well all b would land eventually in 6n+1 or 6n-1 I think I have a hypothesis on this actually want to hear it ?

okay maybe not a perfect hypothesis lol.

Post it. Karsten may be on to something, though. I think 5 is the death knell to my opposition.

well at least for odd b it seems that b^b mod 6 = b mod 6

likely a well known fact with me lol.

though it would determine what k we need to bother to prove/disprove.

actually it looks to work for all numbers !=2 mod 6

[QUOTE=science_man_88;230040]well at least for odd b it seems that b^b mod 6 = b mod 6

likely a well known fact with me lol.

though it would determine what k we need to bother to prove/disprove.[/QUOTE]

n^5 = n (mod 6), so you need only consider the cases mod lcm(5, 6) = 30. If you're restricting to odds, that's 15 cases, which you should be able to check by hand.

what I'm saying is (b^b)%6=b%6 for all b such that b%6 !=2, and that if b%6==2 b^b%6==4 anyways:

for k*5^5+1 = 6*n+1

5^5= 5 mod 6

lcm(5,6) =30 = 5*6 so only every 6th k has to be checked at all.

(7^7)%6 = 1 mod 6

once again lcm(1,6) = 6*1 = 6 every 6th k is all that needs proof/ disproof

(9^9)%6 = 3 mod 6

lcm(3,6) = 2*3 so every 2nd k needs checking.

The case I gave for testing was k=5, so find a prime for 5*b^b+1 for any b>=1!

And here only b == 0 mod 6 have to be tested further because they are not divisible by 2 or 3!

It seems no covering set exists for k=5.

[QUOTE=kar_bon;230083]The case I gave for testing was k=5, so find a prime for 5*b^b+1 for any b>=1!

And here only b == 0 mod 6 have to be tested further because they are not divisible by 2 or 3!

It seems no covering set exists for k=5.[/QUOTE]

doh I forgot about that lol because anything mod 5 6 will give a possible prime which happens every 6th one from k=2 I think if I did the math correct

if all my math right the covering set should be all numbers 0 and 2 mod 6

[CODE](08:36) gp > forstep(k=2,1000,[4,2],if(isprime(k*5^5+1),print(k)))
12[/CODE]

looks like 5 is gone for having no prime lol