A355280 Binary numbers (digits in {0, 1}) with no run of digits with length < 2.

11, 111, 1100, 1111, 11000, 11100, 11111, 110000, 110011, 111000, 111100, 111111, 1100000, 1100011, 1100111, 1110000, 1110011, 1111000, 1111100, 1111111, 11000000, 11000011, 11000111, 11001100, 11001111, 11100000, 11100011, 11100111, 11110000, 11110011, 11111000, 11111100, 11111111

Offset: 1

Author

M. F. Hasler, Oct 17 2022

Keywords

• nonn
• base

Comments

This is the binary representation of the terms in A033015.

The sequence can be seen as a table where row r contains the terms with r digits. Then row r+1 is obtained by from the terms of row r by duplicating their last digit, and from those of row r-1 by appending twice the 1's complement of their last digit. This yields the row lengths given in FORMULA.

Examples

			There can't be a terms with only 1 digit, so the smallest term is a(1) = 11.
The only 3-digit term is a(2) = 111, since in 100 the digit 1 is alone, and in 101 and 110 the digit 0 is alone.
With four digits we must have either no or two digits 0 and they must be at the end (to avoid isolated '1's), i.e., a(3) = 1100 and a(4) = 1111.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10945

Crossrefs

Cf. A033015 (the same terms converted from base 2 to base 10),

Subsequence of A007088 (the binary numbers); A000042 (numbers in base 1) = A002275 \ {0} (repunits) are subsequences; A061851 is the subsequence of palindromes.

Programs

Maple

F:= proc(d) option remember;
   local R,i,j, x0;
   R:= NULL;
   for i from d-2 to 2 by -1 do
     x0:= (10^d - 10^i)/9;
     for j from i-2 to 0 by -1 do
        R:= R, op(map(t -> t + x0, procname(j)))
     od
   od;
   sort([R, (10^d-1)/9])
end proc:
F(0):= [0]; F(1):= [];
seq(op(F[i]),i=2..9); # Robert Israel, May 12 2025

PARI

{is_A355280(n,d=digits(n))=vecmax(d)==1 && is_A033015(fromdigits(d,2))}
A355280(n)=A007088(A033015(n))
concat(apply( {A355280_row(n)=if(n>2, setunion([x*10+x%10|x<-A355280_row(n-1)],[x*100+11*(1-x%10)|x<-A355280_row(n-2)]), n>1, [11],[])}, [1..8])) \\ "Row" of n-digit terms. For (very) large n one should implement memoization instead of this naive recursion.

Python

def A355280_row(n): return [] if n<2 else [11] if n==2 else sorted(
    [x*10+x%10 for x in A355280_row(n-1)] +
    [x*100+11-x%10*11 for x in A355280_row(n-2)]) # M. F. Hasler, Oct 17 2022

Formula

a(n) = A007088(A033015(n)).

The number of terms with n digits is Fibonacci(n-1); the largest such term is A000042(n) = A002275(n).