A138904 Number of rotational symmetries in the binary expansion of a number.

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

Offset: 0

Author

Max Sills, Apr 03 2008, Apr 04 2008

Keywords

• base
• easy
• nonn

Comments

Mersenne numbers of form (2^n - 1) have n rotational symmetries.

For prime length binary expansions these are the only nontrivial symmetries.

For composite length expansions it seems that when the number of symmetries is nontrivial it is equal to a factor of the length. We're working on an explicit formula.

Discovered in the context of random circulant matrices, examining if there's a correlation between degrees of freedom and number of symmetries in the first row.

When combined with A138954, these two sequences should give a full account of the number of redundant rows in a circulant square matrix with at most two distinct values, where a(n) is the encoding of the first row of the matrix into binary such that value a = 1 and value b = 0.

Discovered on the night of Apr 02, 2008 by Maxwell Sills and Gary Doran.

Conjecture: For binary expansions of length n, there are d(n) distinct values that will show up as symmetries, where d is the divisor function. The symmetry values will be precisely the divisors of n.

Example: for binary expansions of length 12, one sees that d(12) = 6 distinct values show up as symmetries (1, 2, 3, 4, 6, 12).

Conjecture: For numbers whose binary expansion has length n which has proper divisors which are all coprime: There will be only one number of length n with n symmetries. That number is 2^n - 1. For each proper divisor d (excluding 1), you can generate all numbers of length n that have n/d symmetries like so: (2^0 + 2^d + 2^2d ... 2^(n-d)) * a, where 2^(d-1) <= a < (2^d) - 1. The rest of the expansions of length n will have only the trivial symmetry.

Also the number of rotational symmetries of the n-th composition in standard order (graded reverse-lexicographic). This composition (row n of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of n, prepending 0, taking first differences, and reversing again. - Gus Wiseman, Apr 19 2020

From Gus Wiseman, Apr 19 2020: (Start)

Aperiodic compositions are counted by A000740.

Aperiodic binary words are counted by A027375.

The orderless period of prime indices is A052409.

Numbers whose binary expansion is periodic are A121016.

Periodic compositions are counted by A178472.

Period of binary expansion is A302291.

Compositions by sum and number of distinct rotations are A333941.

All of the following pertain to compositions in standard order (A066099):

- Length is A000120.

- Necklaces are A065609.

- Sum is A070939.

- Runs are counted by A124767.

- Strict compositions are A233564.

- Constant compositions are A272919.

- Lyndon compositions are A275692.

- Co-Lyndon compositions are A326774.

- Aperiodic compositions are A328594.

- Reversed co-necklaces are A328595.

- Rotational period is A333632.

- Co-necklaces are A333764.

- Reversed necklaces are A333943.

Cf. A000031, A001037, A008965, A019536, A211100, A328595, A328596, A329312, A329313, A329326.

(End).

Examples

			a(10) = 2 because the binary expansion of 10 is 1010 and it has two rotational symmetries (including identity).
		

Links

• Maxwell Sills and Gary Doran, Table of n, a(n) for n = 0..99

Crossrefs

Cf. A136441, A138954.

Programs

Mathematica

Table[IntegerLength[n,2]/Length[Union[Array[RotateRight[IntegerDigits[n,2],#]&,IntegerLength[n,2]]]],{n,100}] (* Gus Wiseman, Apr 19 2020 *)

Formula

a(n) = A070939(n)/A302291(n) = A000120(n)/A333632(n). - Gus Wiseman, Apr 19 2020