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Search: id:A144789
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| A144789 |
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Consider the runs of 0's in the binary representation of n, each of these runs being on the edge of the binary repersentation n and/or being bounded by 1's. a(n) = the length of the shortest such run (with positive length) of 0's in binary n. a(n) = 0 if there are no runs of 0's in binary n. |
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+0
4
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| 0, 1, 0, 2, 1, 1, 0, 3, 2, 1, 1, 2, 1, 1, 0, 4, 3, 1, 2, 1, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 0, 5, 4, 1, 3, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 4, 3, 1, 2, 1, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 0, 6, 5, 1, 4, 2, 1, 1, 3, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 4, 1, 3, 2, 1, 1, 2, 1, 1
(list; graph; listen)
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OFFSET
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1,4
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LINKS
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Leroy Quet, Home Page (listed in lieu of email address)
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EXAMPLE
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20 in binary is 10100. The runs of 0s are as follows: 1(0)1(00). The shortest of these runs contains exactly one 0's So, a(20) = 1.
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MAPLE
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A007814 := proc(n) local nshf, a ; a := 0 ; nshf := n ; while nshf mod 2 = 0 do nshf := nshf/2 ; a := a+1 ; od: a ; end: A144789 := proc(n) option remember ; local lp2, lp2sh, bind ; bind := convert(n, base, 2) ; if add(i, i=bind) = nops(bind) then RETURN(0) ; fi; lp2 := A007814(n) ; if lp2 = 0 then A144789(floor(n/2)) ; else lp2sh := A144789(n/2^lp2) ; if lp2sh = 0 then lp2 ; else min(lp2, lp2sh) ; fi; fi; end: for n from 1 to 140 do printf("%d, ", A144789(n)) ; od: [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 29 2008]
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CROSSREFS
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A087117, A144790
Sequence in context: A085144 A156578 A097230 this_sequence A087117 A029340 A126258
Adjacent sequences: A144786 A144787 A144788 this_sequence A144790 A144791 A144792
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KEYWORD
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base,nonn
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AUTHOR
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Leroy Quet, Sep 21 2008
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EXTENSIONS
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Extended by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 29 2008
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