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

Author

Marc LeBrun, Jul 11 2001

Keywords

• base
• easy
• nonn

Comments

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

a(n) is the position of the leftmost 0 in the binary expansion of n (starting from the least significant bit). For example, 44 = 101100_2 and the leftmost 0 is in position 5, so a(44) = 5. - Chai Wah Wu, Mar 09 2025

Examples

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

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..32767

Crossrefs

Cf. A038572, A048881.

Programs

Mathematica

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

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

Python

def A063250(n): return 0 if (t:=bin(n)[2:].find('0'))==-1 else n.bit_length()-t # Chai Wah Wu, Mar 09 2025

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