Prime Constellations

[CRGreathouse]

Originally Posted by science_man_88

I tried looking for it using pari and my eyes and couldn't find it. I did use x such that gcd(120,x)==1 only.

You couldn't find primorials? Try

primorial(x)=prod(i=1,primepi(x),prime(i))

[science_man_88]

Originally Posted by CRGreathouse

You couldn't find primorials? Try

primorial(x)=prod(i=1,primepi(x),prime(i))

It was the overlap with the residues mod 120 that allow primes, that I couldn't find.

[CRGreathouse]

Originally Posted by science_man_88

It was the overlap with the residues mod 120 that allow primes, that I couldn't find.

I don't know what you're talking about. If you mean the numbers coprime to 120, that's

select(n->gcd(n,120)==1, vector(120,i,i))

Of course there are infinitely many primes congruent to each of these residues mod 120.

[MattcAnderson]

10/31/2013

Thanks for the input, Batalov.

The quadruplets webpage that you referenced,
at the University of Tenisee in Martin shows
the largest quadruplets known,

while the online encyclopedia of integer sequences
lists out the smallest known quadruplets.

Thanks again for the keen eye.

Regards,
Matt

[MattcAnderson]

Thanks again for the input.

As a matter of fact, in regards to prime constellations, new ones can be found.
The pattern B = {0,4,6, ... 76} has 19 elements.
Primes of this form can be found with the expression
x = 30030*n + 29917

I am not saying that every prime fitting expression x will make
x, x+4, x+6, ... x+76
a prime number.

What I am saying is that probably,there is an n which causes the x to make all 17 number prime.

And with enough computing power,
it is feasible.

Regards,
Matt

[MattcAnderson]

Same day ....

My estimate is that x has more than 26 digits.

Regards,
Matt

[Jens K Andersen]

Originally Posted by science_man_88

I tried looking for it using pari

Here is a PARI/GP function to test whether a constellation is admissible:
admissible(v)=forprime(p=2,#v,if(#Set(v%p)==p,return(0)));1

This function returns the inadmissible primes:
admissiblep(v)=local(i=[]);forprime(p=2,#v,if(#Set(v%p)==p,i=concat(i,p)));i

Examples:

? admissible([0,2,4])
%36 = 0
? admissible([0,2,6])
%37 = 1
? admissiblep([0,2,4])
%38 = [3]
? admissiblep([0,2,6])
%39 = []
? admissible([0,4,6,10,12,16,24,30,34,40,42,46,52,54,60,66,70,72,76])
%40 = 1

[Batalov]

Originally Posted by MattcAnderson

10/31/2013

Thanks for the input, Batalov.

The quadruplets webpage that you referenced,
at the University of Tenisee in Martin shows
the largest quadruplets known,

while the online encyclopedia of integer sequences
lists out the smallest known quadruplets.

Thanks again for the keen eye.

Regards,
Matt

I was contrasting the UTM page to Tony Forbes' page, not to the minimal quadruplets page, for which obviously 11,13,17,19 will always be the record.

UTM page is much more alive for quads and trips. (And obviously, I didn't have to have a good eye to notice the quad page change. The currently top entry is simply mine. ;-) ...that would be the reason that I know that Tony's page is a bit out of date.)

[miguelrey]

Originally Posted by MattcAnderson

oops the .zip didn't attach

hi, (sorry i dont speak english)
i demostrated a part of the theory of numbers, but i dont know, if its relevant or not.
this are....
1) if p is p=8k+3 then if: 2ˆ[((p-3)/4)+1] = x (mod p) and if: xˆ2 =-2 (mod p) then p is prime
2) if p is p=8k+7 then if: 2ˆ[((p-3)/4)+1] = x (mod p) and if: xˆ2 = +2 (mod p) then p is prime
3) if p is p=8k+5 then if: 2ˆ[((p-1)/4)] = x (mod p) and if: xˆ2 = -1 (mod p) then p is prime
4: if p is p=8k+1 then if: 2ˆ[((p-1)/4)] = +-1 (mod p) then p is prime or a Fermat number
thank you for your time and waiting your answer, Miguel R.
miguelangelreybonet@yahoo.es

[science_man_88]

Originally Posted by miguelrey

hi, (sorry i dont speak english)
i demostrated a part of the theory of numbers, but i dont know, if its relevant or not.
this are....
1) if p is p=8k+3 then if: 2ˆ[((p-3)/4)+1] = x (mod p) and if: xˆ2 =-2 (mod p) then p is prime
2) if p is p=8k+7 then if: 2ˆ[((p-3)/4)+1] = x (mod p) and if: xˆ2 = +2 (mod p) then p is prime
3) if p is p=8k+5 then if: 2ˆ[((p-1)/4)] = x (mod p) and if: xˆ2 = -1 (mod p) then p is prime
4: if p is p=8k+1 then if: 2ˆ[((p-1)/4)] = +-1 (mod p) then p is prime or a Fermat number
thank you for your time and waiting your answer, Miguel R.
miguelangelreybonet@yahoo.es

Hola Miguel,

Esto no es exactamente lo que este tema se trata. Este tema es acerca de cuando decir p, p+2, p+6 y p+8 podría ser primer por ejemplo.Me identifico con él, posiblemente. Pero voy a tener que mirar en más. Estoy utilizando Google Translate y el final de su correo electrónico para tratar de conseguir algo que se puede leer más fácil. Aunque supongo que debería utilizar el disctionary-español Inglés que tengo. Intenté aprender español de una vez.

[MattcAnderson]

Hi math people,

I want to mention my personal webpage -

https://sites.google.com/site/mattc1anderson/home-1

and especially, under a few layers -

http://oeis.org/A022006

I have put considerable effort into some OEIS entries.
See my prime constellations project.

Regards,
Matt