OFFSET
1,2
COMMENTS
One can reuse a point, but not an edge.
From Hugo van der Sanden, Sep 27 2005: (Start)
The n X n square has #(n) = 2n(n+1) line segments and v_o(n) = 4(n-1) vertices of odd degree. The network is traversable only when v_o(n) <= 2. When not, we must lose sufficient line segments to reduce v_o() to 2.
For odd n >= 3, we have an even number of odd vertices along each edge; we can pair them up and lose the single line segment joining each pair except for our start/end points. So in this case we have a(n) = #(n) - (v_o(n) - 2)/2 = 2n^2 + 3.
For even n >= 2, we have an odd number of odd vertices along each edge. The best we can do is start and end at the two vertices adjacent to one corner; pairing up the rest allows us to lose a single line segment for each of the remaining pairs except for the two vertices adjacent to the diagonally opposite corner, for which we must lose 2 line segments. So in this case we have a(n) = #(n) - ((v_o(n) - 4)/2 + 2) = 2n^2 + 2.
For n < 2, we have no vertices of odd degree, so we cannot save a segment by starting and ending on a pair of them. So we can specify the exact function using a couple of characteristic functions: a(n) = 2n^2 + 1 + c(n > 1) + c(n odd) (I'm assuming a(0) = 1 here, on the grounds that there is a single zero-length path in that case.)
Note that the theoretical maximum is always achievable, e.g. using Fleury's algorithm. (End)
LINKS
Index entries for linear recurrences with constant coefficients, signature (2, 0, -2, 1).
FORMULA
From Colin Barker, May 02 2013: (Start)
Conjecture: a(n) = (9 + (-1)^n - 8*n + 4*n^2)/2 for n > 2.
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n > 6.
G.f.: -x*(x^5 - x^4 + 3*x^3 + 2*x^2 + 2*x + 1) / ((x-1)^3*(x+1)). (End)
MATHEMATICA
Join[{1, 4}, LinearRecurrence[{2, 0, -2, 1}, {10, 21, 34, 53}, 40]] (* Harvey P. Dale, Aug 21 2013 *)
CROSSREFS
KEYWORD
nonn,walk,nice,easy
AUTHOR
Arlin Anderson (starship1(AT)gmail.com)
EXTENSIONS
More terms from Hugo van der Sanden, Sep 27 2005
STATUS
approved