Problems & Puzzles: Problems

Problem 68. More on Brier numbers.

Arkadiusz Wesolowski sent the following puzzle proposal, related to Brier numbers. Regarding this issue, see previous Problems 29, 49, 52 & 58.

A Brier number is an integer k such that both k.2n+1 and k.2n-1 are composite for any value of n.

Let k be a Brier number. Let A be a covering set for the Sierpinski number k and let B be a covering set for the Riesel number k, then A ∪ B = S = {p(1), p(2), ..., p(s)} a set of primes and P = product[p(i)](i = 1...s).
 
Number of primes used in both sets A & B Discoverer P value
1 Cohen & Selfridge (See 1974) 33241542480794255062005345795
2 Arkadiusz Wesolowski (June, 2017) 22366096283725870121540301161955
3 ? ?
4 ? ?


Example by Cohen & Selfridge:

A={3, 5, 13, 17, 97, 257, 241}
B={3, 7, 11, 19, 31, 37, 41, 61, 73, 109, 151, 331}
P=3*5*13*17*97*257*241*7*11*19*31*37*41*61*73*109*151*331=
33241542480794255062005345795 (29 digits)
k=47867742232066880047611079

Example by Arkadiusz:

A = {3, 5, 13, 17, 19, 241, 433, 38737}
B = {3, 7, 11, 19, 31, 37, 41, 61, 73, 109, 151, 331}
P = 3*5*13*17*19*241*433*38737*7*11*31*37*41*61*73* 109*151*331 = 22366096283725870121540301161955 (32 digits)
k=2769569390938829068824114667
Q. Can you find smallest values of P for which "Number of primes used in both sets (A and B)" = 1, 2, 3, 4, etc

Arkadiusz Wesolowski wrote on August 16, 2017_

I have found a new solution.
 
The number of primes used in both sets (A and B) is 2.
 
A = {3, 5, 17, 19, 97, 109, 241, 257, 433}
B = {3, 7, 11, 13, 19, 31, 37, 41, 61, 73, 151, 331}
P = 3*5*17*19*97*109*241*257*433*7*11*13*31*37*41*61*73*151*331 = 
14393587894183912441848314729235 (32 digits)
k = 1631535784434767401179758185111

***

On Aug 19, 2017, Arkadiusz wrote:
Yet another solution to this problem.

 
The number of primes used in both sets (A and B) is 3.

 
A = {3, 5, 11, 13, 17, 181, 241, 331, 433, 631, 1321, 61681}
B = {3, 7, 11, 19, 31, 37, 41, 61, 73, 109, 151, 331}
P = 3*5*11*13*17*181*241*331*433*631*1321*61681*7*19*31*37*41*61*73*109*151 = 
5373117827642170481646844213356125774865 (40 digits)
k = 6002524927312209061963632472348345399

***

On May 26, 2018, Arkadiusz wrote:

I have found another solution.
The number of primes used in both sets (A and B) is 3.

 
A = {3, 7, 11, 13, 19, 31, 37, 41, 61, 73, 109, 151}
B = {3, 5, 11, 17, 19, 97, 181, 241, 257, 331, 433, 631, 1321}
P = 3*7*11*13*19*31*37*41*61*73*109*151*5*17*97*181*241*257*331*433*631*1321 = 2171599914484065886366533930947209990785 (40 digits)
k = 742067290618889036720616945148038913

***

 


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