[QUOTE=MisterBitcoin;504478]Well, at first i guessed I found the largest ck, but. It´s not that easy because n=1 and n=14 are prime; so the guessed number is not the ck.[/quote]

covering.exe has a bug when searching very large k. There is a version bigcovering.exe which fixes this bug - it's available from the same site as covering.exe.

The likely CK for this base is 1429458644604553 which has covering set {19, 37, 61, 307, 1051, 39916801} with period 18. The reason it's so big is that the primes that can appear in the covering set with period d are the primes p such that b has order d mod p; these are prime factors of the cyclotomic number Φ[SUB]d[/SUB](b). If for small d these numbers have lots of small prime factors then there are lots of primes that are likely to appear as factors and so the CK will be small. If there are few small factors then we will either need a big prime in the covering set (making the CK big as well) or a lot of small primes with longer periods (which will probably also make the CK big).

For b = 11!, there are very few small factors:
Φ[SUB]2[/SUB](b) = b+1 = 39916801
Φ[SUB]3[/SUB](b) = b^2+b+1 = 61*26120507576341
Φ[SUB]4[/SUB](b) = b^2+1 = 1593350922240001
Φ[SUB]5[/SUB](b) = b^4+b^3+b^2+b+1 = 761*3336093593961274918315629641
Φ[SUB]6[/SUB](b) = b^2-b+1 = 1051*1516033189651
So it's not a surprise that the CK is huge.

[quote]I´ve only tryed exponent 144 (above is seen using 72; for sure. :P ). Is there any other exponents that I might try?[/QUOTE]

If Φ[SUB]5[/SUB](b) or Φ[SUB]10[/SUB](b) have lots of small factors then it's worth trying periods with 5 as a factor, but that isn't the case here.

hi,
i am starting R2019
CK 304

[QUOTE=10metreh;504502]covering.exe has a bug when searching very large k. There is a version bigcovering.exe which fixes this bug - it's available from the same site as covering.exe.

The likely CK for this base is 1429458644604553 which has covering set {19, 37, 61, 307, 1051, 39916801} with period 18. The reason it's so big is that the primes that can appear in the covering set with period d are the primes p such that b has order d mod p; these are prime factors of the cyclotomic number Φ[SUB]d[/SUB](b). If for small d these numbers have lots of small prime factors then there are lots of primes that are likely to appear as factors and so the CK will be small. If there are few small factors then we will either need a big prime in the covering set (making the CK big as well) or a lot of small primes with longer periods (which will probably also make the CK big).

For b = 11!, there are very few small factors:
Φ[SUB]2[/SUB](b) = b+1 = 39916801
Φ[SUB]3[/SUB](b) = b^2+b+1 = 61*26120507576341
Φ[SUB]4[/SUB](b) = b^2+1 = 1593350922240001
Φ[SUB]5[/SUB](b) = b^4+b^3+b^2+b+1 = 761*3336093593961274918315629641
Φ[SUB]6[/SUB](b) = b^2-b+1 = 1051*1516033189651
So it's not a surprise that the CK is huge.



If Φ[SUB]5[/SUB](b) or Φ[SUB]10[/SUB](b) have lots of small factors then it's worth trying periods with 5 as a factor, but that isn't the case here.[/QUOTE]
3036132198346310 is a small improvement.

[QUOTE=henryzz;505191]3036132198346310 is a small improvement.[/QUOTE]

It would be if it was actually smaller than 1429458644604553 and had a covering set (I haven't found a prime yet, but what is the factor for n=15?)

[QUOTE=10metreh;505195]It would be if it was actually smaller than 1429458644604553 and had a covering set (I haven't found a prime yet, but what is the factor for n=15?)[/QUOTE]

Typically I looked at Riesel by mistake. I also miscounted digits.

[QUOTE=lalera;504610]hi,
i am starting R2019
CK 304[/QUOTE]

What is ck of S2019?

[QUOTE=pepi37;505308]What is ck of S2019?[/QUOTE]
Ok found it
So I start S2019 :)

[QUOTE=pepi37;505329]Ok found it
So I start S2019 :)[/QUOTE]

hi,
S2019, ck is 304

[QUOTE=lalera;505392]hi,
S2019, ck is 304[/QUOTE]

Thanks!

hi,
status update for R2019
range n=1 to 100k done
10 k´s remain
for more info visit lalera.alotspace.com
continuing

[QUOTE=lalera;507347]hi,
status update for R2019
range n=1 to 100k done
10 k´s remain
for more info visit lalera.alotspace.com
continuing[/QUOTE]

For k=4, k=64, and k=144:

Odd n has factor of 5
Even n has algebra factors

For k=100:

Odd n has factor of 101
Even n has algebra factors

Thus R2019 has only 6 k’s remain at n=100K:

84, 114, 204, 242, 296, 302