Other Primes

[rudy235]

Originally Posted by paulunderwood

Congrats to James Winskill for the mega primorial prime: 3267113# - 1 (1,418,398 decimal digits).

Yes, in the last few days we have had two new categories of primes entering the megaprime territory. A Palindromic with 1,234,567 digits and this Primorial with 1,418,398 digits. The next one coming is probably the 3^rd term of a Prime in A.P.

We now have close to 1,125 megaprimes

[Batalov]

Originally Posted by rudy235

...A Palindromic with 1,234,567 digits and ...

How about two of them?

[paulunderwood]

Originally Posted by Batalov

How about two of them?

You mean "two more"?

https://primes.utm.edu/primes/page.php?id=132766
https://primes.utm.edu/primes/page.php?id=132767

[rudy235]

How difficult is to prove a primorial Prime?


3267113# - 1

Verification status (*): InProcess

Is still unproven. I would think that having the primorial +1 100% factored would make proving it a matter of a couple of says. A week in the worse case.

[paulunderwood]

Originally Posted by rudy235

How difficult is to prove a primorial Prime?


3267113# - 1

Verification status (*): InProcess

Is still unproven. I would think that having the primorial +1 100% factored would make proving it a matter of a couple of says. A week in the worse case.

Some numbers require proof attempts at increasing sizes of FFT.

[paulunderwood]

Congrats tp Ryan and Serge for the record Near-rep Digit / Palindrome prime 10^1888529 - 10^944264 - 1

[Batalov]

Originally Posted by paulunderwood

Congrats tp Ryan and Serge for the record Near-rep Digit / Palindrome prime 10^1888529 - 10⁹⁴⁴²⁶⁴ - 1

Yet another custom sieve for such hybrid beasts:
quick sketch:

We are searching for NRP(K,n) = 10²ⁿ⁺¹-K*10ⁿ-1. K can only be 1,2,4,5,7,8. (K=3 has algebraic factorization, which is not needed ...because the whole expression is divisible by 3 when 3|K).

Step 1. Let x=10^n, then NRP(K,n) = 10x²-Kx-1 . I solve this quadratic equation just like in school but x is some Mod(x,p) then sieve by p

Step 2. If quadratic equation has solution (nearly half the time; if it doesn't , nothing to sieve out), then --

Step 3. Solve 10^n = x₁ and 10^n = x₂. This is called znlog() and these values will periodically repeat with period znorder().

Step 4. Sieve out and repeat for 7<= p <= 10^11 or 10^12.

Step 5: remove special cases for p={7,11,13} (this actually removes a huge fraction of candidates with K=2, that's why it is the "thinnest" K)

The trick is to code steps 1, 2 and 3, and to know how.

Step 6. Test. (we test all six number forms in order of size. The fact that K=1 produced the first hit is accidental. With K=1, the number looks a bit more elegant.)

[rudy235]

Originally Posted by rudy235

How difficult is to prove a primorial Prime?


3267113# - 1

Verification status (*): InProcess

Is still unproven. I would think that having the primorial +1 100% factored would make proving it a matter of a couple of days. A week in the worse case.

After 20 days (10/17/21) it was proven prime.

[sweety439]

Originally Posted by Batalov

Step 5: remove special cases for p={7,11,13} (this actually removes a huge fraction of candidates with K=2, that's why it is the "thinnest" K)

You mean that the Nash weight (or difficulty) for K=2 (999...9997999...999) is very low?

[Batalov]

Originally Posted by sweety439

You mean that the Nash weight ...

Dare you to define it (for these six sequences), but yes.

Originally Posted by sweety439

... (or difficulty) for K=2 (999...9997999...999) is very low?

Dare you to define it (for these six sequences), but yes.

[pepi37]

If I may ask how many candidates remain after that ?