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%I #48 Nov 24 2022 04:21:02
%S 1,3,3,7,5,7,7,15,9,11,11,15,13,15,15,31,17,19,19,23,21,23,23,31,25,
%T 27,27,31,29,31,31,63,33,35,35,39,37,39,39,47,41,43,43,47,45,47,47,63,
%U 49,51,51,55,53,55,55,63,57,59,59,63,61,63,63,127,65,67,67,71,69,71
%N Replace all trailing 0's with 1's in binary representation of n.
%C a(k+1) = smallest number greater than k having in its binary representation exactly one 1 more than k has; A000120(a(n)) = A063787(n). - _Reinhard Zumkeller_, Jul 31 2010
%C a(n) is the least m >= n-1 such that the Hamming distance D(n-1,m) = 1. - _Vladimir Shevelev_, Apr 18 2012
%H Reinhard Zumkeller, <a href="/A086799/b086799.txt">Table of n, a(n) for n = 1..10000</a>
%H Ralf Stephan, <a href="/somedcgf.html">Some divide-and-conquer sequences ...</a>
%H Ralf Stephan, <a href="/A079944/a079944.ps">Table of generating functions</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BinaryCarrySequence.html">Binary Carry Sequence</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/OddPart.html">Odd Part</a>
%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>
%F a(n) = n + 2^A007814(n) - 1.
%F a(n) is odd; a(n) = n iff n is odd.
%F a(a(n)) = a(n); A007814(a(n)) = a(n); A000265(a(n)) = a(n).
%F A023416(a(n)) = A023416(n) - A007814(n) = A086784(n).
%F A000120(a(n)) = A000120(n) + A007814(n).
%F a(2^n) = a(A000079(n)) = 2*2^n - 1 = A000051(n+1).
%F a(n) = if n is odd then n else a(n/2)*2 + 1.
%F a(n) = A006519(n) + n - 1. - _Reinhard Zumkeller_, Feb 02 2007
%F a(n) = n OR n-1 (bitwise OR of consecutive numbers). - _Russ Cox_, May 15 2007
%F a(2*n) = A038712(n) + 2*n. - _Reinhard Zumkeller_, Aug 07 2011
%F a((2*n-1)*2^p) = 2^(p+1)*n-1, p >= 0. - _Johannes W. Meijer_, Feb 01 2013
%F Sum_{k=1..n} a(k) ~ n^2/2 + (1/(2*log(2)))*n*log(n) + (3/4 + (gamma-1)/(2*log(2)))*n, where gamma is Euler's constant (A001620). - _Amiram Eldar_, Nov 24 2022
%e a(20) = a('10100') = '10100' + '11' = '10111' = 23.
%p nmax:=70: for p from 0 to ceil(simplify(log[2](nmax))) do for n from 1 to ceil(nmax/(p+2)) do a((2*n-1)*2^p) := 2^(p+1)*n-1 od: od: seq(a(n), n=1..nmax); # _Johannes W. Meijer_, Feb 01 2013
%t Table[BitOr[(n + 1), n], {n, 0, 100}] (* _Vladimir Joseph Stephan Orlovsky_, Jul 19 2011 *)
%o (C) int a(int n) { return n | (n-1); } // _Russ Cox_, May 15 2007
%o (Haskell)
%o a086799 n | even n = (a086799 $ div n 2) * 2 + 1
%o | otherwise = n
%o -- _Reinhard Zumkeller_, Aug 07 2011
%o (PARI) a(n)=bitor(n,n-1) \\ _Charles R Greathouse IV_, Apr 17 2012
%o (Python)
%o def a(n): return n | (n-1)
%o print([a(n) for n in range(1, 71)]) # _Michael S. Branicky_, Jul 13 2022
%Y Cf. A000051, A000079, A000120, A000265, A001620, A006519, A007814, A023416, A038712, A063787, A086784.
%Y Cf. also A007088, A179857, A220466.
%K nonn,base
%O 1,2
%A _Reinhard Zumkeller_, Aug 05 2003
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