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A063250
Number of binary right-rotations (iterations of A038572) to reach fixed point.
11
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.
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
Sequence in context: A368282 A285099 A306754 * A348967 A285308 A276543
KEYWORD
base,easy,nonn
AUTHOR
Marc LeBrun, Jul 11 2001
STATUS
approved