A336953 -id:A336953 - cp's OEIS frontend

A366140 Fixed points of the binary rotations A336953 and A366139: numbers k >= 0 such that A336953(k) = A366139(k) = k.

0, 1, 2, 3, 6, 7, 8, 10, 12, 15, 20, 25, 30, 31, 36, 42, 45, 48, 54, 60, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 127, 128, 136, 144, 152, 160, 168, 170, 176, 184, 192, 200, 204, 208, 216, 224, 232, 240, 248, 255, 261, 270, 279, 288, 297, 306, 315, 324, 333

Offset: 1

Paolo Xausa, Sep 30 2023

If a number is a fixed point of A336953, then it's also a fixed point of A366139, and vice versa.

k is a term iff A302291(k)|k.

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A302291, A336953, A366139.

A366140Q[n_]:=FromDigits[RotateLeft[IntegerDigits[n,2],n],2]==n;
Select[Range[0,500],A366140Q]

A366139 In binary representation, rotate the digits of n left n places.

Original entry on oeis.org

0, 1, 2, 3, 1, 6, 6, 7, 8, 3, 10, 13, 12, 11, 11, 15, 1, 6, 20, 25, 20, 11, 26, 29, 12, 25, 21, 15, 7, 30, 30, 31, 2, 12, 40, 49, 36, 11, 26, 60, 10, 52, 42, 23, 50, 45, 43, 55, 48, 35, 11, 30, 13, 58, 54, 47, 35, 15, 46, 61, 60, 59, 59, 63, 1, 6, 20, 56, 17, 98

Offset: 0

Paolo Xausa, Sep 30 2023

			a(4) = 1 because 4 in base 2 is 100 and 100 rotated left 4 times is 100 -> 001 -> 010 -> 100 -> 001 = 1.
a(5) = 6 because 5 in base 2 is 101 and 101 rotated left 5 times is 101 -> 011 -> 110 -> 101 -> 011 -> 110 = 6 in base 10.

Paolo Xausa, Table of n, a(n) for n = 0..16383

Cf. A006257, A007088, A366140 (fixed points).

Cf. A336953 (rotate right).

A366139[n_]:=FromDigits[RotateLeft[IntegerDigits[n,2],n],2];
Array[A366139,100,0]

def A366139(n): return int((s:=bin(n))[(m:=n%n.bit_length()+2):]+s[2:m],2) if n else 0 # Chai Wah Wu, Oct 03 2023

cp's OEIS Frontend

A366140 Fixed points of the binary rotations A336953 and A366139: numbers k >= 0 such that A336953(k) = A366139(k) = k.

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