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%I #37 Aug 07 2023 19:56:45
%S 5,9,13,17,21,25,29,33,37,41,45,49,53,57,61,65,69,73,77,81,85,89,93,
%T 97,101,105,109,113,117,121,125,129,133,137,141,145,149,153,157,161,
%U 165,169,173,177,181,185,189,193,197,201,205,209,213,217,221,225
%N Numbers whose binary expansion ends 01.
%C These are the numbers for which zeta(2*x+1) needs just 3 terms to be evaluated. - _Jorge Coveiro_, Dec 16 2004
%C The binary representation of a(n) has exactly the same number of 0's and 1's as the binary representation of a(n+1). - _Gil Broussard_, Dec 18 2008
%C Number of monomials in n-th power of x^4+x^3+x^2+x+1. - _Artur Jasinski_, Oct 06 2008
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).
%F a(n) = 8*n-a(n-1)-2 (with a(1)=5). - _Vincenzo Librandi_, Nov 18 2010
%F From _Colin Barker_, Jun 24 2013: (Start)
%F a(n) = 2*a(n-1)-a(n-2).
%F G.f.: -x*(x-5) / (x-1)^2. (End)
%F E.g.f.: exp(x)*(1 + 4*x) - 1. - _Stefano Spezia_, Feb 02 2023
%p seq( 4*x+1, x=1..100 );
%t a = {}; k = x^4 + x^3 + x^2 + x + 1; m = k; Do[AppendTo[a, Length[m]]; m = Expand[m*k], {n, 1, 100}]; a (* _Artur Jasinski_, Oct 06 2008 *)
%t Select[Range[2,250],Take[IntegerDigits[#,2],-2]=={0,1}&] (* or *) LinearRecurrence[{2,-1},{5,9},70] (* _Harvey P. Dale_, Aug 07 2023 *)
%o (PARI) a(n)=4*n+1 \\ _Charles R Greathouse IV_, Oct 07 2015
%Y Essentially same as A016813.
%K nonn,easy
%O 1,1
%A _N. J. A. Sloane_
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