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
***
|