login
Perfect numbers minus 1.
5

%I #33 Nov 06 2018 08:20:36

%S 5,27,495,8127,33550335,8589869055,137438691327,2305843008139952127,

%T 2658455991569831744654692615953842175,

%U 191561942608236107294793378084303638130997321548169215

%N Perfect numbers minus 1.

%C Conjecture: every odd prime is a factor of at least one number in this sequence. - _J. Lowell_, Apr 17 2014

%C a(n) written in base 2 is A138831(n) which is also a member of A138148 (cyclops numbers with binary digits only), assuming there are no odd perfect numbers. - _Omar E. Pol_, Oct 24 2016 [clarified on Nov 03 2018].

%H G. C. Greubel, <a href="/A135627/b135627.txt">Table of n, a(n) for n = 1..15</a>

%F a(n) = A000396(n) - 1.

%t (# (# + 1))/2 & /@ Select[2^Range[100] - 1, PrimeQ] - 1 (* _G. C. Greubel_, Oct 24 2016, based on _Harvey P. Dale_ code from A000396 *)

%Y Cf. A000396, A065091, A138148, A134808, A138831.

%K nonn

%O 1,1

%A _Omar E. Pol_, Nov 27 2007