OFFSET
3,1
COMMENTS
There are n vertices and n midpoints. The total number of lines connecting vertices and midpoints is 2*n*(n-2).
LINKS
FORMULA
Where c is the number of cycles in each of the 2n symmetries, a(n) = Sum_{2^c}/(2n).
a(n) = (1/2) * ( 2^floor((n*(2*n-3)-3)/2) + 2^floor((n*(2*n-3)-2)/2) + (1/n) * Sum_{d|n} phi(d)*2^((2*n - 3 - d mod 2)*n/d) ). - Andrew Howroyd, Mar 02 2026
EXAMPLE
For n = 3, a(3) = 20, since there are 20 distinct ways you can connect vertices and midpoints without using the same edge, including the empty and full configurations.
MATHEMATICA
A393552[n_] := (2^Quotient[# - 3, 2] + 2^Quotient[# - 2, 2] + DivisorSum[n, EulerPhi[#]*2^((2*n - 3 - Mod[#, 2])*n/#) &]/n)/2 & [n*(2*n - 3)];
Array[A393552, 10, 3] (* Paolo Xausa, Mar 09 2026 *)
PROG
(PARI)
r(n) = {my(m=2*n*(n-2)); if(n%2, 2^((m+n-1)/2), 3*2^((m+n)/2-2))}
c(n) = {my(m=2*n*(n-2)); sumdiv(n, d, eulerphi(d)*2^if(n%2, m/d, (m-n)/d + gcd(d, 2)*n/d))/n}
a(n) = {(c(n) + r(n))/2} \\ Andrew Howroyd, Feb 21 2026
CROSSREFS
KEYWORD
nonn
AUTHOR
Rhys Feltman, Feb 20 2026
STATUS
approved
