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a(n) = (2^A002326(n)-1)/(2*n+1).
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%I #14 Jun 04 2020 08:16:41

%S 1,1,3,1,7,93,315,1,15,13797,3,89,41943,9709,9256395,1,31,117,

%T 1857283155,105,25575,381,91,178481,42799,5,84973577874915,19065,4599,

%U 4885260612740877,18900352534538475,1,63,1101298153654301589

%N a(n) = (2^A002326(n)-1)/(2*n+1).

%C a(n) = 1 <=> n is in A000225 <=> n = 2^k - 1 with k = 0, 1, 2, ... - _M. F. Hasler_, Sep 20 2017

%H Robert Israel, <a href="/A165781/b165781.txt">Table of n, a(n) for n = 0..1672</a>

%p A002326 := proc(n) if n = 0 then 1 ; else numtheory[order](2,2*n+1) ; end if ; end proc:

%p A165781 := proc(n) (2^A002326(n)-1)/(2*n+1) ; end proc:

%p seq(A165781(n),n=0..60) ; # _R. J. Mathar_, Nov 16 2009

%t a[n_] := (2^MultiplicativeOrder[2, 2n+1]-1)/(2n+1);

%t a /@ Range[0, 40] (* _Jean-François Alcover_, Jun 04 2020 *)

%o (PARI) a(n)=(2^znorder(Mod(2,n=2*n+1))-1)/n \\ _M. F. Hasler_, Sep 20 2017

%Y Cf. A002326, A053446, A000225.

%K easy,nonn

%O 0,3

%A _Ctibor O. Zizka_, Sep 26 2009

%E Sign in definition and offset corrected by _R. J. Mathar_, Nov 16 2009