

A175946


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.


4



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, 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, 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
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OFFSET

1,4


COMMENTS

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


LINKS

Paul Tek, Table of n, a(n) for n = 1..10000


EXAMPLE

From R. J. Mathar, May 28 2011 (Start):
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.
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.
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)


MATHEMATICA

takelist[l_, t_] := Module[{lent, term}, Set[lent, Length[t]]; Table[l[[t[[y]]]], {y, 1, lent}]]
frombinrep[x_] := FromDigits[Flatten[Table[Table[If[OddQ[n], 1, 0], {d, 1, x[[n]]}], {n, 1, Length[x]}]], 2]
binrep[x_] := repcount[IntegerDigits[x, 2]]
onebinrep[x_]:=Module[{b}, b=binrep[x]; takelist[b, Range[1, Length[b], 2]]]
zerobinrep[x_]:=Module[{b}, b=binrep[x]; takelist[b, Range[2, Length[b], 2]]]
Table[frombinrep[zerobinrep[n]], {n, START, END}]


CROSSREFS

Sequence in context: A216954 A124801 A124926 * A115378 A120060 A143295
Adjacent sequences: A175943 A175944 A175945 * A175947 A175948 A175949


KEYWORD

base,nonn


AUTHOR

Dylan Hamilton, Oct 28 2010


STATUS

approved



