OFFSET
1,1
COMMENTS
Related to Diophantine equations of the form (2^x-1)*(3^y-1) = n*z^2.
LINKS
Sofia Lacerda, C++ program (github).
MathOverflow, Solve this Diophantine equation (2^x-1)(3^y-1)=2z^2
MATHEMATICA
Select[Range[10^6], PrimeQ[#] && CoprimeQ[MultiplicativeOrder[2, #], MultiplicativeOrder[3, #]] &] (* Amiram Eldar, May 11 2021 *)
PROG
(C++) // See Lacerda link.
(PARI) isok(p) = isprime(p) && (gcd(znorder(Mod(2, p)), znorder(Mod(3, p))) == 1); \\ Michel Marcus, May 11 2021
(Python)
from sympy.ntheory import n_order
from sympy import gcd, nextprime
A344202_list, p = [], 5
while p < 10**9:
if gcd(n_order(2, p), n_order(3, p)) == 1:
A344202_list.append(p)
p = nextprime(p) # Chai Wah Wu, May 12 2021
CROSSREFS
KEYWORD
nonn,more,hard
AUTHOR
Sofia Lacerda, May 11 2021
EXTENSIONS
a(3)-a(5) from Michel Marcus, May 11 2021
a(6)-a(8) from Amiram Eldar, May 11 2021
a(9) from Daniel Suteu, May 16 2021
a(10) from Sofia Lacerda, Jul 07 2021
a(11) from Sofia Lacerda, Aug 03 2021
STATUS
approved