A123466 Write the positive integer n in binary. Subdivide the binary n into runs each consisting entirely of 0's or of 1's, where the runs alternate between those of 1's and those of 0's. The sequence gives those numbers n such that there is at least one run of 1's of the same length as at least one run of 0's.
2, 5, 10, 11, 12, 13, 18, 19, 20, 21, 22, 23, 25, 26, 29, 34, 37, 38, 40, 41, 42, 43, 44, 45, 46, 47, 50, 51, 52, 53, 56, 58, 61, 66, 69, 70, 71, 74, 75, 76, 77, 78, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 98, 100, 101, 102, 103, 104
Offset: 1
Examples
25 written in binary is 11001. The runs are (11)(00)(1). Since at least one run of 1's (the leftmost run here) is the same length as at least one run of 0's (the only run of 0's here), 25 is included in this sequence.
Links
- Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Crossrefs
Cf. A033015.
Programs
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PARI
is(n)=my(ones=List(),zeros=List()); if(n%2, listput(ones, valuation(n+1,2)); n>>=ones[1]); while(n, listput(zeros, valuation(n,2)); n>>=zeros[#zeros]; listput(ones, valuation(n+1,2)); n>>=ones[#ones]); #setintersect(vecsort(Vec(ones),,8), vecsort(Vec(zeros),,8))>0 \\ Charles R Greathouse IV, Mar 29 2013
Formula
a(n) ~ n. - Charles R Greathouse IV, Mar 29 2013
Extensions
a(16) to a(27) from Ray G. Opao, Jan 08 2009
a(28)-a(64) from Lars Blomberg, Dec 09 2011
Comments