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A092339
Number of adjacent identical digits in the binary representation of n.
7
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.
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
KEYWORD
nonn,base
AUTHOR
Ralf Stephan, Mar 18 2004
STATUS
approved