A318928 Runs-resistance of binary representation of n.
1, 2, 1, 3, 2, 3, 1, 3, 3, 2, 4, 2, 4, 3, 1, 3, 3, 5, 4, 4, 2, 5, 4, 3, 4, 4, 3, 3, 4, 3, 1, 3, 3, 5, 3, 3, 5, 4, 3, 4, 5, 2, 4, 3, 4, 5, 4, 3, 3, 3, 2, 4, 4, 3, 3, 2, 3, 4, 3, 3, 4, 3, 1, 3, 3, 5, 3, 3, 5, 3, 4, 3, 3, 5, 6, 4, 5, 3, 3, 4, 5, 4, 4, 4, 2, 5, 4, 5, 5, 4, 5, 5, 4, 5, 4
Offset: 1
Examples
11 in binary is [1, 0, 1, 1], which has runs of lengths [1, 1, 2], which has runs of lengths [2, 1], which has runs of lengths [1, 1], which has a single run of length [2]. This took four steps, so a(11) = 4.
Links
- N. J. A. Sloane, Table of n, a(n) for n = 1..10000
- Claude Lenormand, Deux transformations sur les mots, Preprint, 5 pages, Nov 17 2003. Apparently unpublished. This is a scanned copy of the version that the author sent to me in 2003.
- N. J. A. Sloane, Transforms
Crossrefs
Programs
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Maple
with(transforms); # compute Lenormand's "resistance" of a list resist:=proc(a) local ct,i,b; if whattype(a) <> list then ERROR("input must be a list"); fi: ct:=0; b:=a; for i from 1 to 100 do if nops(b)=1 then return(ct); fi; b:=RUNS(b); ct:=ct+1; od; end; a:=[1]; for n from 2 to 100 do b:=convert(n,base,2); r:=resist(b); a:=[op(a),r]; od: -
Mathematica
Table[If[n == 1, 1, Length[NestWhileList[Length/@Split[#] &, IntegerDigits[n, 2], Length[#] > 1 &]] - 1], {n, 50}] (* Gus Wiseman, Nov 25 2019 *)
Extensions
a(1) corrected by N. J. A. Sloane, Sep 20 2018
Comments