Problems & Puzzles: Puzzles

Puzzle 415. Sets of consecutive primes such that... J.M. Bergot observes that {3,5,7} are consecutive primes, 3+5 can be divided by 2, 3+5+7 can be divided by 3.  Furthermore {47, 53, 59, 61} are consecutive primes, 47+53 is a multiple of 2, 47+53+59 is a multiple of 3 and 47+53+59+61 is a multiple of 4. I [C.R.] have produced the smallest set of k consecutive primes having the property described by Bergot, for k= 5 to 11. Including the two sets mentioned by Bergot for k=3 & 4, these are the following:  k First prime in the set  3       3  4       47  5       1531  6       4073  7       5081  8       537661  9       5538947  10      5981567  11      148871869 Q1. Can you extend this table? Q2.  Bergot asks if you devise interesting properties for these sets?

---

Contributions came from Adam Stinchcombe & Dan Dima.

***

Adam wrote:

If you believe the extended Dirichlet conjecture (arbitrary strings of linear functions contain infinitely many instances of simultaneous primes) then there are sequences of length at least up to 15 (and probably longer).  E.g., if 360360n+{19, 43, 61, 73, 79, 97, 139, 193, 223, 233, 281, 311, 367, 373, 433, 467} is a set of numbers that have the required modular relations.  All that is left is that the numbers have to be consecutive primes.  If the Dirichlet conjecture is true, they could all be primes, and if n is large enough, gap estimates would *suggest* that they would therefore be consecutive primes (frequently).  That is, if n is large enough so that a gap size (360360n+223) - (360360n+193) = 30 is unusual, then if they are prime, they are probably consecutive primes.  I tested this sequence out to about n=5.7 million and only found strings of length<=10 (first prime about 2*10^12).

***

Dan Dima found the smallest solution for k=12: 5545986967

***

 

---