login
Numbers k such that ((2^(2k) - 1) mod 2k) - (2^(2k-1) mod 2k) = 1.
1

%I #20 Jun 28 2024 09:29:43

%S 1,3,5,7,11,13,15,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,

%T 85,89,91,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,

%U 173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281,283,293,307

%N Numbers k such that ((2^(2k) - 1) mod 2k) - (2^(2k-1) mod 2k) = 1.

%C Apparently, the sequence contains 1, odd primes and the elements of A020136. - _R. J. Mathar_, Jan 09 2011

%H Robert Israel, <a href="/A179458/b179458.txt">Table of n, a(n) for n = 1..10000</a>

%p select(n -> (2 &^ (2*n)-1 mod 2*n)-(2 &^(2*n-1) mod 2*n) = 1, [$1..1000]); # _Robert Israel_, Oct 25 2017

%o (PARI) isok(n) = (((2^(2*n)-1) % (2*n)) - (2^(2*n-1) % (2*n)) == 1) \\ _Michel Marcus_, Jul 25 2013

%Y Cf. A001567.

%K nonn

%O 1,2

%A _Juri-Stepan Gerasimov_, Jan 07 2011