OFFSET
1,1
COMMENTS
If 2^(k-1) == 1 (mod k) and 2^(2^(k-1)-1) == 1 (mod k), then 2^(2^(k-1)-1) == 1 (mod k^2). In fact, all such pseudoprimes are strong pseudoprimes to base 2.
Other terms; 951481 = 271*3511, 2215441 = 631*3511, 28758601 = 8191*3511, ... are not Fermat pseudoprimes to base 2, where 3511 is a Wieferich prime. The Wieferich prime 1093 cannot be a factor of these numbers (see A374953).
MATHEMATICA
q[k_] := Module[{m = MultiplicativeOrder[2, k^2]}, PowerMod[2, k-1, m] == 1]; Select[Range[1, 10^6, 2], CompositeQ[#] && q[#] &] (* Amiram Eldar, Sep 17 2024 *)
PROG
(PARI) is(k) = (k > 1) && k % 2 && !isprime(k) && Mod(2, znorder(Mod(2, k^2)))^(k-1) == 1; \\ Amiram Eldar, Sep 17 2024
CROSSREFS
KEYWORD
nonn,new
AUTHOR
Thomas Ordowski, Sep 17 2024
EXTENSIONS
More terms from Amiram Eldar, Sep 17 2024
STATUS
approved