[quote=kar_bon;230021]Why? It's a question accordingly to the results here and I don't know if anybody has solved/proved this before.[/quote]
Sort of a k-b-b analogue to the Sierpinski and Riesel conjectures. We can call it the Bonath Conjecture. :smile:

[QUOTE=Karsten]Why? It's a question accordingly to the results here and I don't know if anybody has solved/proved this before.
[/QUOTE]

Because the fact that a*n + 1 always has an infinite amount of primes shoots that conjecture to hell.

You could try the other way;

Karsten's Conjecture: There is a k such that k * b[sup]b[/sup] + 1 is never prime for all b>=1.

[QUOTE=3.14159;230026]Because the fact that a*n + 1 always has an infinite amount of primes shoots that conjecture to hell.[/QUOTE]

The same amount of primes got k*2^n+1 or k*2^n-1 and those values I think of are well known!

Proof: Finding a potential candidate k, and giving a rigorous proof that k * b[sup]b[/sup] + 1 is never prime for all b>=1.

Potential Refutation: Giving a rigorous proof that shows that every k has at least [B]one[/B] prime for any b>=1.

Any search results on this one?

A briefly promising candidate: 8.

Prime found at 17: 6617922095090694113417.

Another; 17.

Debunked at 210; 17 * 210[sup]210[/sup] + 1 is prime.

21 * 990[sup]990[/sup] + 1 is prime. Trying again!

[QUOTE=3.14159;230031]21 * 990[sup]990[/sup] + 1 is prime. Trying again![/QUOTE]

can we calculate what values don't hit 6n+1 or 6n-1 ?

[QUOTE=science_man_88]can we calculate what values don't hit 6n+1 or 6n-1 ?
[/QUOTE]

Easy. Any value that is 3 mod 6. Divisible by three.

Ex: 8 * 11[sup]11[/sup] + 1 = 7988726777109 = 3 * 13 * 40429 * 5066639.

Or;

If k and b are both 2 mod 3, k * b[sup]b[/sup] + 1 is divisible by 3.

Okay, the above is false.

32 * 47[sup]47[/sup] + 1 has a smallest factor of 701.

[QUOTE=3.14159;230031]21 * 990[sup]990[/sup] + 1 is prime. Trying again![/QUOTE]

Why not using the Database?

And you're searching the false direction: k=5 is the first value with no prime of the form k*b^b+1 and b<1000!

The next values with no prime for b<1000 are k=29, 35, 41, 53, ...

And:

5*b^b+1 is composite for:
- b == 1 mod 2 (factor 2)
- b == 2 , 4 mod 6 (factor 3)

So a possible b-value has to be b == 0 mod 6.

Exactly. I'm looking for a prime.

Let's get rid of 5, 29, 41, and 53.

b=2 is impossible as 2^2*k-/+1 lands in 4n+1 and 4n+3.