

A138904


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


5



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
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OFFSET

0,4


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^(nd)) * a, where 2^(d1) <= a < (2^d)  1. The rest of the expansions of length n will have only the trivial symmetry.


LINKS

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


EXAMPLE

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


CROSSREFS

Cf. A136441, A138954.
Sequence in context: A281119 A317176 A318812 * A196660 A135222 A285706
Adjacent sequences: A138901 A138902 A138903 * A138905 A138906 A138907


KEYWORD

base,easy,nonn


AUTHOR

Max Sills, Apr 03 2008, Apr 04 2008


STATUS

approved



