## Conjecture

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• Hello! Let p is prime and q=k*p+1 is prime too, k=2(mod 4). Then q divides 2^p-1 with a probability 1/k. I guess this conjecture is already known. Does
Message 1 of 30 , Jun 17, 2001
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Hello!

Let p is prime and q=k*p+1 is prime too, k=2(mod 4). Then q divides 2^p-1 with a
probability 1/k.

I guess this conjecture is already known. Does somebody know about author? Or

Thanks,

Andrey
• Hi Andrey, ... It has been noticed before, and the shape looks about right, but.... For it to be correct the sum of 1/k; (k=2, k+=2, k
Message 2 of 30 , Jun 18, 2001
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Hi Andrey,

Andrey Kulsha wrote:

> Hello!
>
> Let p is prime and q=k*p+1 is prime too, k=2(mod 4). Then q divides 2^p-1 with a
> probability 1/k.

It has been noticed before, and the shape looks about right, but....

For it to be correct the sum of 1/k; (k=2, k+=2, k< (2^p-1)/p ) and q prime must
equal 1.
I would also expect the products of q/k to equal 2^p-1.

If you rotate the problem slightly and look at it "perversely" (converse, inverse, reverse
etc. already have precise technical meanings :-) and view it as
Given a prime q, for which n does it first divide 2^n-1?
... the approximation 1/k doesn't work.

Having said that, it is a good enough approximation, and I haven't got anything
better :-)

Cheers,
Paul Landon

>
>
> I guess this conjecture is already known. Does somebody know about author? Or
>
> Thanks,
>
> Andrey
• For every (2n,k) pair, k,n positive integers, there exists a prime of the form x^k+2n and also of the form x^k-2n (although the x are probably different for
Message 3 of 30 , Dec 3, 2002
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For every (2n,k) pair, k,n positive integers, there exists a prime of the
form

x^k+2n

and also of the form

x^k-2n

(although the x are probably different for each form).

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths/
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
• ... Do you really think there exists a prime for (2n,k)=(8,3) ??? For which values of x is x^3+8 prime? :-)
Message 4 of 30 , Dec 3, 2002
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--- In primenumbers@y..., "Jon Perry" <perry@g...> wrote:
> For every (2n,k) pair, k,n positive integers, there exists a
> prime of the form
>
> x^k+2n

Do you really think there exists a prime for (2n,k)=(8,3) ???

For which values of x is x^3+8 prime? :-)
• ... x=-1
Message 5 of 30 , Dec 3, 2002
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> For which values of x is x^3+8 prime? :-)
x=-1
• ... My oops. For which values of x is x^3+512 prime? I think that s composite for all integer x. :-)
Message 6 of 30 , Dec 3, 2002
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> Jack Brennen wrote:
> > For which values of x is x^3+8 prime? :-)
> x=-1

My oops.

For which values of x is x^3+512 prime?

I think that's composite for all integer x. :-)
• Also x^3+216, I guess?
Message 7 of 30 , Dec 3, 2002
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Also x^3+216, I guess?
• ... That depends on whether you consider (-7)^3+216 == -127 to be a prime.
Message 8 of 30 , Dec 3, 2002
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> Also x^3+216, I guess?

That depends on whether you consider

(-7)^3+216 == -127

to be a prime.
• ... So, for other than obvious exceptions, the conjecture is true? Jon Perry perry@globalnet.co.uk http://www.users.globalnet.co.uk/~perry/maths/
Message 9 of 30 , Dec 4, 2002
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>For which values of x is x^3+512 prime?
>I think that's composite for all integer x. :-)

So, for other than obvious exceptions, the conjecture is true?

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths/
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
• ... Yes, it is true for all numbers except for those that it is not true for :-) __________________________________________________ Virus checked by
Message 10 of 30 , Dec 5, 2002
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> So, for other than obvious exceptions, the conjecture is true?

Yes, it is true for all numbers except for those that it is not true for :-)

__________________________________________________
Virus checked by MessageLabs Virus Control Centre.
• For n 5, if A(n)={a_i} define the set of a_i n=m Jon Perry perry@globalnet.co.uk
Message 11 of 30 , Apr 11, 2003
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For n>5, if A(n)={a_i} define the set of a_i<n such that n+a_i is prime,
then A(n)=A(m) <=> n=m

Jon Perry
perry@...
http://www.users.globalnet.co.uk/~perry/maths/
BrainBench MVP for HTML and JavaScript
http://www.brainbench.com
• Conjecture: Every prime number Q can be written as Q = absolute(N+-M), where N,M = natural numbers, coprime and products of powers of primes pi^ai (i =
Message 12 of 30 , Mar 1 3:40 AM
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Conjecture:

Every prime number Q can be written as Q = absolute(N+-M), where N,M =
natural numbers, coprime and products of powers of primes pi^ai (i =
1,2,3, ai >= 1), the pi complete up to pn, and Q < (p(n+1))². One of
N,M may as well be = 1.

Ex.: Q = 2*5*11 +- 3*7 = 131, 89. Both prime because < 13²=169.

Ex.: Q = 2*3*5 +- 1 = 31, 29. Both prime because < 7²=49.

Ex.: Q = 2³*7 +- 3²*5 = 101, 11. Both prime because < 11²=121.

Werner
• Hi, John Renze s PARI chg.gp script has just finished a test of 4955-digit number 4^8230-3 = 2^16460-2-1 (as some people prefer). It is the largest prime
Message 13 of 30 , Mar 2 4:44 PM
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Hi,

John Renze's PARI chg.gp script has just finished a test of 4955-digit
number 4^8230-3 = 2^16460-2-1 (as some people prefer). It is the largest
prime number b^n-(b-1) I know.

I have no verifying program, so I'd be very much obliged if someone would
provide me with such program or would do an independet test. I'll provide
her/him with n, F, and G numbers or/and with the output of chg.gp. The
test has taken some days on my laptop since (from the output):
Number to be tested has 4955 digits.
Modulus has 1321 digits.
Modulus is 26.64255144% of n

Regards to all, especially to
\\ [...] John Renze
\\ [...] David Broadhurst, Greg Childers and the PrimeForm community.

Wojtek

PS
This result will be included in my page
http://perta.fizyka.amu.edu.pl/pnq/prp1.html

WsF

===============================================
Wojciech Florek (WsF)
Adam Mickiewicz University, Faculty of Physics
ul. Umultowska 85, 61-614 Poznan, Poland

Phone: (++48-61) 8295033 fax: (++48-61) 8257758
email: Wojciech.Florek@...
• Sometime ago I had conjectured that all the prime factors of a Carmichael number cannot be Mersenne. For that I need the following lemma: *Thread*:
Message 14 of 30 , May 28, 2009
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Sometime ago I had conjectured that all the prime factors of a Carmichael
number cannot be Mersenne. For that I need the following lemma: *Thread*:
View Single Post [image: Old] 01 Jun 05, 07:39 PM
#*14*<http://www.mersenneforum.org/showpost.php?p=55271&postcount=14>
maxal <http://www.mersenneforum.org/member.php?u=1643>

[image: maxal's Avatar] <http://www.mersenneforum.org/member.php?u=1643>

Posts: 145
Join Date: Feb 2005
[image: Default] *Pomerance's proof*
------------------------------
I've got a permission from Carl Pomerance to reproduce his proof that every
Carmichael number is Devaraj number. I present an extended version of the
proof which applies to r-factor Carmichael numbers for arbitrary r (the
original proof is for r=3).

Recall that n is Carmichael number iff (p-1) divide (n-1) for every prime p
dividing n.

Let n=p1*p2*...*pr be Carmichael number, where p1,p2,...,pr are prime.
To show that n is Devaraj number it's enough to prove that
(p1-1)*(n-1)^(r-2)/(p2-1)...(pr-1) is integer.

(p1-1)*(n-1)^(r-2)/(p2-1)...(pr-1) is integer iff for every prime q its
maximum degree dividing the numerator is greater or equal than its maximum
degree dividing the denominator.

Let q be any prime number and let a,a1,a2,...,ar be maximum non-negative
integers such that q^a divides n-1, q^a1 divides (p1-1), ..., q^ar divides
(pr-1). Since n is Carmichael number, a1<=a, a2<=a, ..., ar<=a. (*)
Then the maximum degree of q dividing the numerator (p1-1)*(n-1)^(r-2) is
a1+(r-2)*a while the maximum degree of q dividing the numerator
(p2-1)...(pr-1) is a2+a3+...+ar.
We will show that a1 >= min(a2,a3,...,ar).

Assume that a1 < min(a2,a3,...,ar).
Let p1=k1*q^a1 + 1, p2 = k2*q^a2 + 1, ..., pr=kr*q^ar + 1 where k1, k2, ...,
kr are not divisible by q. Then
n = p1*p2*...*pr = (k1*q^a1 + 1)*(k2*q^a2 + 1)*...*(kr*q^ar + 1) = 1 +
k1*q^a1 + k2*q^a2 + ... + kr*q^ar + C where C includes all summands
divisible by at least q^(a1+min(a2,a3,...,ar)).
Hence, n-1 = k1*q^a1 + k2*q^a2 + ... + kr*q^ar + C is divisible by at most
q^a1 meanning a=a1, a contradiction to (*). Therefore, a1 >=
min(a2,a3,...,ar).

It is easy to see that a1 >= min(a2,a3,...,ar) together with (*) implies
a1+(r-2)*a >= a2+a3+...+ar.
Q.E.D.
------------------------------
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maxal<http://www.mersenneforum.org/private.php?do=newpm&u=1643> Find
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Mersenne. I hope to prove it in the near future; for this I need
the above lemma :

A.K.Devaraj

[Non-text portions of this message have been removed]
• A CONJECTURE Let f(x) =a^b^x + c where a, b,c & x belong to N ( a, b and c fixed.). Then a^(b^ (x+k* Eulerphi(Eulerphi(f(x)) + c is congruent to 0 (mod f(x)).
Message 15 of 30 , Jun 11, 2009
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A CONJECTURE

Let f(x) =a^b^x + c where a, b,c & x belong to N ( a, b and c fixed.).

Then a^(b^ (x+k* Eulerphi(Eulerphi(f(x)) + c is congruent to 0 (mod f(x)).
Here k belongs to N.

Sketch proof: This is a corollary of �Euler's Generalisation of Fermat's
Theorem- a further generalisation. (Hawaii International Conference,
2004-ISSN 1550 3747).

A.K. Devaraj

[Non-text portions of this message have been removed]
• ... Nesting brackets isn t that hard a skill. Please acquire it. ... A conjecture has no proof, sketchy or otherwise. A theorem has a proof. ... The
Message 16 of 30 , Jun 11, 2009
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> A CONJECTURE
>
> Let f(x) =a^b^x + c where a, b,c & x belong to N ( a, b
> and c fixed.).
>
> Then a^(b^ (x+k* Eulerphi(Eulerphi(f(x)) + c is congruent
> to 0 (mod f(x)).

Nesting brackets isn't that hard a skill. Please acquire it.

> Here k belongs to N.
>
> Sketch proof:

A conjecture has no proof, sketchy or otherwise. A theorem has a proof.

> This is a corollary of “Euler's Generalisation of Fermat's
> Theorem- a further generalisation. (Hawaii International
> Conference, 2004-ISSN 1550 3747).

The International Conference of People Who Want to Pay Large Amounts of Money in Order to Claim That They've Presented a Paper at a Conference?

However, I'm pleased to see that at the above "conference" someone presented a paper on "Factorization of Intergers[sic]"; it would be a shame to see such a wonderfully illucid concept as intergers fade away. (Or maybe it was just a typo, that would be a shame.)

Phil
• Dear Group Members,   The following is an extension of ideas recently presented for research.  Part of the story of this particular sequence, also (I might
Message 17 of 30 , Sep 16, 2012
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Dear Group Members,

The following is an extension of ideas recently presented for research.  Part of the story of this particular sequence, also (I might add), has something to do with the plausibility of Cancer VZ as a signal source.  Cancer VZ is a Scuti-alpha (formerly, 'dwarf Cepheid') type of star.

Conjecture:  The number of values of the form 2^a*3^b for which the number of times each digit from 0 through 9 appears exactly a prime number of times is 40.

The largest I have so far is 2^1997*3^135, with an exceptionally long wait.  However, this conjecture a) is in a computing environment I am not sure I should trust, b) is not based upon any real heuristic reasoning of finitude, and c) has not even been discussed with anybody else.  It took me a while to even remember or figure out what I had programmed.

The computing environment is probably sound to data but possibly not to the question of whether certain programs have been asked to cease, and even the former is in question.  So, if anybody would like to tackle a check, some real reasoning, and/or an attempt at continuing this sequence, it would be welcome.  I do have data, so if a choice is made to submit data to OEIS and the 40th term is right, it would be appropriate to join me as co-author or for me to add a confirmer when I submit.

Yours Truly, James G. Merickel

[Non-text portions of this message have been removed]
• ... http://www.micosmos.com/geos/nc/NC_0857.pdf I do not think that variable stars care much about the number of human fingers. David
Message 18 of 30 , Sep 16, 2012
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James Merickel <moralforce120@...> wrote:

> Conjecture: The number of values of the form 2^a*3^b for
> which the number of times each digit from 0 through 9
> appears exactly a prime number of times is 40.

> something to do with the plausibility of Cancer VZ
> as a signal source

http://www.micosmos.com/geos/nc/NC_0857.pdf

I do not think that variable stars care much
about the number of human fingers.

David
• The largest I have so far is 2^1997*3^135, with an exceptionally long wait. n=2^1742*3^4350 is a larger solution, the distribution of the digits: [241, 251,
Message 19 of 30 , Sep 16, 2012
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"The largest I have so far is 2^1997*3^135, with an exceptionally long
wait."

n=2^1742*3^4350 is a larger solution, the distribution of the digits: [241,
251, 257, 269, 251, 257, 241, 281, 281, 271]

And it is very likely that your conjecture is false, for e2,e3<N the
expected number of solutions is at least c*N^2/log(N)^10, for c>0. Moreover
it should be false even for the sequence of 2^e2.

[Non-text portions of this message have been removed]
• Dr. Broadhurst, There is a physics text in a Springer Verlag Series on plausible physics for manipulating the Sun (I do not own this particular book or know
Message 20 of 30 , Sep 16, 2012
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There is a physics text in a Springer Verlag Series on plausible physics for manipulating the Sun (I do not own this particular book or know its title offhand, but it does exist; so what a star cares about is not what I was thinking (and what I was thinking was digressive, but I'll use your response for my placement of a specific correction: The type of star is called 'delta-Scuti', not 'alpha-')).  Your focus on my digressive note is ... noted.  It was not the primary topic.
Jim Merickel

Date: Sunday, September 16, 2012, 10:33 AM

James Merickel <moralforce120@...> wrote:

> Conjecture: The number of values of the form 2^a*3^b for
> which the number of times each digit from 0 through 9
> appears exactly a prime number of times is 40.

> something to do with the plausibility of Cancer VZ
> as a signal source

http://www.micosmos.com/geos/nc/NC_0857.pdf

I do not think that variable stars care much
about the number of human fingers.

David

[Non-text portions of this message have been removed]
• Okay.    Thanks, Mr. Gerbicz.  I retract as a reasonable conjecture, without checking.  This is somewhere  There was or was not a check of term?  This
Message 21 of 30 , Sep 16, 2012
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Okay.

Thanks, Mr. Gerbicz.  I retract as a reasonable conjecture, without checking.  This is somewhere  There was or was not a check of term?  This one is the following term, as far as your computation goes?  This sequence, as I have it, is coincidence-laden from any perspective and more strongly so from my own.  And my highschool class number was the count of 0s and 7s of the term you have added.

JGM

--- On Sun, 9/16/12, Robert Gerbicz <robert.gerbicz@...> wrote:

From: Robert Gerbicz <robert.gerbicz@...>
Date: Sunday, September 16, 2012, 11:26 AM

"The largest I have so far is 2^1997*3^135, with an exceptionally long
wait."

n=2^1742*3^4350 is a larger solution, the distribution of the digits: [241,
251, 257, 269, 251, 257, 241, 281, 281, 271]

And it is very likely that your conjecture is false, for e2,e3<N the
expected number of solutions is at least c*N^2/log(N)^10, for c>0. Moreover
it should be false even for the sequence of 2^e2.

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

[Non-text portions of this message have been removed]
• -- In primenumbers@yahoogroups.com, ... I don t know that book, but I have read about the IRAS upper limit on nearby Dyson spheres:
Message 22 of 30 , Sep 16, 2012
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James Merickel <moralforce120@...> wrote:

> There is a physics text in a Springer Verlag Series on plausible
> physics for manipulating the Sun

IRAS upper limit on nearby Dyson spheres:

http://arxiv.org/ftp/arxiv/papers/0811/0811.2376.pdf

>> This Dyson Sphere search has looked at a significant fraction of the IRAS LRS sources with temperatures under 600 ºK. Since IRAS covered 96% of the sky this is essentially a whole-sky search. Indeed this search may be one of the only SETI/cosmic archaeology whole-sky searches conducted so far. Unlike many radio and optical SETI searches this one does not require purposeful intent to communicate on the part of the originating source of the signature of intelligence. <<

In any case, it still seems to me to be silly to link
such ideas to base-10 representations of integers,
as seemed to be implied by James' obiter dictum.

David
• Dr. Broadhurst: It was not meant to be the main topic, but since you have continued on it...the assumption of the use of Dyson spheres is an assumption of
Message 23 of 30 , Sep 16, 2012
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It was not meant to be the main topic, but since you have continued on it...the assumption of the use of Dyson spheres is an assumption of energy greediness in a near-zone.  We should not assume too much about intelligence beyond Earth.  We only know us.  I am grappling with three possibilities for life's origination: 1) Pre-bigbang, 2) Pre-metalization, and 3) post-metalization but before and affecting Earth's life.  These base-10 coincidences are the foundation of this.  You have the right to your opinion.  It's more my area than yours, though.
Jim Merickel
P.S. I will present the data on the sequence in question as I have it tomorrow, along with the code I use to generate it a second time.  Right now I cannot include the term given by Mr. (or Dr.) Gentzen as the 41st term, though I assume that's what was meant right now.  I will test it from my computer, but cannot get that far quickly to rule out an intermediate term.

Date: Sunday, September 16, 2012, 4:26 PM

James Merickel <moralforce120@...> wrote:

> There is a physics text in a Springer Verlag Series on plausible
> physics for manipulating the Sun

IRAS upper limit on nearby Dyson spheres:

http://arxiv.org/ftp/arxiv/papers/0811/0811.2376.pdf

>> This Dyson Sphere search has looked at a significant fraction of the IRAS LRS sources with temperatures under 600 ºK. Since IRAS covered 96% of the sky this is essentially a whole-sky search. Indeed this search may be one of the only SETI/cosmic archaeology whole-sky searches conducted so far. Unlike many radio and optical SETI searches this one does not require purposeful intent to communicate on the part of the originating source of the signature of intelligence. <<

In any case, it still seems to me to be silly to link
such ideas to base-10 representations of integers,
as seemed to be implied by James' obiter dictum.

David

[Non-text portions of this message have been removed]
• Three more terms, and no more for e2,e3
Message 24 of 30 , Sep 17, 2012
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Three more terms, and no more for e2,e3<40000 (if we write n=2^e2*3^e3)
2^26887*3^13438
2^35399*3^26448
2^18350*3^38015

[Non-text portions of this message have been removed]
• ... At least this definitively rules out the Conjecture in its initial form. (To leave it open, change 40 to a finite number .) Anyway, as would say a
Message 25 of 30 , Sep 17, 2012
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--- In primenumbers@yahoogroups.com, Robert Gerbicz <robert.gerbicz@...> wrote:
>
> Three more terms, and no more for e2,e3<40000 (if we write n=2^e2*3^e3)
> 2^26887*3^13438
> 2^35399*3^26448
> 2^18350*3^38015

At least this definitively rules out the Conjecture in its initial form.
(To leave it open, change "40" to "a finite number".)
Anyway, as would say a Russian colleague of mine, 40 is not a number(*).
More surprisingly, it seems that 42 isn't the answer, either.

(* Although I vaguely recall some people stating that 10^40 was a "fundamental" ratio occuring in several instances when looking at sizes and/or masses in the known universe... no link to dwarf Cepheid type stars though.)

For the records, I tentatively submitted the draft http://oeis.org/draft/A216854 to store these results.
(Not sure other editors of OEIS will not delete rather than "approve" & publish this...)

Regards,
Maximilian
• No, Max.  I have already (in another mailing on this topic) conceded it should not be finite.  It would be nice to have that there are 44 values with
Message 26 of 30 , Sep 18, 2012
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No, Max.  I have already (in another mailing on this topic) conceded it should not be finite.  It would be nice to have that there are 44 values with exponents under 44000 (or 44444), though.  It's just cutesy stuff related to coincidences though.

Robert, I have checked these terms.

I will not have time for this or anything else here.  I slept when I got home last night.  It's more important to change that home than do this.

--- On Mon, 9/17/12, maximilian_hasler <maximilian.hasler@...> wrote:

From: maximilian_hasler <maximilian.hasler@...>
Date: Monday, September 17, 2012, 3:34 PM

--- In primenumbers@yahoogroups.com, Robert Gerbicz <robert.gerbicz@...> wrote:
>
> Three more terms, and no more for e2,e3<40000 (if we write n=2^e2*3^e3)
> 2^26887*3^13438
> 2^35399*3^26448
> 2^18350*3^38015

At least this definitively rules out the Conjecture in its initial form.
(To leave it open, change "40" to "a finite number".)
Anyway, as would say a Russian colleague of mine, 40 is not a number(*).
More surprisingly, it seems that 42 isn't the answer, either.

(* Although I vaguely recall some people stating that 10^40 was a "fundamental" ratio occuring in several instances when looking at sizes and/or masses in the known universe... no link to dwarf Cepheid type stars though.)

For the records, I tentatively submitted the draft http://oeis.org/draft/A216854 to store these results.
(Not sure other editors of OEIS will not delete rather than "approve" & publish this...)

Regards,
Maximilian

[Non-text portions of this message have been removed]
• Got it, Max, but just focused on the early part of it to the exclusion of the end.  Sorry I did not read more carefully.  Your sequence is only the beginning
Message 27 of 30 , Sep 18, 2012
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Got it, Max, but just focused on the early part of it to the exclusion of the end.  Sorry I did not read more carefully.  Your sequence is only the beginning of what's available.  Mine are the exponents.  It seems that what is best is to keep mine and possibly also yours (which cannot have 44 numbers in the visible 'front part' ofr the sequence at OEIS, but actually link full data.  If Robert has that, that would be good.  I cannot move my data in such a large way to anywhere on the web right now.  Incompetence is part, but only actually a rather small cause of that.
JimMe

--- On Mon, 9/17/12, maximilian_hasler <maximilian.hasler@...> wrote:

From: maximilian_hasler <maximilian.hasler@...>
Date: Monday, September 17, 2012, 3:34 PM

--- In primenumbers@yahoogroups.com, Robert Gerbicz <robert.gerbicz@...> wrote:
>
> Three more terms, and no more for e2,e3<40000 (if we write n=2^e2*3^e3)
> 2^26887*3^13438
> 2^35399*3^26448
> 2^18350*3^38015

At least this definitively rules out the Conjecture in its initial form.
(To leave it open, change "40" to "a finite number".)
Anyway, as would say a Russian colleague of mine, 40 is not a number(*).
More surprisingly, it seems that 42 isn't the answer, either.

(* Although I vaguely recall some people stating that 10^40 was a "fundamental" ratio occuring in several instances when looking at sizes and/or masses in the known universe... no link to dwarf Cepheid type stars though.)

For the records, I tentatively submitted the draft http://oeis.org/draft/A216854 to store these results.
(Not sure other editors of OEIS will not delete rather than "approve" & publish this...)

Regards,
Maximilian

[Non-text portions of this message have been removed]
• Very well.  Do that.  My reading has been that all my terms for exponents are correct and in order through the 44th terms.  It may only be the case for
Message 28 of 30 , Sep 19, 2012
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Very well.  Do that.  My reading has been that all my terms for exponents are correct and in order through the 44th terms.  It may only be the case for (at least) the 41st term--you may wish to consider 42nd through 44th tentatively correct or likely wrong.  I'm in no position to check those (next several days).  Just add them in a note or hold off until they are checked.  The problem: It appears that Robert is saying that he is checking in sequence through a maximum exponent value.  The first 40 were found 'correctly', and it shouldn't matter for at least the 41st.
--- On Wed, 9/19/12, Maximilian Hasler <maximilian.hasler@...> wrote:

From: Maximilian Hasler <maximilian.hasler@...>
To: "James Merickel" <moralforce120@...>
Date: Wednesday, September 19, 2012, 12:00 PM

I can put your name as co-author in the three sequences already published since Monday
(see FORMULA).

Maximilian

On Wed, Sep 19, 2012 at 2:52 PM, James Merickel <moralforce120@...> wrote:

Max:
I'm changing the titles to include your A-number.  It is a sequence I had been planning to do myself.  So, all three should link across to two others (to start with, but there's that skew-coincidence also that is going to be expanded upon as well, and probably other stuff).

JimMe

--- On Mon, 9/17/12, maximilian_hasler <maximilian.hasler@...> wrote:

From: maximilian_hasler <maximilian.hasler@...>
Date: Monday, September 17, 2012, 3:34 PM

--- In primenumbers@yahoogroups.com, Robert Gerbicz <robert.gerbicz@...> wrote:
>
> Three more terms, and no more for e2,e3<40000 (if we write n=2^e2*3^e3)
> 2^26887*3^13438
> 2^35399*3^26448
> 2^18350*3^38015

At least this definitively rules out the Conjecture in its initial form.
(To leave it open, change "40" to "a finite number".)
Anyway, as would say a Russian colleague of mine, 40 is not a number(*).
More surprisingly, it seems that 42 isn't the answer, either.

(* Although I vaguely recall some people stating that 10^40 was a "fundamental" ratio occuring in several instances when looking at sizes and/or masses in the known universe... no link to dwarf Cepheid type stars though.)

For the records, I tentatively submitted the draft http://oeis.org/draft/A216854 to store these results.
(Not sure other editors of OEIS will not delete rather than "approve" & publish this...)

Regards,
Maximilian

[Non-text portions of this message have been removed]
• The other day I was browsing through Riesel and opened it at Appendix 3, Legendre’s symbol, x^2 º a mod p; a thought came into my mind from where who
Message 29 of 30 , Jul 6, 2014
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The other day I was browsing through Riesel and opened it at Appendix
3, Legendre’s symbol,
x^2 º a mod p; a thought came into my mind from where who knows!!
It was just silly, that it was equivalent to N = pq.

Conjecture

An odd p < s (s the integer square root) will be a factor of N a
composite odd number if and only if

s^2 = - (N - s^2) mod p .

I can prove the main part but not that p can be either a prime or a
composite.

Proof

N = pq

s = flr(sqr( N))

Redefine p and q as follows:-

p = s - t ( t is an integer - 0 < t < s )

q = s + t + k ( k is an integer and will always be of the form of
either (0 mod 4) or (2 mod 4) depending on N

Substituting the redefined p and q , back into N = pq

N = ( s - t)( s + t + k)

N = s^2 + ks - kt - t^2 (the st values cancel out)

Solve for t^2

t^2 = k(s - t) - (N - s^2)

Taking the modulus of both sides using (s - t) gives

t^2 = 0 - ( N - s^2) (mod ( s - t))

t^2 mod(s - t) = s^2 mod ( s- t) (any difference is a multiple of ( s
- t))

This gives

s^2 = - ( N - s^2) mod p.

QED

Example

I will use a number from Riesel (page 147 of my edition)

N = 13199

s = 114

p = 67

(114^2) mod 67 = 65

(N - s^2) = 203 => 203 mod 67 = 2

This is 67 - 2 = 65

I have looked at some of the RSA numbers that have been solved and they
also confirm the above.

A number that shows that p can be composite is 1617

N = 1617

s = 40

p = 33

(40^2) mod 33 = 16

N - s^2 = 1617 - 1600 = 17

p - 17 = 33 - 17 = 16

If the residues to a number N, are found using the N and then (N-s^2)
they give 2 different sets; example using 12007001

Residues for N =? (-1 2 3 5 7 11 13 23 29 37 43 71 73 89 97)

Residues for (N - s^2) => (-1 2 5 23 31 43 53 59 61 67 71 89 97)

Intersection of the 2 sets gives (-1 2 5 23 43 71 89 97)

I am not sure that this will be of any real benefit, but who knows.

Ron
• ... This is trivially equivalent to N = 0 mod p, whatever s might be. So the conjecture is that N = 0 mod p if and only if p divides N. This is true, but
Message 30 of 30 , Jul 7, 2014
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> s^2 = - (N - s^2) mod p .

This is trivially equivalent to N = 0 mod p, whatever s might be.
So the "conjecture" is that N = 0 mod p if and only if p
divides N. This is true, but hardly noteworthy.

David

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