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

0,8

Comments

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

References

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

Links

Table of n, a(n) for n=0..101.

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

Example

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.

Prog

(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

Crossrefs

Cf. A005811.

Sequence in context: A208568 A193804 A180424 * A079693 A117444 A257145

Adjacent sequences: A092336 A092337 A092338 * A092340 A092341 A092342

Keyword

nonn,base

Author

Ralf Stephan, Mar 18 2004

Status

approved