A question about primes of a particular form

[enzocreti]

Can a pg(k) be divisible by 5555?

[enzocreti]

an can exist a pg(k) divisible by 1111?

[science_man_88]

Originally Posted by enzocreti

an can exist a pg(k) divisible by 1111?

if not, then the former question is also no.

[enzocreti]

Originally Posted by science_man_88

if not, then the former question is also no.

1111=11*101


so probably the analysis of the remainders rules out that

[science_man_88]

Originally Posted by enzocreti

1111=11*101


so probably the analysis of the remainders rules out that

if you could do the math, why'd you ask ...

[Dr Sardonicus]

Originally Posted by enzocreti

Can a pg(k) be divisible by 5555?

Work out criteria for divisibility by 5, 11, and 101, and you'll be able to answer this yourself.

[enzocreti]

Originally Posted by Dr Sardonicus

Work out criteria for divisibility by 5, 11, and 101, and you'll be able to answer this yourself.

A mathexchange user found that no pg(k) is divisible by either 1321 or 3191 or 3541. 1321, 3191 and 3541 are all emirps. And the reverse of 1321, 3191, 3541 that is 1231, 1913, 1453 are all primes of the form x^2+23y^2

[enzocreti]

A033217 see the OEIS sequence

[LaurV]

That may be coincidental. Again, there is nothing special about these numbers.

Also, there are no other "non-divisors" in sight (I tested all primes to 10k, they all divide some pg(x), this test takes just few minutes, what i implemented is a very efficient way to test, you can tel me why, and what I have done in the code; you can directly copy/paste this to a pari/gp window.

pg(d=1321)=
{
    a=1;
    b=1;
    m=21;
    t=10;
    z=10;
    n=1;
    while((r=(a*m+t)%d)!=0,
        a=((a<<1)+1)%d;
        b=(b<<1)+1;
        if(b>z,
            z=10*z;
            m=(10*m-9)%d;
            t=(10*t)%d
        );
        n++;
        printf("...%d...   %c",n,13)
        /* ; write("ff.out",d," : ",n,", ",r) */
    );
    n
}

p=3;
 while(1,print([p=nextprime(p+1),pg(p)]))

[enzocreti]

Originally Posted by LaurV

That may be coincidental. Again, there is nothing special about these numbers.

Also, there are no other "non-divisors" in sight (I tested all primes to 10k, they all divide some pg(x), this test takes just few minutes, what i implemented is a very efficient way to test, you can tel me why, and what I have done in the code; you can directly copy/paste this to a pari/gp window.

pg(d=1321)=
{
    a=1;
    b=1;
    m=21;
    t=10;
    z=10;
    n=1;
    while((r=(a*m+t)%d)!=0,
        a=((a<<1)+1)%d;
        b=(b<<1)+1;
        if(b>z,
            z=10*z;
            m=(10*m-9)%d;
            t=(10*t)%d
        );
        n++;
        printf("...%d...   %c",n,13)
        /* ; write("ff.out",d," : ",n,", ",r) */
    );
    n
}

p=3;
 while(1,print([p=nextprime(p+1),pg(p)]))

What does the code do? How does it work?

[enzocreti]

the mersenne number 131071 does not divide any pg(k)