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%I #17 Dec 01 2021 03:52:41
%S 0,1,0,3,1,1,0,7,3,2,1,3,1,1,0,15,7,6,3,4,2,2,1,7,3,2,1,3,1,1,0,31,15,
%T 14,7,12,6,6,3,8,4,5,2,4,2,2,1,15,7,6,3,4,2,2,1,7,3,2,1,3,1,1,0,63,31,
%U 30,15,28,14,14,7,24,12,13,6,12,6,6,3,16,8,9,4,11,5,5,2,8,4,5,2,4,2,2,1
%N List the run lengths of n's binary runs of zeros, then interpret this list as lengths of runs of alternating ones and zeros in binary.
%C A variant of A175945, where in the first (analyzing) step not the runs of 1's but the runs of 0's determine the list of run lengths. The second (synthesizing) step is the same in both sequences. - _R. J. Mathar_, May 28 2011
%H Paul Tek, <a href="/A175946/b175946.txt">Table of n, a(n) for n = 1..10000</a>
%e From _R. J. Mathar_, May 28 2011: (Start)
%e N=16 is 10000 in binary which has one run of 4 zeros, and its run-length encoding is 4. This is interpreted as one run of 4 one's, 1111, which back to decimal is a(16)=15.
%e N=14 is 1110 in binary which has one run of 1 zero, and its run-length encoding is 1. This is interpreted as one run of 1 one, 1, which is in decimal a(14)=1.
%e N=19 is 10011 in binary which has one run of 2 zeros, and its run-length encoding is 2. This is interpreted as one run of 2 ones, 11, which back to decimal is a(19)=3. (End)
%t takelist[l_, t_] := Module[{lent, term},Set[lent, Length[t]]; Table[l[[t[[y]]]], {y, 1, lent}]]
%t frombinrep[x_] := FromDigits[Flatten[Table[Table[If[OddQ[n], 1, 0], {d, 1, x[[n]]}], {n, 1, Length[x]}]], 2]
%t binrep[x_] := repcount[IntegerDigits[x, 2]]
%t onebinrep[x_]:=Module[{b},b=binrep[x];takelist[b,Range[1,Length[b],2]]]
%t zerobinrep[x_]:=Module[{b},b=binrep[x];takelist[b,Range[2,Length[b],2]]]
%t Table[frombinrep[zerobinrep[n]], {n,START,END}]
%K base,nonn
%O 1,4
%A _Dylan Hamilton_, Oct 28 2010
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