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a(n) = 2^n - A103529(n).
5

%I #18 May 08 2020 06:07:50

%S 2,1,2,1,4,3,2,5,8,7,6,1,12,11,10,13,16,15,14,9,4,19,18,21,24,23,22,

%T 17,28,27,26,29,32,31,30,25,20,3,34,37,40,39,38,33,44,43,42,45,48,47,

%U 46,41,36,51,50,53,56,55,54,49,60,59,58,61,64,63,62,57

%N a(n) = 2^n - A103529(n).

%H Nathaniel Johnston, <a href="/A103530/b103530.txt">Table of n, a(n) for n = 1..1000</a>

%H David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, Sloping binary numbers: a new sequence related to the binary numbers [<a href="http://neilsloane.com/doc/slopey.pdf">pdf</a>, <a href="http://neilsloane.com/doc/slopey.ps">ps</a>].

%H David Applegate, Benoit Cloitre, Philippe Deléham and N. J. A. Sloane, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL8/Sloane/sloane300.html">Sloping binary numbers: a new sequence related to the binary numbers</a>, J. Integer Seq. 8 (2005), no. 3, Article 05.3.6, 15 pp.

%F Also, miraculously, a(n+1) = 2^n - A102371(n) for n >= 1.

%e a(7) = 2^7 - A103529(7) = 128 - 126 = 2.

%p a:=proc(n)local k,t: if(n=1)then return 2:fi: t:= 2^(n-1) + (n-1): for k from 1 to n-1 do if(k = (n-1) mod 2^k)then t:=t-2^k: fi: od: return t: end: seq(a(n),n=1..80); # _Nathaniel Johnston_, Apr 30 2011

%K nonn,easy

%O 1,1

%A _N. J. A. Sloane_, _David Applegate_ and _Philippe Deléham_, Mar 22 2005