people who could help me when I have questions and
willing to help me find the answers. For example I have recieved the
following fowarded message which I can not answer. I wonder if any
of you can. If not who and where can I ask?



Can you answer the following?

John

---------
From: John W. Nicholson <reddwarf2956@...>
To: caldwell@...
Subject: Questions
Sent: Friday, January 28, 2005 7:24 PM
I have many question. ;-)

On your page with Fortunate numbers:
http://primes.utm.edu/glossary/page.php?sort=FortunateNumber

Is there a term for prime numbers not in both Fortunate number and
the
Fortunate number list? I am thinking of calling them unfortunate
prime numbers.
Simular, but for primes that show up more than once, like 61 called
very
fortunate numbers. Are there an infinete number of both types?

  Is there a relationship between composite pair of p#+/-1 and the
size of the
gap mentioned? With both of these numbers composite there is a large
gap
between the less Fortunate number and the Fortunate number. How big
is it?

I know there is a relationship between the gap with p#+/-1 and the p!
+/-1 gap.
I mean if p#+/-1 is prime, then there is a gap >= the gap of size p.
Because p!
has a gap after p!+1 of at least p, realize that n!/n# = integer.
And For any N, the sequence (N + 1)! + 2, (N + 1)! + 3, ..., (N +
1)! + N + 1
  the gap is of size n.
From: Paulo Ribenboim's book The Little Book of Big Primes. page 142

In your paper on http://primes.utm.edu/references/refs.cgi?
long=CG2000 states
some proofs, namely theorems 2.4 and 2.5, which relates to p# or
some number <
p#, right? I mean there is a prime q which has the same congruences
to p! as to
p#.

Can you state theorems 2.4 and 2.5 on web pages connected
with "factorial
primes"? You might want to link the primorial prime pages to it to
the
Fortunate numbers?

Also, you might add to the page
http://primes.utm.edu/glossary/page.php?sort=PrimeGaps a comment
about the gap
stated above. I know that it is not the smallest gap  because 89 is
the first
the gap which follows the sum of two gaps larger than the n! gap.
With "sum of
two gaps" I mean while looking at largest number of the smallest
factor of
composites of the gap. This may sound confusing, so restating.

For a gap, d, for a prime number p_n to p_(n+1), there are d numbers
which are
composite to primes smaller than p_n and p_(n+1). Each of these
composites have
a factor f_d which is the smallest factor. The largest f_d in the
gap makes the
final gap size d by summing the gap that was before and after. This
was for the
d after p_n there is another one for d before p_n too. Is there a
previously
stated gap function with these terms? If not, why not?

More questions with http://primes.utm.edu/references/refs.cgi?
long=CG2000

In Theorem 2.4 is
i) n divides 1!-1 and 0!-1
Why is this statement needed? I mean (1!-1)/n = 0 and (0!-1)/n = 0,
right?


Table 4:
  1) What reason are the other primes < 499 not listed?
  2) Why are there pairs? (My guess is that there is a relationship
with
quadratic congruences.)
  3) Is there a proof which states that the individual pairing (the
ones in
which p = n - k -1) is always even-even or odd-odd but the total
pairing does
not have to be all even or all odd? Why are they not mixed?
  4) With my guess with quadratic congruences, what is relationship
of the signs
of the units added with quadratic congruences?
  5) No squares with pairings. Why? Also no n^m? Why, again?
  6) Is there a correlation between the factor and the sign of the
unit?
  7) Is there a correlation between the factorial and the sign of the
unit?
  8) Is there a correlation between the k value and the sign of the
unit?
  9) Are there tables of factors of factorial numbers n!+/-1
somewhere? (at
least to 4000, to get pass the Wilson and Wieferich primes 563,
1093, and 3511,
with and without the pairing)
10) With twins, how does the above questions relate?
11) I see with 16!+1 both 137 and 139 are factors, does this mean
anything with
twins? What about 16!+1 pairs 120!-1 and 122!-1?
12) I see with prime 61 both 16!+1 and 18!+1 are divisible
factorials, does
this mean anything with twins? Again, with the factorial pairs
respectfully?
13) What about with: Theorem: (Clement 1949)
The integers n, n+2, form a pair of twin primes if and only if
  4[(n-1)!+1] = -n (mod n(n+2))?
How about with this theorem with the theorems 2.4 and 2.5 and the
one here
http://www.primepuzzles.net/problems/prob_002.htm?

From your page: http://primes.utm.edu/glossary/page.php?
sort=TwinPrime and the
book stated above too.
14) What about Wilson primes, is there anything relating them to
(mod n^2)?

Can the theorem state in Q13) be combined with Proth's? Maybe some
other prime test, like Fermat's little theorem or a strong psp test?

Can you make a web page on the inclusion/exclusion principle
relitive to primes? You could show why all attempts so far with the
twin primes conjecture has failed.

"Erdos proved that there exist at least one prime of the form 4k+1
and at least one prime of the form 4k+3 between n and 2n for all n >
6."
This was stated at:
Eric W. Weisstein. "Modular Prime Counting Function." From MathWorld-
-A Wolfram Web Resource.
http://mathworld.wolfram.com/ModularPrimeCountingFunction.html

Does anyone know where is this proof is?
Second, how does this NOT prove Mark's question?

I mean when can with n > 6 and m^2 >6,
n < p1 < m^2 < (m+1)^2 = m^2 + 2*m + 1 < p2 < 2*n,
happen without a prime inside p1 < m^2 < (m+1)^2 = m^2 + 2*m + 1 <
p2?

Note n > 6 and m^2 >6:

If  m^2 < n < 2*n <(m+1)^2 then a primes exist inside the squares by
Erdos's and by Bertrand's.

If n < m^2 < 2*n < (m+1)^2 and both Erdos primes are n < p1 < p2 <
m^2 then a prime exist inside the squares by at least Bertrand's
Postulate for prime p3.

If n < m^2 < 2*n < (m+1)^2 and a Erdos prime is m^2 < p1 < p2 < 2*n
or p1 < m^2 < p2 < 2*n then a prime exist inside the squares.

If  m^2 < n < (m+1)^2 < 2*n and a Erdos prime is m^2 < p1 < p2 < 2*n
or m^2 < p1 < 2*n < p2 then a prime exist inside the squares.

If m^2 < n <(m+1)^2 < 2*n and both prime are between (m+1)^2 and 2*n
then Bertrand's Postulate holds.

If m < (m+1)^2 < n < 2*n  or n < 2*n < m < (m+1)^2, then we have bad
unrelated sets in this examination, But a prime p1 < m^2 or (m+1)^2
< p2 will be near enough to m^2 or (m+1)^2 so that p3 can be a
solution by Bertrand's (or by Bertrand's used multiple times).

Note that n*(n+1) can be subtituded in the logic above because of
the two nearby primes.


As for the also part, Mark see the these pages:

http://mathworld.wolfram.com/AndricasConjecture.html
http://mathworld.wolfram.com/OmegaConstant.html

Look at the generalization of Andrica's conjecture considers the
equation

p_(n+1)^x - p_(n)^x = 1

I know that the sign has been changed for your case, but it may give
you insight.

Nick


--- In primenumbers@yahoogroups.com, "Mark Underwood"
<mark.underwood@s...> wrote:
>
>
> It's true that there is at least one prime between n^2 and (n+1)
^2,
> although I can't recall if it has been proven.
>
>
> Now n*(n+1) lies between n^2 and (n+1)^2, and we observe a tighter
> result, namely that for n>1 there is always a prime between
>
> n^2 and n*(n+1)
>
> and also between
>
> n*(n+1) and (n+1)^2
>
> (This is yet another way to divide all odd primes into two groups!)
>
>
> ******************
>
>
> Also, I was toying with a minimum  value of c such that there is
> always a prime between n^c and (n+1)^c.
>
> I'm not totally certain but pretty sure that the lowest value of c
is
> about 1.5683. Some lower values come very close but no cigar.
>
>
> For example, there is one prime between 12^1.5683 and 13^1.5683.
That
> is, 53 lies between 49.258 and 55.846.
>
>
> Let s(n) be the number of primes between n^1.5683 and (n+1)
^1.5683.
>
>
> s(n) for n=1 to n=14 is 1,2,1,1,1,2,1,2,1,1,2,1,2,1.
>
> s(n) = 1 for 18 values of n from 1 to 100.
>
> s(n) = 1 for 33 values of n from 1 to 737. After that s(n) is
always
> greater than 1.
>
>
> Mark

The Erdos proof is available at Ask Dr. Math - not that I understand it!

Bob

nick_honolson <nick_honolson@...> wrote:

"Erdos proved that there exist at least one prime of the form 4k+1
and at least one prime of the form 4k+3 between n and 2n for all n >
6."
This was stated at:
Eric W. Weisstein. "Modular Prime Counting Function." From MathWorld-
-A Wolfram Web Resource.
http://mathworld.wolfram.com/ModularPrimeCountingFunction.html

Does anyone know where is this proof is?
Second, how does this NOT prove Mark's question?

I mean when can with n > 6 and m^2 >6,
n < p1 < m^2 < (m+1)^2 = m^2 + 2*m + 1 < p2 < 2*n,
happen without a prime inside p1 < m^2 < (m+1)^2 = m^2 + 2*m + 1 <
p2?

Note n > 6 and m^2 >6:

If  m^2 < n < 2*n <(m+1)^2 then a primes exist inside the squares by
Erdos's and by Bertrand's.

If n < m^2 < 2*n < (m+1)^2 and both Erdos primes are n < p1 < p2 <
m^2 then a prime exist inside the squares by at least Bertrand's
Postulate for prime p3.

If n < m^2 < 2*n < (m+1)^2 and a Erdos prime is m^2 < p1 < p2 < 2*n
or p1 < m^2 < p2 < 2*n then a prime exist inside the squares.

If  m^2 < n < (m+1)^2 < 2*n and a Erdos prime is m^2 < p1 < p2 < 2*n
or m^2 < p1 < 2*n < p2 then a prime exist inside the squares.

If m^2 < n <(m+1)^2 < 2*n and both prime are between (m+1)^2 and 2*n
then Bertrand's Postulate holds.

If m < (m+1)^2 < n < 2*n  or n < 2*n < m < (m+1)^2, then we have bad
unrelated sets in this examination, But a prime p1 < m^2 or (m+1)^2
< p2 will be near enough to m^2 or (m+1)^2 so that p3 can be a
solution by Bertrand's (or by Bertrand's used multiple times).

Note that n*(n+1) can be subtituded in the logic above because of
the two nearby primes.


As for the also part, Mark see the these pages:

http://mathworld.wolfram.com/AndricasConjecture.html
http://mathworld.wolfram.com/OmegaConstant.html

Look at the generalization of Andrica's conjecture considers the
equation

p_(n+1)^x - p_(n)^x = 1

I know that the sign has been changed for your case, but it may give
you insight.

Nick


--- In primenumbers@yahoogroups.com, "Mark Underwood"
<mark.underwood@s...> wrote:
>
>
> It's true that there is at least one prime between n^2 and (n+1)
^2,
> although I can't recall if it has been proven.
>
>
> Now n*(n+1) lies between n^2 and (n+1)^2, and we observe a tighter
> result, namely that for n>1 there is always a prime between
>
> n^2 and n*(n+1)
>
> and also between
>
> n*(n+1) and (n+1)^2
>
> (This is yet another way to divide all odd primes into two groups!)
>
>
> ******************
>
>
> Also, I was toying with a minimum  value of c such that there is
> always a prime between n^c and (n+1)^c.
>
> I'm not totally certain but pretty sure that the lowest value of c
is
> about 1.5683. Some lower values come very close but no cigar.
>
>
> For example, there is one prime between 12^1.5683 and 13^1.5683.
That
> is, 53 lies between 49.258 and 55.846.
>
>
> Let s(n) be the number of primes between n^1.5683 and (n+1)
^1.5683.
>
>
> s(n) for n=1 to n=14 is 1,2,1,1,1,2,1,2,1,1,2,1,2,1.
>
> s(n) = 1 for 18 values of n from 1 to 100.
>
> s(n) = 1 for 33 values of n from 1 to 737. After that s(n) is
always
> greater than 1.
>
>
> Mark





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

Thank you Nick, that was helpful! Erdos figured prominently as an
amazing 'networker' in the book I recently read called The Music of
the Primes. And that Wolfram site is amazing.

Mike, I see now that even lesser minds can think alike to greater
minds! One wonders how much of this stuff has been hashed through
even hundreds of years go. Oh well the joy is in the (re)discovery.

You are waxing philosophical when you say

"There is no special "reason" preventing more primes from turning up -
it's just the way the cookie crumbles sometimes."

Perhaps, yet maybe *God* knows the 'reason' why cookies crumble
exactly the way they do :-)


Mark







--- In primenumbers@yahoogroups.com, "nick_honolson"
<nick_honolson@y...> wrote:
>
> "Erdos proved that there exist at least one prime of the form 4k+1
> and at least one prime of the form 4k+3 between n and 2n for all n
>
> 6."
> This was stated at:
> Eric W. Weisstein. "Modular Prime Counting Function." From
MathWorld-
> -A Wolfram Web Resource.
> http://mathworld.wolfram.com/ModularPrimeCountingFunction.html
>
> Does anyone know where is this proof is?
> Second, how does this NOT prove Mark's question?
>
> I mean when can with n > 6 and m^2 >6,
> n < p1 < m^2 < (m+1)^2 = m^2 + 2*m + 1 < p2 < 2*n,
> happen without a prime inside p1 < m^2 < (m+1)^2 = m^2 + 2*m + 1 <
> p2?
>
> Note n > 6 and m^2 >6:
>
> If  m^2 < n < 2*n <(m+1)^2 then a primes exist inside the squares
by
> Erdos's and by Bertrand's.
>
> If n < m^2 < 2*n < (m+1)^2 and both Erdos primes are n < p1 < p2 <
> m^2 then a prime exist inside the squares by at least Bertrand's
> Postulate for prime p3.
>
> If n < m^2 < 2*n < (m+1)^2 and a Erdos prime is m^2 < p1 < p2 <
2*n
> or p1 < m^2 < p2 < 2*n then a prime exist inside the squares.
>
> If  m^2 < n < (m+1)^2 < 2*n and a Erdos prime is m^2 < p1 < p2 <
2*n
> or m^2 < p1 < 2*n < p2 then a prime exist inside the squares.
>
> If m^2 < n <(m+1)^2 < 2*n and both prime are between (m+1)^2 and
2*n
> then Bertrand's Postulate holds.
>
> If m < (m+1)^2 < n < 2*n  or n < 2*n < m < (m+1)^2, then we have
bad
> unrelated sets in this examination, But a prime p1 < m^2 or (m+1)^2
> < p2 will be near enough to m^2 or (m+1)^2 so that p3 can be a
> solution by Bertrand's (or by Bertrand's used multiple times).
>
> Note that n*(n+1) can be subtituded in the logic above because of
> the two nearby primes.
>
>
> As for the also part, Mark see the these pages:
>
> http://mathworld.wolfram.com/AndricasConjecture.html
> http://mathworld.wolfram.com/OmegaConstant.html
>
> Look at the generalization of Andrica's conjecture considers the
> equation
>
> p_(n+1)^x - p_(n)^x = 1
>
> I know that the sign has been changed for your case, but it may
give
> you insight.
>
> Nick
>
>
> --- In primenumbers@yahoogroups.com, "Mark Underwood"
> <mark.underwood@s...> wrote:
> >
> >
> > It's true that there is at least one prime between n^2 and (n+1)
> ^2,
> > although I can't recall if it has been proven.
> >
> >
> > Now n*(n+1) lies between n^2 and (n+1)^2, and we observe a
tighter
> > result, namely that for n>1 there is always a prime between
> >
> > n^2 and n*(n+1)
> >
> > and also between
> >
> > n*(n+1) and (n+1)^2
> >
> > (This is yet another way to divide all odd primes into two
groups!)
> >
> >
> > ******************
> >
> >
> > Also, I was toying with a minimum  value of c such that there is
> > always a prime between n^c and (n+1)^c.
> >
> > I'm not totally certain but pretty sure that the lowest value of
c
> is
> > about 1.5683. Some lower values come very close but no cigar.
> >
> >
> > For example, there is one prime between 12^1.5683 and 13^1.5683.
> That
> > is, 53 lies between 49.258 and 55.846.
> >
> >
> > Let s(n) be the number of primes between n^1.5683 and (n+1)
> ^1.5683.
> >
> >
> > s(n) for n=1 to n=14 is 1,2,1,1,1,2,1,2,1,1,2,1,2,1.
> >
> > s(n) = 1 for 18 values of n from 1 to 100.
> >
> > s(n) = 1 for 33 values of n from 1 to 737. After that s(n) is
> always
> > greater than 1.
> >
> >
> > Mark

Anybody up to checking some calculations??
Sure would be appreciated.

See http://www.opertech.com/primes/residues.html
in particular the value of C2 (think it is a little large)

Tom

Hi, Mark,

An interesting expression.

Let  np ( n ) = n^n + (n+1)^(n+1) .
I confirm that  n = 1, 2, and 3  are the only known primes of this form.

If  n mod 6  =  4,
then  3  and  ( n^2 + n + 1 ) / 3  (twice)  are divisors.

If  n = 415 ,  then  np ( n )  has only 2 prime factors
and one of them is  29 .

You might want to look at posts on this subject to NMBRTHRY, e.g.,
http://listserv.nodak.edu/scripts/wa.exe?A2=ind0112&L=nmbrthry&P=R403
and
http://listserv.nodak.edu/scripts/wa.exe?A2=ind0209&L=nmbrthry&P=R1218

Also, you might want to try to fill in the gap left by the reply of
Ismael Jimenez Calvo in explaining the production of the interesting
factors cited.

There are some sequences in the On-line Encyclopedia.  E.g.,
http://www.research.att.com/projects/OEIS?Anum=A056187.

Cheers.

Walter Nissen

Hi,

If a solution to GC was not couched in formal mathematics but was
let's say based on observation and a little arithmetic, easy to
understand and short; would that be acceptable for peer review as a
solution and ultimately publication.

I am aware that the defintion of "easy to understand" is (as all
definitions) relative; however please bare with me on that definition.

I understand that such a format may be acceptable for peer review
(especially if it were short and easy to understand), but not for
publication; however I would be concerned that if that were the case,
then someone could translate my work into formal mathematics and then
claim the solution as their own.

Or extract seminal ideas from the work and integrate it more formally
into their own presentation.

I am aware that expressing such misgivings may be interpreted by some
as paranoid.

I am also aware that any post which assumes a solution will be
considered with some disdain and condescension by many.

However the question is asked.

Chris

Chris

On Friday 04 March 2005 09:50, you wrote:
> Hi,
>
> If a solution to GC was not couched in formal mathematics but was
> let's say based on observation and a little arithmetic, easy to
> understand and short; would that be acceptable for peer review as a
> solution and ultimately publication.

Why don't you go ahead and submit it? Supposing you actually have a valid
proof of Goldbach's, I'm pretty sure any journal editor would go out of their
way to format your paper and have it published in his journal (instead of
someone else's).

> I am aware that the defintion of "easy to understand" is (as all
> definitions) relative; however please bare with me on that definition.

s/bare/bear/

> I understand that such a format may be acceptable for peer review
> (especially if it were short and easy to understand), but not for
> publication; however I would be concerned that if that were the case,
> then someone could translate my work into formal mathematics and then
> claim the solution as their own.

See above.

> Or extract seminal ideas from the work and integrate it more formally
> into their own presentation.
>
> I am aware that expressing such misgivings may be interpreted by some
> as paranoid.

It is. If you had posted it here someone would already have debunked it into
oblivion. But that's what your afraid of, right? When you try to submit your
pseudo-proof to a journal and it gets rejected, you're going to tell yourself
that it was because you didn't `couch it in formal mathematics,' right?

Face it, YOU DON'T HAVE A VALID PROOF.

> I am also aware that any post which assumes a solution will be
> considered with some disdain and condescension by many.

That's expected. How can you be so pretentious as to think that you have a
proof that all math geniuses missed, using nothing but `a little arithmetic,
easy to understand and short'? The very idea is laughable.

Andrew Wiles remarked that he wanted to prove FLT ever since his childhood;
luckily he didn't insist on using high-school-level mathematics only,
otherwise we'd be still waiting for a proof.

> However the question is asked.

And our time is wasted, again.

Décio


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

I UNDERSTOOD THE LOGIC BUT JUST CANT FINISH IT...
IT IS SIMILAR TO THE FROBENIUS PROBLEM ONLY THAT THIS QUESTION DEALS
WITH THREE NUMBERS
>
>
> VIVEK AND AISHWARYA LIMITED SELLS BISCUITS.
> THEY SELL BISCUITS IN PACKETS OF 56, 95 AND 145.
> (THAT MEANS IT IS POSSIBLE TO BUY 56+95=151 BISCUITS BUT NOT 200
> BISCUITS EXACTLY)
> I WENT TO THEIR SHOP EXPECTING AISHWARYA TO BE SELLING BISCUITS. I
> HAD PLANNED TO PURCHASE A LARGE QUANTITY OF BISCUITS SO THAT I
COULD
> IMPRESS AISHWARYA BEING A GOOD CUSTOMER.
> BUT WHEN I REACHED THERE I SAW VIVEK AT THE COUNTER.
> I COULD NOT POSSIBLY GO OUT WITHOUT BUYING ANYTHING FROM THE SHOP
AS
> IT WOULD CREATE A BAD IMPRESSION.
> SO I DECIDED TO ORDER THE LARGEST QUANTITY OF BISCUITS THAT VIVEK
> WOULD NOT BE ABLE TO PROVIDE ME WITH.
>
> AND I WOULD GO OUT WITHOUT HAVING TO BUY ANYTHING.
> HOW MUCH SHOULD I ORDER?
>
>
> (according to me-
> if you can find x in
>  56a+95b+145c= x
> where x is the largest possible prime number according to the given
> constraints you would have solved the question)

WHAT ARE THE MINIMUM NUMBER OF FACTORS THAT THE EXPRESSION
n^4-5*(n^2)+4
WILL HAVE IF N IS A SET OF PRIME NUMBERS GREATER THAN 500

On Friday 04 March 2005 12:28, you wrote:
> WHAT ARE THE MINIMUM NUMBER OF FACTORS THAT THE EXPRESSION
> n^4-5*(n^2)+4
> WILL HAVE IF N IS A SET OF PRIME NUMBERS GREATER THAN 500

It helps if you factor n^4 - 5n^2 + 4 as (n-2)(n-1)(n+1)(n+2). Right off the
bat you already ensure that it has 4 factors, but you can do better. If n is
a prime, then it's either == 1 or 2 mod 3; in the former case, both n-1 and
n+2 are divisible by 3, and in the latter, both n-2 and n+1 are divisible by
3. Also, obviously n-1 and n+1 are even, but furthermore, one of these is
divisible by 4, so that (n-1)(n+1) = n^2 - 1 is divisible by 8. So you have
five factors already, 2*2*2*3*3. Finally, and again by the pigeonhole
principle and the fact that n is prime, one of the factors of your polynomial
are divisible by 5. So the values of your polynomial are always divisible by
2*2*2*3*3*5 when n is a prime > 5.

This is the best we can do regarding factors of 2, 3 and 5. For instance, the
polynomial evaluated at n = 509 is divisible by 8 but not 16, 9 but not 27
and 5 but not 25.

Since (for large enough n) none of this is going to fully cancel out any of
the original factors of the polynomial, there's going to be at least 4 extra
factors. The first n > 500 for which this lower bound is reached is n = 787,
since

n-2 = 785 = 5*157,
n-1 = 786 = 2*3*131,
n+1 = 788 = 2^2*197,
n+2 = 789 = 3*263.

The values n = 1229, 1669, 2083, 3371, 3461, 8221, 9013, ... also have only 4
cofactors upon division by 2*2*2*3*3*5. This PARI/GP script can be used to
compute the remaining values (just replace minp and maxp by whatever bounds
interest you):

forprime(n=minp,maxp,
   f = factorint(eval(n^4 - 5*n^2 + 4)/(2^3*3^2*5));
   k=sum(i=1,matsize(f)[1],f[i,2]);
   if(k==4,print(n" "f" "k));
)

With some modifications, you can use it to study other quantities related to
your problem.

To sum it up, the minimum number of factors with multiplicity is 10. I'm
fairly sure the minimum number of distinct factors is 7, but I haven't proved
it -- it's up to you if you're interested.

Please observe that these results are not valid for small n, though; consider
n = 7 (3 distinct factors), 11 (4 distinct factors), 13 (5 distinct factors),
37 (6 distinct factors). Considering multiplicity, n = 7 has 8 factors and n
= 19 has 9 factors. This happens because my assumption above (that none of
the cofactors would be fully canceled out by the obligatory factors of
2*2*2*3*3*5) was violated. The largest n for which the assumption is violated
is n = 59, since

n-2 = 57 = 3*19
n-1 = 58 = 2*29
n+1 = 60 = 2^2*3*5 (this is where the assumption breaks down)
n+2 = 61 is prime.

However, you said you were interested in n > 500.

Décio


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

I think the below is true, but I can't prove or disprove the
statement. Please help.

If 3^(3^n)+3^(((3^n)+1)/2)+1 is prime then is 3^(3^n)-3^(((3^n)+1)/2)
+1 also prime, and vice-versa? I can't find any counter examples.

true for n=0,1,2 up to 10(Till where I checked).

Note:- The above terms are factors of 3^(3^(n+1))+1, hence will have
special factors only.

Also, if 2*3^n+1 is prime then it divides these kind of numbers.

I am also interested in sieving for these numbers, try to find the
smallest factor for each, does anyone have any software available,
or could write me one?


Thanks,
Harsh Aggarwal

Define sequence a(n+1)=n*f(n)+a(n), a(1)=2, f(n) is nth term of
fibonacci sequence. The first few terms are 2,3,7,13,25,73...........
Notice that all terms except 25 are primes. I think that other terms
of the sequence are even so I think there will be no more primes. What
do you think?

On Friday 04 March 2005 15:33, you wrote:
> Define sequence a(n+1)=n*f(n)+a(n), a(1)=2, f(n) is nth term of
> fibonacci sequence. The first few terms are 2,3,7,13,25,73...........
> Notice that all terms except 25 are primes. I think that other terms
> of the sequence are even so I think there will be no more primes. What
> do you think?

There seems to be a definite pattern to the parity of n*f(n). Try to work from
there and see what you can prove about your sequence.

Décio


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

On Friday 04 March 2005 15:01, you wrote:
> I think the below is true, but I can't prove or disprove the
> statement. Please help.
>
> If 3^(3^n)+3^(((3^n)+1)/2)+1 is prime then is 3^(3^n)-3^(((3^n)+1)/2)
> +1 also prime, and vice-versa? I can't find any counter examples.
>
> true for n=0,1,2 up to 10(Till where I checked).

Law of small numbers. It's hard to tell if this is true because you can't
check much further than that -- n = 15 is about the limit given current
technology.

Of course, for all I can tell, there could be a proof of this, but I wouldn't
be jumping to conclusions with such a small dataset.

> Note:- The above terms are factors of 3^(3^(n+1))+1, hence will have
> special factors only.
>
> Also, if 2*3^n+1 is prime then it divides these kind of numbers.
>
> I am also interested in sieving for these numbers, try to find the
> smallest factor for each, does anyone have any software available,
> or could write me one?

I could in exchange for some CPU-time for my primo-factorial sieving and
PRP-testing tasks (:

Décio


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

SHOUTING, SHOUTING+1 and SHOUTING-1.  Add these
together, subtract 2, and multiply by the square root
of infinity to get your final result.

J.


--- rajeevbat <rajeevbat@...> wrote:
>
> WHAT ARE THE MINIMUM NUMBER OF FACTORS THAT THE
> EXPRESSION
> n^4-5*(n^2)+4
> WILL HAVE IF N IS A SET OF PRIME NUMBERS GREATER
> THAN 500
>
>
>
>




__________________________________
Celebrate Yahoo!'s 10th Birthday!
Yahoo! Netrospective: 100 Moments of the Web
http://birthday.yahoo.com/netrospective/

At 07:50 AM 3/4/2005, chrisdarroch wrote:

>Hi,
>
>If a solution to GC was not couched in formal mathematics but was
>let's say based on observation and a little arithmetic, easy to
>understand and short; would that be acceptable for peer review as a
>solution and ultimately publication.

If it was a genuine proof, then yes.  It would be monumental.  The problem
is, though, that it is very unlikely that there is such a proof.

>I am also aware that any post which assumes a solution will be
>considered with some disdain and condescension by many.

Well, there are many people who have thought that they have proven it, but
none of the proposed "proofs" have been valid.

There is a Law of Small Numbers in statistics,
but you seem to be using the term incorrectly;
just dropping it in where it makes no sense.

Milton L. Brown
miltbrown at earthlink.net



> [Original Message]
> From: Décio Luiz Gazzoni Filho <decio@...>
> To: <primenumbers@yahoogroups.com>
> Date: 3/4/2005 11:23:19 AM
> Subject: Re: [PrimeNumbers] 3+ question
>
>
> On Friday 04 March 2005 15:01, you wrote:
> > I think the below is true, but I can't prove or disprove the
> > statement. Please help.
> >
> > If 3^(3^n)+3^(((3^n)+1)/2)+1 is prime then is 3^(3^n)-3^(((3^n)+1)/2)
> > +1 also prime, and vice-versa? I can't find any counter examples.
> >
> > true for n=0,1,2 up to 10(Till where I checked).
>
> Law of small numbers. It's hard to tell if this is true because you can't
> check much further than that -- n = 15 is about the limit given current
> technology.
>
> Of course, for all I can tell, there could be a proof of this, but I
wouldn't
> be jumping to conclusions with such a small dataset.
>
> > Note:- The above terms are factors of 3^(3^(n+1))+1, hence will have
> > special factors only.
> >
> > Also, if 2*3^n+1 is prime then it divides these kind of numbers.
> >
> > I am also interested in sieving for these numbers, try to find the
> > smallest factor for each, does anyone have any software available,
> > or could write me one?
>
> I could in exchange for some CPU-time for my primo-factorial sieving and
> PRP-testing tasks (:
>
> Décio
>
>
> [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
>
>
>
>
>
>

Milton Brown wrote:

> There is a Law of Small Numbers in statistics,
> but you seem to be using the term incorrectly;
> just dropping it in where it makes no sense.

Googling "Law of Small Numbers" shows something in good agreement with Decio's
use of the term, e.g.:
http://primes.utm.edu/glossary/page.php?sort=LawOfSmall

Milton's posts on the other hand, seem to often make no sense.

Harsh Aggarwal wrote:

> I think the below is true, but I can't prove or disprove the
> statement. Please help.
>
> If 3^(3^n)+3^(((3^n)+1)/2)+1 is prime then is 3^(3^n)-3^(((3^n)+1)/2)+1
> also prime, and vice-versa? I can't find any counter examples.
>
> true for n=0,1,2 up to 10(Till where I checked).

Only true for n=0 if 1 is considered a prime. Then n = 0, 1, 2 gives 6 small
primes:  (7, 1), (37, 19), (19927, 19441).
There are no other primes for n<=10. The number of digits approximately
triples for each n. This growth is so fast that there are probably no more
primes at all. That means Harsh's guess is actually likely to be true, but in
an uninteresting way.

PS: Do not use the default primeform Fermat 3-prp test on these base 3
expressions.

--
Jens Kruse Andersen

--- In primenumbers@yahoogroups.com, "Jens Kruse Andersen"
<jens.k.a@g...>wrote:
>Milton Brown wrote:
>
>>There is a Law of Small Numbers in statistics,
>>but you seem to be using the term incorrectly;
>>just dropping it in where it makes no sense.
>
>Googling "Law of Small Numbers" shows something in good agreement
>with Decio's use of the term, e.g.:
>http://primes.utm.edu/glossary/page.php?sort=LawOfSmall
>
>Milton's posts on the other hand, seem to often make no sense.

Cosigned, since the "Law of Small Numbers" makes perfect since in
this case. Looking at the first 3 tiny pairs of numbers could lead
someone to believe that there is some law "causing" this to happen.
Well, there "might" be such a law, but most likely it is due to
blind luck, and such tiny numbers which are statistically rich in
primes.

One could also look at the "plus" form, and make a statement
(although wrong), that since n=0 and n=1 and n=2 are all prime,
then all will be prime (note a Great one, Fermat made such a wild
claim in the past).

Both of the above claims are cases where the "Law of Small Numbers"
causes someone to make a false (or likely false) claim.  However,
for the claim of if "plus" prime, then "minus" prime, (if the
stipulation of n>1 is added to the original question posed),
"might" be true, due to the fact that it is likely there are no
additional primes of either form.  There is no "law" (that I am
aware of), but simple blind luck happened to form these pairs.

>Harsh Aggarwal wrote:
>
>>I think the below is true, but I can't prove or disprove the
>>statement. Please help.
>>
>>If 3^(3^n)+3^(((3^n)+1)/2)+1 is prime then is 3^(3^n)-3^(((3^n)
+1)/2)+1
>>also prime, and vice-versa? I can't find any counter examples.
>>
>>true for n=0,1,2 up to 10(Till where I checked).
>
>Only true for n=0 if 1 is considered a prime. Then n = 0, 1, 2
gives 6
>small primes:  (7, 1), (37, 19), (19927, 19441).
>There are no other primes for n<=10. The number of digits
approximately
>triples for each n. This growth is so fast that there are probably
no
>more primes at all. That means Harsh's guess is actually likely to
be
>true, but in an uninteresting way.
>
>PS: Do not use the default primeform Fermat 3-prp test on these
base 3
>expressions.

I don't know about that PS statement.  It might be good to have some
of the "newer" prime hunters use default Fermat-3 tests.  Frequently
a wrong answer (especially one which might smack someone right in the
forehead) will help teach people willing to learn, better than a
correct answer.

>Jens Kruse Andersen

chrisdarroch <chrisdarr2@...> wrote:
Hi,

If a solution to GC was not couched in formal mathematics but was
let's say based on observation and a little arithmetic, easy to
understand and short; would that be acceptable for peer review as a
solution and ultimately publication.

I am aware that the defintion of "easy to understand" is (as all
definitions) relative; however please bare with me on that definition.

I understand that such a format may be acceptable for peer review
(especially if it were short and easy to understand), but not for
publication; however I would be concerned that if that were the case,
then someone could translate my work into formal mathematics and then
claim the solution as their own.

Or extract seminal ideas from the work and integrate it more formally
into their own presentation.

I am aware that expressing such misgivings may be interpreted by some
as paranoid.

I am also aware that any post which assumes a solution will be
considered with some disdain and condescension by many.

However the question is asked.

Chris

Chris


Publish and be damned!


Bob


---------------------------------
Celebrate Yahoo!'s 10th Birthday!
  Yahoo! Netrospective: 100 Moments of the Web

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

On Sat, 5 Mar 2005, Bob Gilson wrote:
> If a solution to GC was not couched in formal mathematics but was
> let's say based on observation and a little arithmetic, easy to
> understand and short; would that be acceptable for peer review as a
> solution and ultimately publication.
> ...
> I am also aware that any post which assumes a solution will be
> considered with some disdain and condescension by many.


The disdain, at least on my part, is not for assuming a solution,
but for the foolish games those who think they have one often play.
I get tired of "I got the solution, where do I collect money",
"I have a solution, but it is secret", and of course the
insulting "I have a solution and it is simple" which implies
thousands of mathematicians over hundreds of years have missed
the simple.  Insulting and unlikely.

To avoid some disdain do this: take the time and effort to read what
others have done so you can put your "proof" in context and in
the language of your audience.  Then attempt to publish in a
reputible manner.

If too lazy to do that, then try this: write it up as neatly
as you can and post it publically with a note "I think this is a
proof of _______, do you see any problems with it?"  Of course
if it is too outlandishly written "using the z-spectral
reasoning only I understand we get..." or "adding my zero-point
axioms we can prove all unproven theorem" you will get disdain.

Finally, always listen more than you speak.  If most of the posts
on a list on any one subject are from you; it is time to stop talking
about it.

Chris.

Most nuts are incapable of taking correction--if no one can give you a
reasonable correction, then up you nut index by three.  If you
alone have a right answer an the whole world misses the simple truth,
up your nut inex by 8 (10 is the max).  For each incorrect
proof that you retracted appropriately--subtract one from the nut
index.

Hello,

Chris speaks wisdom, take it from a former 8.  Anyone can claim
proof.  Someone who actually has proof does not need to make any
claims whatsoever, they simply have to ask others what is wrong with
it and they will take up the challenge of finding out.  If it's truly
that short and clearly written, it will be digestible and
understandable to many members of this list.  The list has an archive
and world-wide membership as witness.  You may be a nut and you
simply don't know it yet.  You can cure yourself of the condition by
learning as much as you can about your subject matter and about the
realities of the academic mathematical world into which you seek
acceptance.  Make no claims, ask questions.  Any proof, or any
mathematical idea for that matter, is worthless if it is unknown.  If
it's incorrect, it's worthless.  If you don't put it out there for
scrutiny, it's worthless whether it's correct or not.  And
ultimately, it's not your call on whether it is valid proof or not,
it must pass the scrutiny of everyone else in the world, and you have
to start somewhere.  Consider how long it will take you in isolation
to learn enough about math and logic to be able to critically attack
your own paper.  Just a little effort along these lines sometimes
goes a long way toward prevention of self-embarrassment and
spontaneous nut index inflation.

But your safeguard is the presentation, make no claims, ask
questions, thank those who take the time to help you and learn from
them how to save them time on future endeavors.  I guarantee it's
100% O.K., even admirable, to be wrong - UNLESS, you're wrong about
being right.  So simply do not go there.  If no one from this group
can find anything wrong with a posted proof, it will be a rarity
indeed, an event to remember in and of itself.

If you are a mathematician trapped in a nut body, you must and will
educate yourself as the means to escape, otherwise you'll remain a
nut.  How do you educate yourself? - Well start asking questions
about what you see in numbers.  Until you start asking questions and
seeking out answers, you shall remain trapped, and there ye' shall be.

-Dick Boland

> Most nuts are incapable of taking correction--if no one can give
you a
> reasonable correction, then up you nut index by three.  If you
> alone have a right answer an the whole world misses the simple
truth,
> up your nut inex by 8 (10 is the max).  For each incorrect
> proof that you retracted appropriately--subtract one from the nut
> index.

Chris Caldwell <caldwell@...> wrote:
On Sat, 5 Mar 2005, Bob Gilson wrote:
> If a solution to GC was not couched in formal mathematics but was
> let's say based on observation and a little arithmetic, easy to
> understand and short; would that be acceptable for peer review as a
> solution and ultimately publication.
> ...
> I am also aware that any post which assumes a solution will be
> considered with some disdain and condescension by many.


The disdain, at least on my part, is not for assuming a solution,
but for the foolish games those who think they have one often play.
I get tired of "I got the solution, where do I collect money",
"I have a solution, but it is secret", and of course the
insulting "I have a solution and it is simple" which implies
thousands of mathematicians over hundreds of years have missed
the simple. Insulting and unlikely.

To avoid some disdain do this: take the time and effort to read what
others have done so you can put your "proof" in context and in
the language of your audience. Then attempt to publish in a
reputible manner.

If too lazy to do that, then try this: write it up as neatly
as you can and post it publically with a note "I think this is a
proof of _______, do you see any problems with it?" Of course
if it is too outlandishly written "using the z-spectral
reasoning only I understand we get..." or "adding my zero-point
axioms we can prove all unproven theorem" you will get disdain.

Finally, always listen more than you speak. If most of the posts
on a list on any one subject are from you; it is time to stop talking
about it.

Chris.

Most nuts are incapable of taking correction--if no one can give you a
reasonable correction, then up you nut index by three. If you
alone have a right answer an the whole world misses the simple truth,
up your nut inex by 8 (10 is the max). For each incorrect
proof that you retracted appropriately--subtract one from the nut
index.


Hey Chris

I was merely replying to what chrisdarroch wrote, and commented "Publish and be
damned"

So please do not attribute to me that which I did not say.

Many thanks

Bob Gilson








__________________________________________________
Do You Yahoo!?
Tired of spam?  Yahoo! Mail has the best spam protection around
http://mail.yahoo.com

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

richard042@... wrote:

Hello,

Chris speaks wisdom, take it from a former 8.  Anyone can claim
proof.  Someone who actually has proof does not need to make any
claims whatsoever, they simply have to ask others what is wrong with
it and they will take up the challenge of finding out.  If it's truly
that short and clearly written, it will be digestible and
understandable to many members of this list.  The list has an archive
and world-wide membership as witness.  You may be a nut and you
simply don't know it yet.  You can cure yourself of the condition by
learning as much as you can about your subject matter and about the
realities of the academic mathematical world into which you seek
acceptance.  Make no claims, ask questions.  Any proof, or any
mathematical idea for that matter, is worthless if it is unknown.  If
it's incorrect, it's worthless.  If you don't put it out there for
scrutiny, it's worthless whether it's correct or not.  And
ultimately, it's not your call on whether it is valid proof or not,
it must pass the scrutiny of everyone else in the world, and you have
to start somewhere.  Consider how long it will take you in isolation
to learn enough about math and logic to be able to critically attack
your own paper.  Just a little effort along these lines sometimes
goes a long way toward prevention of self-embarrassment and
spontaneous nut index inflation.

But your safeguard is the presentation, make no claims, ask
questions, thank those who take the time to help you and learn from
them how to save them time on future endeavors.  I guarantee it's
100% O.K., even admirable, to be wrong - UNLESS, you're wrong about
being right.  So simply do not go there.  If no one from this group
can find anything wrong with a posted proof, it will be a rarity
indeed, an event to remember in and of itself.

If you are a mathematician trapped in a nut body, you must and will
educate yourself as the means to escape, otherwise you'll remain a
nut.  How do you educate yourself? - Well start asking questions
about what you see in numbers.  Until you start asking questions and
seeking out answers, you shall remain trapped, and there ye' shall be.

-Dick Boland

> Most nuts are incapable of taking correction--if no one can give
you a
> reasonable correction, then up you nut index by three.  If you
> alone have a right answer an the whole world misses the simple
truth,
> up your nut inex by 8 (10 is the max).  For each incorrect
> proof that you retracted appropriately--subtract one from the nut
> index.

Dear Richard

Perhaps you can persuade chrisdarroch to publish his "proof" so that Decio can
have a field day. On the other hand...

Bob Gilson









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