A063037 Numbers without 3 consecutive equal binary digits.

0, 1, 2, 3, 4, 5, 6, 9, 10, 11, 12, 13, 18, 19, 20, 21, 22, 25, 26, 27, 36, 37, 38, 41, 42, 43, 44, 45, 50, 51, 52, 53, 54, 73, 74, 75, 76, 77, 82, 83, 84, 85, 86, 89, 90, 91, 100, 101, 102, 105, 106, 107, 108, 109, 146, 147, 148, 149, 150, 153, 154, 155, 164, 165, 166

Offset: 1

Author

Lior Manor, Jul 05 2001

Keywords

• easy
• nonn

Comments

Complement of A136037; intersection of A003796 and A003726. - Reinhard Zumkeller, Dec 11 2007

From Emeric Deutsch, Jan 27 2018: (Start)

Also 0 together with the indices of the compositions that have no parts larger than 2. For the definition of the index of a composition see A298644.

For example, 105 is in the sequence since its binary form is 1101001 and the composition [2,1,1,2,1] has no parts larger than 2.

On the other hand, 132 is not in the sequence since its binary form is 10000100 and the composition [1,4,1,2] has a part larger than 2.

The command c(n) from the Maple program yields the composition having index n. (End)

The sequence contains A000045(n+1) positive terms with binary length n. - Rémy Sigrist, Sep 30 2022

Examples

			The binary representation of 9 (1001) has no 3 consecutive equal digits.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000045, A000975, A166535, A286262.

Programs

Maple

isA063037 := proc(n)
    local bdgs,rep,d,i ;
    if n = 0 then
        return true;
    end if;
    bdgs := convert(n,base,2) ;
    rep := 1;
    d := op(1,bdgs) ;
    for i from 2 to nops(bdgs) do
        if op(i,bdgs) = op(i-1,bdgs) then
            rep := rep+1 ;
        else
            rep :=1 ;
        end if ;
        if rep > 2 then
            return false;
        end if;
    end do:
    return true ;
end proc:
for n from 0 to 50 do
    if isA063037(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, Dec 18 2013
# Second Maple program:
Runs := proc (L) local j, r, i, k: j := 1: r[j] := L[1]: for i from 2 to nops(L) do if L[i] = L[i-1] then r[j] := r[j], L[i] else j := j+1: r[j] := L[i] end if end do: [seq([r[k]], k = 1 .. j)] end proc: RunLengths := proc (L) map(nops, Runs(L)) end proc: c := proc (n) ListTools:-Reverse(convert(n, base, 2)): RunLengths(%) end proc: A := {0}: for n to 250 do if `subset`(convert(c(n), set), {1, 2}) then A := `union`(A, {n}) else end if end do: A; # most of this Maple program is due to W. Edwin Clark. - Emeric Deutsch, Jan 27 2018

Mathematica

Select[Range[0, 168], AllTrue[Length /@ Split@ IntegerDigits[#, 2], # < 3 &] &] (* Michael De Vlieger, Aug 20 2017 *)

PARI

{ n=0; for (m=0, 10^9, x=m; t=1; b=2; while (x>0, d=x-2*(x\2); x\=2; if (d==b, c++; if (c==3, t=x=0), b=d; c=1)); if (t, write("b063037.txt", n++, " ", m); if (n==1000, break)) ) } \\ Harry J. Smith, Aug 16 2009

PARI

a(n) = { n--; if (n<=1, return (n), my (s=1); for (i=1, oo, my (f=fibonacci(i+1)); if (nRémy Sigrist, Sep 30 2022

Python

from itertools import count, islice
def A063037_gen(startvalue=0): # generator of terms >= startvalue
    return filter(lambda n: not ('000' in (s:=bin(n)[2:]) or '111' in s), count(max(0,startvalue)))
A063037_list = list(islice(A063037_gen(),20)) # Chai Wah Wu, Oct 04 2022

Formula

It appears (but has not yet been proved) that the terms of {a(n)} can be computed recursively as follows. Let {c(i)} be defined for i >= 4 by c(i) = 2c(i-1) + 1, if i is a multiple of 3, else c(i) = 2c(i-1) - 1, with c(4) = 1. I.e., {c(i)} = {1,1,3,5,9,19,37,73,147,...}, for i=4,5,6,... . Let a(1)=1, a(2)=2, a(3)=3. For n > 3, choose k so that F(k)-2 < n <= F(k+1)-2, where F(k) denotes the k-th Fibonacci number (A000045). Then a(n) = c(k) + 2a(F(k)-2) - a(2F(k)-n-3). This has been verified for n up to 1100. - John W. Layman, May 26 2009

Extensions

Missing "less than" sign supplied in the conjectured recurrence (thanks to Franklin T. Adams-Watters for pointing this out) by John W. Layman, Nov 09 2009