A092339 Number of adjacent identical digits in the binary representation of n.

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

Offset: 0

Author

Ralf Stephan, Mar 18 2004

Keywords

• nonn
• base

Comments

In binary: number of 00 blocks plus number of 11 blocks. (Note: the blocks can overlap. See the example below.)

Examples

			60 in binary is 111100, it has 4 blocks of adjacent digits, so a(60)=4.
Equally, 60's binary Gray code expansion is A003188(60)=34, 100010 in binary, which contains four zeros.
		

References

• J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 84.

Crossrefs

Cf. A005811.

Programs

PARI

a(n)=local(v); v=binary(n); sum(k=1, length(v)-1, v[k]==v[k+1])

PARI

a(n)=if(n<1,0,if(n%2==0,a(n/2)+(n>0&&(n/2)%2==0),a((n-1)/2)+((n-1)/2)%2))

Scheme

(define (A092339 n) (A080791 (A003188 n))) ;; Antti Karttunen, Jul 05 2013

Formula

Recurrence: a(2n) = a(n) + [n even], a(2n+1) = a(n) + [n odd].

a(n) = A014081(n) + A056973(n).

For n>0, A227185(n) = a(n)+1.

a(n) = A080791(A003188(n)) [because the sequence gives the number of nonleading zeros in binary Gray code expansion of n] - Antti Karttunen, Jul 05 2013