Mersenne Numbers Known from Number Practice to Be Composite

[LaurV]

Sorry, busy here around, I just looked for this thread, which you should read carefully, especially from post #32 onward, till the end.

[User140242]

Originally Posted by LaurV

Sorry, busy here around, I just looked for this thread, which you should read carefully, especially from post #32 onward, till the end.

In addition you can calculate the possible remainders and automatically you can find the possible factors:

tf_c(p,fromk=0,tok=10000)=
{
    PrimesBase=[2,3,5,7,11];
    n_PB=5;
    PrimesBaseProd=2;
    for(i=1,n_PB,PrimesBaseProd*=PrimesBase[i];);
    bW=2*p*PrimesBaseProd;
    RW=List();
    listput(RW,1);
    for(i=1,PrimesBaseProd-1,for(j=1,n_PB,;if(Mod(1+2*p*i,PrimesBase[j])^1==0,break;);if ((j==n_PB)&&(Mod(1+2*p*i,PrimesBase[j])^1!=0)&&(Mod(1+2*p*i,8)^1==1||Mod(1+2*p*i,8)^1==7),listput(RW,1+2*p*i););););
    nR=length(RW);
    for(k=fromk, tok, for(i=1,nR,q=RW[i]+bW*k; if(Mod(2,q)^p==1, if(q>1,print("M"p" has a factor: "q); break;));));
}

[LaurV]

Why would you do that? It will take a lot of time to build that list, and it will find a lot of unneeded composite factors which you need to factor later to find the real prime factors.

[User140242]

Originally Posted by LaurV

Why would you do that? It will take a lot of time to build that list, and it will find a lot of unneeded composite factors which you need to factor later to find the real prime factors.

You can use n_PB as a parameter and if you check the number of remainders that nR uses they are 2 if n_PB = 1, 4 if n_PB = 2, 16 if n_PB = 3, 96 if n_PB = 4 and 960 if n_PB = 5.


That is, it is the same job as creating the table but you can perform an automatic operation as needed.


Of course with this method it checks (tok-fromk)*nR elements instead with forstep it checks (tok-fromk)/nR elements but the factors are the same only the size of the interval changes.


Finally, multiples of small factors can be easily eliminated for each row.



Here you find an example with 2 classes and to speed up we use two vectors for the two classes:
https://gist.github.com/user140242/0...d2738788bb1cae

[User140242]

Originally Posted by User140242

Finally, multiples of small factors can be easily eliminated for each row.

To extend the example for classes greater than 2 for each class we must use this:

It is possible to eliminate the multiples of the small factors p_j>PrimesBase[n_PB] with p_j<bW and p_j!=p and p_j!=RW[n]
for each RW[i] take a vector FACTOR of size dim_step+1 with dim_step=(tok-fromk) initially set to True

all multiples of p_j can be set to False in this way:

r_t1=Euclidean_Diophantine(bW,p_j); //function return y in  Diophantine equation  bW * x + p_j * y  = 1
r_t=(r_t1*RW[i])%bW;
if (r_t>0)
    r_t-=bW;
C2=(-bW+p_j)*r_t/bW;
C1=r_t-bW+p_j;
for (m =bW+C1+C2; m<=dim_step && m>=0; m += p_j)
    FACTOR[m] = false;

and then use it in this way to check for possible factors:

for(k=0;k<=dim_step;k++)
    if (FACTOR[k])
    {
        D=RW[i]+bW*k;
        ....
    }

[Dobri]

Originally Posted by Dr Sardonicus

...
For p == 1 (mod 4), this automatically rules out q = 2*k*p + 1 for k = 12*m + 1, 12*m + 5, or 12*m + 9 since 2*k*p + 1 == 3 (mod 8) for such k.

Thanks, this nicely eliminates the cases for k = 12m + 1, 12m + 5, and 12m + 9 indeed.

Originally Posted by Dobri

...
2(12m+8)p + 1,
...
2(12m+11)p + 1, m = 0, 1, 2,...,

cannot be factors of M1277.

However, the cases for k = 12m + 8 and 12m + 11 still need a theoretical interpretation.
I do not have a counterexample with p mod 6 = 5 that would result in a prime q = 2kp + 1 for k = 12m + 8 or 12m + 11 that divides 2^p - 1.
Currently, I am checking the known prime factors in the GIMPS database for a counterexample.
Such a counterexample with p mod 6 = 5 could be rare or does not exist.

[charybdis]

Originally Posted by Dobri

However, the cases for k = 12m + 8 and 12m + 11 still need a theoretical interpretation.
I do not have a counterexample with p mod 6 = 5 that would result in a prime q = 2kp + 1 for k = 12m + 8 or 12m + 11 that divides 2^p - 1.
Currently, I am checking the known prime factors in the GIMPS database for a counterexample.
Such a counterexample with p mod 6 = 5 could be rare or does not exist.

These have the simplest explanation of all. If p and k are both 2 mod 3 then 2kp+1 is divisible by 3 and so can't be prime.

[Dobri]

Originally Posted by charybdis

These have the simplest explanation of all. If p and k are both 2 mod 3 then 2kp+1 is divisible by 3 and so can't be prime.

Thanks, this flawlessly eliminates the cases for k = 12m + 8 and 12m + 11 indeed.
Also, divisibility by 3 is observed for k = 12m + 5 since 2(12m + 5)(6n + 5) + 1 gives the coefficient 2×5×5 + 1 = 51 which is divisible by 3.
Despite the equivalence between 2 mod 3 and 5 mod 6 (and between 1 mod 3 and 1 mod 6) for prime number considerations, I still use 'mod 6' for uniformity with a few other threads.

What I am trying to evaluate is the additive dependence between the values of k for several factors for a given Mersenne number.
On the basis of quite limited observations for k up to 11 (see posts #29 and #30), the values of k for another factor are obtained as follows:
- the sum of a prime number (or 1) and 2ⁿ such as 5 + 2² = 3² and 1 + 2³ = 3²;
- the sum of 1 and a prime number giving 2ⁿ such as 1 + 3 = 2² and 1 + 7 = 2³;
- the sum of two prime numbers giving 2ⁿ such as 3 + 5 = 2³; or
- the sum of 2ⁿ and a prime number giving a prime number such as 2² + 3 = 7 and 2³ + 3 = 11.

The exhaustive computations for k = 12 are ongoing. The custom source code was modified to collect data not only for p < 10⁹ but also for p < 2³⁰.

[Dobri]

For k = 12, all possible pairs of two factors for a given Mersenne number are observed with Δk = 1, 3, 4, 5, 7, 8, 9, and 11, where Δk = k₂ - k₁ with k₂ = 12 and k₂ > k₁. No cases of three factors for a given Mersenne number are found.

For k = 13, only cases with Δk = 1, 4, 9, and 12 for a given Mersenne number are found, where Δk = k₂ - k₁ with k₂ = 13 and k₂ > k₁. Here 1 + 12 = 13 and 4 + 9 = 13.

[User140242]

Originally Posted by Dobri

For k = 12, all possible pairs of two factors for a given Mersenne number are observed with Δk = 1, 3, 4, 5, 7, 8, 9, and 11, where Δk = k₂ - k₁ with k₂ = 12 and k₂ > k₁. No cases of three factors for a given Mersenne number are found.

For k = 13, only cases with Δk = 1, 4, 9, and 12 for a given Mersenne number are found, where Δk = k₂ - k₁ with k₂ = 13 and k₂ > k₁. Here 1 + 12 = 13 and 4 + 9 = 13.

It is possible to find the values ​​of Δk like this:

if you use a basis bW=2*p*k we have

for k = 12 = 2*2*3

q = r + bW*d where r are the possible remainders with 0 <= r <bW if we consider that q = 1 mod 2p then r = 1 mod 2p

hence r = 1 + 2p*b with 0 <= b < k

so for example if k = 12 and p = 29 the possible remainders are

(1, 59, 117, 175, 233, 291, 349, 407, 465, 523, 581, 639)

from these remainders you have to remove those that have factors in common with bW or remove the multiples of 3 since r = 1 mod 2p

and since bW = 0 mod 8 we must consider that it must be r = 7 mod 8 or r = 1 mod 8 therefore remain

(1, 175, 233, 407) so if you take the indices we find Δk or we consider (r-1)/2/p

(0,3,4,7)

for p = 31 we find (1, 311, 497, 559) or (0, 5, 8, 9)

for p = 37 we find (1, 223, 593, 815) or (0, 3, 8, 11)

for p = 47 we find (1, 95, 377, 847) or (0, 1, 4, 9)

as mentioned in a previous post there are 4 cases based on the value of p%4 and p%6

if k = 13 then bW is not divisible by 8 it is not convenient to use it in any case the procedure does not change

surely you have forgotten some cases for example p=37 and q=223 then k1 = 3 and Δk = 13-3 = 10

[Dobri]

Originally Posted by User140242

...

surely you have forgotten some cases for example p=37 and q=223 then k1 = 3 and Δk = 13-3 = 10

No cases have been forgotten.
In the quoted example, 2³⁷-1 = 137438953471, then 13743895347 / 223 = 616318177, and k₂ = (616318177-1)/(2×37) = 8328624, thus Δk = 8328624 - 3 = 8328621 = 3×7×396601, which is unrelated to k = 13.