Square Riesel numbers < 3896845303873881175159314620808887046066972469809^2

[Citrix]

Looking at https://oeis.org/A270896

Is there a similar sequence on the riesel side?

[VBCurtis]

Are you asking whether we have a list for which n² -1 is prime?

[science_man_88]

Originally Posted by VBCurtis

Are you asking whether we have a list for which n² -1 is prime?

the linked sequence has this definition:

"Values of n for which n^2 is a Sierpiński number."

the equivalent on the Riesel numbers is:

"Values of n for which n^2 is a Riesel number."

in theory.

[Batalov]

Originally Posted by M. Filaseta et al, 2008

Theorem 11
There are infinitely many squares that are Riesel numbers. One such Riesel number is 3896845303873881175159314620808887046066972469809^2.

The first sentence of this theorem is not new; it is a consequence of the work of Y.-G. Chen:
Y.-G. Chen, On integers of the forms k^r − 2ⁿ and k^r2ⁿ + 1, J. Number Theory 98 (2003), 310–319.

M. Filaseta et al., On Powers Associated with Sierpinski Numbers, Riesel Numbers and Polignac’s Conjecture, Journal of Number Theory, Volume 128, Issue 7, July 2008, Pages 1916-1940.

This is one of the papers linked in OEIS A101036

[VBCurtis]

Originally Posted by science_man_88

the linked sequence has this definition:

"Values of n for which n^2 is a Sierpiński number."

the equivalent on the Riesel numbers is:

"Values of n for which n^2 is a Riesel number."

in theory.

I would not have needed to ask if I'd just looked up the def'n of Sierpinski and Riesel numbers; after all the CRUS work, I'd forgotten that base 2 is default. I was conflating coefficient and base, among other mistakes...

[LaurV]

[thinking]
The part with "under the unproved assumption..." which is mentioned in the OEIS sequence linked by Serge, occurred to me when I was looking at the sequence linked by the OP (where it is not mentioned). I was thinking, "how the hack do they know those are all Sierpinski numbers?". Because if we could prove that, we won't need to do all the CRUS effort we do now, would we?
[/thinking]

[VBCurtis]

Originally Posted by LaurV

[thinking]
The part with "under the unproved assumption..." which is mentioned in the OEIS sequence linked by Serge, occurred to me when I was looking at the sequence linked by the OP (where it is not mentioned). I was thinking, "how the hack do they know those are all Sierpinski numbers?". Because if we could prove that, we won't need to do all the CRUS effort we do now, would we?
[/thinking]

It's easy to show a number is Sierpinski; just exhibit the covering set. It's rather a pain to show one is the *lowest* possible, as CRUS has demonstrated for years now.

[LaurV]

Well... not.

This affirmation assumes that all S' numbers have a small covering set. If we would know that, we don't need to do anything for CRUS. This is what I was saying. But we don't know that, and that is why all the effort for CRUS. We can show a number is S' easily, by showing its covering set assuming it has one which is small enough/finite/whatever.

The "small note" was mentioned in the OEIS in the R' side sequence (linked by Serge) but not in the S' side (linked by OP).

We know some S' or R' numbers (those with a small covering set), but we don't know if those are ALL (of course, under a limit. I am not talking about infinity here).

[science_man_88]

Originally Posted by LaurV

Well... not.

This affirmation assumes that all S' numbers have a small covering set. If we would know that, we don't need to do anything for CRUS. This is what I was saying. But we don't know that, and that is why all the effort for CRUS. We can show a number is S' easily, by showing its covering set assuming it has one which is small enough/finite/whatever.

The "small note" was mentioned in the OEIS in the R' side sequence (linked by Serge) but not in the S' side (linked by OP).

We know some S' or R' numbers (those with a small covering set), but we don't know if those are ALL (of course, under a limit. I am not talking about infinity here).

true we don't know if these are all but we can know if any of the values divide by a small prime very fast for most inputs.

for k*2^n-1 :

k=2 mod 3 n = 1 mod 2 creates a value divisible by 3
k=1 mod 3 n = 0 mod 2 also creates divisibility by 3
k= 0 mod 3 never divisible by 3 for the final number, regardless of n.

...