OFFSET
1,4
COMMENTS
For n>=3, a(n) is the number of topological configurations (up to cyclic shifts and reversal) of n points and n lines, where the points lie at the vertices of a convex cyclic n-gon and the lines are the perpendicular bisectors of its sides. Counting such configurations without quotienting out by cyclic shifts and reversal gives the sequence A028243.
a(n) is also the number of equivalence classes (up to cyclic shifts and reversal) of 2n-tuples composed of n 0's and n 1's which have an interlacing signature. The signature of a 2n-tuple (v_1,...,v_{2n}) is the n-tuple (s_1,...,s_n) defined by s_i=v_i+v_{i+n}. The signature is called interlacing if after deleting the 1's, there are letters remaining and the remaining 0's and 2's are alternating.
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..1000
P. Melotti, S. Ramassamy and P. Thévenin, Points and lines configurations for perpendicular bisectors of convex cyclic polygons, arXiv:2003.11006 [math.CO], 2020.
FORMULA
From Andrew Howroyd, Dec 25 2021: (Start)
a(n) = (b(n) + (Sum_{d|n, n/d==1 (mod 2)} phi(n/d)*((3^d - (-1)^d)/2 - 2^d))/n)/2 where b(n) = 1 + 3^(n/2-1) for even n and 0 otherwise.
G.f.: (1/2)*(x^2/(1-x^2) + x^2/(1-3*x^2)) + (1/4)*Sum_{k>=0} phi(2*k+1)*log(B(x^(2*k+1)))/(2*k+1) where B(x) = (1+x)*(1-2*x)^2/(1-3*x).
(End)
EXAMPLE
For n=3, drawing the three perpendicular bisectors of a triangle divides the plane into 6 regions. Three of these regions contain one vertex of the triangle and the other three contain none. Up to cyclic shifts and reversal, the only possible configuration is (nonempty, nonempty, empty, empty, nonempty, empty), thus a(3)=1.
For n=3, the only 6-tuple (up to cyclic shift and reversal) which has interlacing signature is (1,1,0,0,1,0). Its signature is (1,2,0).
For n=4, the a(4)=5 equivalence classes of 8-tuples with interlacing signature are (0,1,0,1,0,1,0,1), (0,0,0,1,0,1,1,1), (0,1,0,1,0,0,1,1), (0,1,1,1,0,0,1,0) and (0,0,1,1,0,1,1,0).
PROG
(PARI) \\ here c(n) is up to rotations only.
c(n)={(n%2==0) + sumdiv(n, d, if(n/d%2==1, eulerphi(n/d)*((3^d - (-1)^d)/2 - 2^d)))/n}
a(n)={(c(n) + if(n%2==0, 3^(n/2-1)))/2} \\ Andrew Howroyd, Dec 25 2021
(PARI)
seq(n)=Vec((x^2/(1-x^2) + x^2/(1-3*x^2))/2 + sum(k=0, (n-1)\2, my(d=2*k+1); eulerphi(d)*log((1+x^d)*(1-2*x^d)^2/(1-3*x^d) + O(x*x^n))/d)/4, -n) \\ Andrew Howroyd, Dec 25 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Sanjay Ramassamy, Dec 23 2021
EXTENSIONS
Terms a(11) and beyond from Andrew Howroyd, Dec 25 2021
STATUS
approved