%I #8 Feb 25 2019 08:25:22
%S 2,3,4,6,9,7,8,12,10,11,13,14,17,15,16,24,18,19,20,22,37,23,25,28,26,
%T 27,29,30,33,31,32,48,34,35,36,38,41,39,40,44,42,43,45,46,53,47,49,56,
%U 50,51,52,54,69,55,57,60,58,59,61,62,65,63,64,96,66,67,68
%N Next larger integer with same number of runs of 1's in its binary representation as n.
%C Number of runs of 1's in binary representation is given by A069010.
%C Each nonnegative number either appears in this sequence or in A002450.
%F a(A023758(n)) = A023758(n+1) for any n > 1.
%F a(A043682(n)) = A043682(n+1) for any n > 0.
%F a(A043683(n)) = A043683(n+1) for any n > 0.
%F a(A043684(n)) = A043684(n+1) for any n > 0.
%F a(A043685(n)) = A043685(n+1) for any n > 0.
%F a(A043686(n)) = A043686(n+1) for any n > 0.
%e The first terms, in decimal and in binary, are:
%e n a(n) bin(n) bin(a(n))
%e -- ---- ------ ---------
%e 1 2 1 10
%e 2 3 10 11
%e 3 4 11 100
%e 4 6 100 110
%e 5 9 101 1001
%e 6 7 110 111
%e 7 8 111 1000
%e 8 12 1000 1100
%e 9 10 1001 1010
%e 10 11 1010 1011
%e 11 13 1011 1101
%e 12 14 1100 1110
%e 13 17 1101 10001
%e 14 15 1110 1111
%e 15 16 1111 10000
%e 16 24 10000 11000
%o (PARI) r1(n) = my (c=0); while (n, my (v=valuation(n+(n%2),2)); if (n%2, c++); n\=2^v); c
%o a(n) = my (r=r1(n)); for (k=n+1, oo, if (r==r1(k), return (k)))
%Y Cf. A002450, A023758, A043682, A043683, A043684, A043685, A043686, A057168, A069010.
%K nonn,base
%O 1,1
%A _Rémy Sigrist_, Feb 15 2019