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%I #11 May 08 2020 06:08:47
%S 0,1,1,1,0,2,1,1,0,1,1,1,1,2,1,1,0,1,1,1,0,2,1,1,0,1,1,2,1,2,1,1,0,1,
%T 1,1,0,2,1,1,0,1,1,1,1,2,1,1,0,1,1,1,0,2,1,1,0,1,2,2,1,2,1,1,0,1,1,1,
%U 0,2,1,1,0,1,1,1,1,2,1,1,0,1,1,1,0,2,1,1,0,1,1,2,1,2,1,1,0,1,1,1,0,2,1,1,0,1
%N Number of k >= 1 such that k+n == 0 mod 2^k.
%C Number of terms in the summation in the formula for A102370(n).
%C Also, a(n) is the number of 1's in (A103185(n) written in base 2).
%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 a(2^k + y) = a(y) + 1 if y = 2^k - k - 1, = a(y) otherwise (where 0 <= y <= 2^k - 1).
%p f:=proc(n) local t1,l; t1:=0; for l from 1 to n do if n+l mod 2^l = 0 then t1:=t1+1; fi; od: t1; end;
%Y Cf. A102370, A103185, A105035 (records).
%K nonn
%O 0,6
%A _N. J. A. Sloane_, Apr 02 2005
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