|
%I #11 Nov 07 2013 10:27:27
%S 0,1,2,4,5,7,8,10,14,20,29,38,49,62,82,94,98,155,172,349,853,1307,
%T 1768,2902,5249,5344,5638,6194,7238
%N Number of 0's in A127962(n).
%C Apparently n such that (2^(2*n+3)+1)/3 is prime. - _James R. Buddenhagen_, Apr 14 2011
%F a(n) = A023416(A000979(n)). - _Michel Marcus_, Nov 07 2013
%t b = {}; Do[c = 1 + Sum[2^(2n - 1), {n, 1, x}]; If[PrimeQ[c], AppendTo[b, c]], {x, 0, 1000}]; a = {}; Do[AppendTo[a, FromDigits[IntegerDigits[b[[x]], 2]]], {x, 1, Length[b]}]; d = {}; Do[AppendTo[d, DigitCount[a[[x]], 10, 0]], {x, 1, Length[a]}]; d (*Artur Jasinski*)
%Y Cf. A127962, A127961, A000979, A000978, A124400, A126614, A127955, A127956, A127957, A127958, A127936.
%K nonn
%O 1,3
%A _Artur Jasinski_, Feb 09 2007
%E a(22)-a(29) from _Vincenzo Librandi_, Mar 31 2012
|
