A063250 Number of binary right-rotations (iterations of A038572) to reach fixed point.

0, 0, 1, 0, 2, 2, 1, 0, 3, 3, 3, 3, 2, 2, 1, 0, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 2, 2, 1, 0, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 3, 2, 2, 1, 0, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 5, 5, 5, 5, 5, 5, 5, 5, 5

Offset

0,5

Comments

a(n) = 0 when n is a fixed point of form 2^k-1 left-rotation analog appears to be same as A048881.

Links

Table of n, a(n) for n=0..104.

Formula

If n+1 is a power of 2 then a(n)=0 otherwise a(n) = 1 + a(floor(n/2)).

Conjectured g.f.: 1/(1-x) * Sum_{j>=0} x^(2^j) - (1-x^(2^j)) * Sum_{k>=1} x^((2^j)*(2^k-1)). - Mikhail Kurkov, Sep 29 2019

Example

a(11)=3 since under right-rotation 11 -> 13 -> 14 -> 7 and 7 is a fixed point.

Mathematica

Table[Length[FixedPointList[FromDigits[RotateRight[IntegerDigits[ #, 2]], 2]&, n]]-2, {n, 0, 110}] (* Harvey P. Dale, Dec 23 2011 *)

Prog

(Python)
def a(n):
    if n<2: return 0
    b=bin(n)[2:]
    s=0
    while "0" in b:
        N=int(b[-1] + b[:-1], 2)
        s+=1
        b=bin(N)[2:]
    return s
print([a(n) for n in range(105)]) # Indranil Ghosh, May 25 2017

Crossrefs

Cf. A038572, A048881.

Sequence in context: A368282 A285099 A306754 * A348967 A285308 A276543

Adjacent sequences: A063247 A063248 A063249 * A063251 A063252 A063253

Keyword

base,easy,nonn

Author

Marc LeBrun, Jul 11 2001

Status

approved