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A038374
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Length of longest contiguous block of 1's in binary expansion of n.
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25
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1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 2, 2, 3, 4, 1, 1, 1, 2, 1, 1, 2, 3, 2, 2, 2, 2, 3, 3, 4, 5, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 2, 2, 3, 4, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 5, 6, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 2, 2, 3, 4, 1, 1, 1, 2, 1, 1, 2, 3, 2, 2, 2
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OFFSET
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1,3
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LINKS
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FORMULA
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May be defined by the recurrence given in A245196, taking G(n)=n+1 (n>=0) and m=1. - N. J. A. Sloane, Jul 25 2014
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EXAMPLE
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a(157) = 3 because 157 in base 2 is 10011101 and longest contiguous block of 1's is of length 3.
May be arranged into blocks of lengths 1, 2, 4, 8, 16, ...:
1,
1, 2,
1, 1, 2, 3,
1, 1, 1, 2, 2, 2, 3, 4,
1, 1, 1, 2, 1, 1, 2, 3, 2, 2, 2, 2, 3, 3, 4, 5,
1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 2, 2, 3, 4, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 5, 6,
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MAPLE
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A038374 := proc(n) local nshft, thisr, resul; nshft := n ; resul :=0 ; thisr :=0 ; while nshft > 0 do if nshft mod 2 <> 0 then thisr := thisr+1 ; else resul := max(resul, thisr) ; thisr := 0 ; fi ; nshft := floor(nshft/2) ; od ; resul := max(resul, thisr) ; RETURN(resul) ; end : for n from 1 to 80 do printf("%d, ", A038374(n)) ; od : # R. J. Mathar, Jun 15 2006
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MATHEMATICA
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Table[Max[Length/@DeleteCases[Split[IntegerDigits[n, 2]], _?(MemberQ[ #, 0] &)]], {n, 120}] (* Harvey P. Dale, Jun 10 2013 *)
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PROG
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(Haskell)
import Data.List (unfoldr, group)
a038374 = maximum . map length . filter ((== 1) . head) . group .
unfoldr (\x -> if x == 0 then Nothing else Just $ swap $ divMod x 2)
(PARI) a(n)=if (n==0, return (0)); n>>=valuation(n, 2); if(n<2, return(n)); my(e=valuation(n+1, 2)); max(e, a(n>>e)) \\ Charles R Greathouse IV, Jan 12 2014; edited by Michel Marcus, Apr 14 2019
(Python)
from itertools import groupby
def a(n): return max(len(list(g)) for k, g in groupby(bin(n)[1:]) if k=='1')
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CROSSREFS
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KEYWORD
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base,easy,nonn,tabf
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AUTHOR
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STATUS
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approved
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