A196096 Occurrences of '11' in base 3 expansion of n.

0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 2, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 2, 3, 2, 1, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 2, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 0

Author

Jonathan Vos Post, Sep 27 2011

Keywords

• easy
• nonn

Comments

Occurrences of '11' in A007089(n). This is to base 3 and A007089 as A014081 is to base 2 A007088.

First occurrence of k>0 = 4, 13, 40, 121, 364, ..., = A003462(k). - Robert G. Wilson v, Sep 27 2011

Examples

			a(4) = 1 because 4 in base 3 is "11" which has one instance of "11".
a(13) = 2 because the number 13 in base 3 is "111" which has two substrings of "11".
		

Links

• T. D. Noe, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A007089, A014081.

Programs

Maple

A196096 := proc(n)
        local a,dgs3 ;
        a := 0 ;
        dgs3 := convert(n,base,3) ;
        for i from 1 to nops(dgs3)-1 do
                if op(i,dgs3)=1 and op(i+1,dgs3)=1 then
                        a := a+1 ;
                end if;
        end do;
        a ;
end proc:
seq(A196096(n),n=0..80) ; # R. J. Mathar, Sep 28 2011

Mathematica

f[n_] := Count[ Partition[ IntegerDigits[ n, 3], 2, 1], {1, 1}]; Array[f, 100, 0] (* Robert G. Wilson v, Sep 27 2011 *)