Primes of the form b^N + 1 are not hard to come by. For example with exponent N=2, we have 1² + 1, 2² + 1, 4² + 1, 6² + 1, 10² + 1, 14² + 1, 16² + 1, 20² + 1, etc.

But what if we want 1 < b < N? Then only the following are known:

• 
  
• 2⁴ + 1
  
• 2⁸ + 1 and 4⁸ + 1
  
• 2¹⁶ + 1
  
• 30³² + 1
  
• 
  
• 120¹²⁸ + 1
  
• 
  
• 46⁵¹² + 1
  
• 824¹⁰²⁴ + 1
  
• 150²⁰⁴⁸ + 1
  
• 1534⁴⁰⁹⁶ + 1
  
• 
  
• 
  
• 
  
• 48594⁶⁵⁵³⁶ + 1
  
• 62722¹³¹⁰⁷² + 1 and 130816¹³¹⁰⁷² + 1
  
• 24518²⁶²¹⁴⁴ + 1 and 40734²⁶²¹⁴⁴ + 1 and 145310²⁶²¹⁴⁴ + 1
  
• 75898⁵²⁴²⁸⁸ + 1 and 341112⁵²⁴²⁸⁸ + 1 and 356926⁵²⁴²⁸⁸ + 1 and 475856⁵²⁴²⁸⁸ + 1
  
• 919444^1048576 + 1
  

That's twenty-one primes, where we have counted 65537 twice (2¹⁶ + 1 and 4⁸ + 1). I guess 65537 is the only prime that can be written as b^N + 1 with 1 < b < N in more than one way?

The next bullet in the list is unknown! Run GFN21 to help determine if there are any primes here. The bullet after that is going to be GFN22+DYFL.

To be explicit, GFN21 needs to reach b=2097152 to determine how many primes in bullet #21. And GFN22 needs to catch up with DYFL, and DYFL needs to reach incredible b=4194304, with no "holes" (if DYFL jumps because of a new World Record outside PrimeGrid, then GFN22 must fill the hole) to complete bullet #22.

There is my OEIS entry A277967 where I count the number of primes in each bullet. Help extend it.

A very interesting sequence JeppeSN!

GFN19 seems to have been incredibly lucky to end up with 4 terms with b<524288. Only about a 2% chance of that based on Yves' calculator http://yves.gallot.pagesperso-orange.fr/primes/stat.html.

GFN20 was SO close to being 2 rather than 1 as well. The second known prime is 1059094^1048576+1, which is only 10518 past the cutoff.

I love searches like this because the goal is reached whether we find one prime, ten primes, or no primes. As soon as we pass b=2097152 with all b below that complete the sequence can be updated. With the current sieve depth on GFN21, a rough estimate is that we have about 275000 terms left to check, which means about 550000 tasks. I propose a GFN21 challenge next year to help push that forward!

Gets my vote.

Theoretical problem: Should A277967(n) tend to zero as n grows without bound? Can we give a qualified guess even if a proof is out of reach? See Yves Gallot's pages Results and Statistics.

Theoretical problem: Should A277967(n) tend to zero as n grows without bound? Can we give a qualified guess even if a proof is out of reach? See Yves Gallot's pages Results and Statistics.

After having consulted Yves, I now believe A277967 is "balanced" in the sense that limsup A277967(n) = +∞, and liminf A277967(n) = 0. And the "expected" value of each term is near 1.0. Of course, this is a conjecture at most, not something we can prove. /JeppeSN