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A318928
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Runs-resistance of binary representation of n.
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65
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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
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OFFSET
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1,2
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COMMENTS
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Following Lenormand (2003), we define the "runs-resistance" of a finite list L to be the number of times the RUNS transformation must be applied to L in order to reduce L to a list with a single element.
Here it is immaterial whether we read the binary representation of n from left to right or right to left.
The RUNS transformation must be applied at least once, in order to obtain a list, so a(n) >= 1.
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LINKS
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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 Apparently unpublished.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.
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EXAMPLE
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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.
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MAPLE
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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:
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MATHEMATICA
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Table[If[n == 1, 1, Length[NestWhileList[Length/@Split[#] &, IntegerDigits[n, 2], Length[#] > 1 &]] - 1], {n, 50}] (* Gus Wiseman, Nov 25 2019 *)
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CROSSREFS
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Ignoring the first digit gives A329870.
Compositions counted by runs-resistance are A329744.
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KEYWORD
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nonn,base,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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