A087117 Number of zeros in the longest string of consecutive zeros in the binary representation of n.

1, 0, 1, 0, 2, 1, 1, 0, 3, 2, 1, 1, 2, 1, 1, 0, 4, 3, 2, 2, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 0, 5, 4, 3, 3, 2, 2, 2, 2, 3, 2, 1, 1, 2, 1, 1, 1, 4, 3, 2, 2, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 0, 6, 5, 4, 4, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 4, 3, 2, 2, 2, 1, 1, 1, 3, 2, 1, 1, 2, 1, 1, 1, 5, 4, 3, 3, 2, 2

Offset: 0

Author

Reinhard Zumkeller, Aug 14 2003

Keywords

• nonn
• base
• easy

Comments

The following four statements are equivalent: a(n) = 0; n = 2^k - 1 for some k > 0; A087116(n) = 0; A023416(n) = 0.

The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. Then a(k) is the maximum part of this composition, minus one. The maximum part is A333766(k). - Gus Wiseman, Apr 09 2020

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A023416, A007088, A007814.

Cf. A025480, A038374, A090046, A090047, A090048, A090049, A090050.

Positions of zeros are A000225.

Positions of terms <= 1 are A003754.

Positions of terms > 0 are A062289.

Positions of first appearances are A131577.

The version for prime indices is A252735.

The proper maximum is A333766.

The version for minimum is A333767.

Maximum prime index is A061395.

All of the following pertain to compositions in standard order (A066099):

- Length is A000120.

- Sum is A070939.

- Runs are counted by A124767.

- Strict compositions are A233564.

- Constant compositions are A272919.

- Runs-resistance is A333628.

- Weakly decreasing compositions are A114994.

- Weakly increasing compositions are A225620.

- Strictly decreasing compositions are A333255.

- Strictly increasing compositions are A333256.

Cf. A029931, A048793, A061395, A087117, A228351, A328594, A333217, A333218, A333219, A333767, A333768.

Programs

Haskell

import Data.List (unfoldr, group)
a087117 0       = 1
a087117 n
  | null $ zs n = 0
  | otherwise   = maximum $ map length $ zs n where
  zs = filter ((== 0) . head) . group .
       unfoldr (\x -> if x == 0 then Nothing else Just $ swap $ divMod x 2)
-- Reinhard Zumkeller, May 01 2012

Maple

A087117 := proc(n)
    local d,l,zlen ;
    if n = 0 then
        return 1 ;
    end if;
    d := convert(n,base,2) ;
    for l from nops(d)-1 to 0 by -1 do
        zlen := [seq(0,i=1..l)] ;
        if verify(zlen,d,'sublist') then
            return l ;
        end if;
    end do:
    return 0 ;
end proc; # R. J. Mathar, Nov 05 2012

Mathematica

nz[n_]:=Max[Length/@Select[Split[IntegerDigits[n,2]],MemberQ[#,0]&]]; Array[nz,110,0]/.-\[Infinity]->0 (* Harvey P. Dale, Sep 05 2017 *)

PARI

h(n)=if(n<2, return(0)); my(k=valuation(n,2)); if(k, max(h(n>>k), k), n++; n>>=valuation(n,2); h(n-1))
a(n)=if(n,h(n),1) \\ Charles R Greathouse IV, Apr 06 2022

Formula

a(n) = max(A007814(n), a(A025480(n-1))) for n >= 2. - Robert Israel, Feb 19 2017

a(2n+1) = a(n) (n>=1); indeed, the binary form of 2n+1 consists of the binary form of n with an additional 1 at the end - Emeric Deutsch, Aug 18 2017

For n > 0, a(n) = A333766(n) - 1. - Gus Wiseman, Apr 09 2020