OFFSET
1,2
COMMENTS
For the ratio a(n+1)/a(n) there appears to be an oblique asymptote.
For determining the sum of the arch lengths for one solution the following is a quick way of finding the sum for one solution.
Given semi-meander top arches:
___________
/ _______ \
/ / /\ \ \
/\ /\ / / /\ //\\ \ /\\
arch lengths= 1 1 6 4 1 2 1 1 total length = 17
Write the above semi-meander in terms of the number of covering arches from the top down.
Number of covering arches: 0 1 2 3
number of arches: 3 2 2 1
Sum multiplying (cover+1) * (number of arches) = (0+1)*(3) + (1+1)*(2) + (1+2)*(2) + (1+3)*(1) = 17.
For one semi-meander solution with n top arches the following formula will generate the length of the arches in next generation (n+1) when the exterior arch splitting algorithm is used. Each individual solution in generation n must use this formula to get the total lengths of all semi-meanders in generation n + 1.
r = number or returns (noncovered arches in generation n)
t = total length of arches for a solution with n top arches
n = number of top arches
total arches in generation n+1 for one solution = (r * t + 4) + (r - 2) (n + 2)
semi-meander with n arches:
split split split _____ __ __
/\ / /\\ /\ //\\ / \
/\ //\\ => / //\\\ //\\ ///\\\ //\ \ /\
//\\ /\ ///\\\ /\ /\ //\ ///\\\\, ///\\\ ////\\\\, ///\\/\\ //\\ /\
lengths 2 1 1 3 2 1 =10 1 1 5 1 3 2 1 3 2 1 4 3 2 1 4 2 1 1 2 1 1 = 42
formula (3 * 10 + 4) + (3 - 2) * (6 + 2) = 42
EXAMPLE
For n = 3, a(3) = 8: /\ /\
/\ //\\, //\\ /\
arch lengths: 1+2+1 2+1+1 sum = 8.
CROSSREFS
KEYWORD
nonn
AUTHOR
Roger Ford, Aug 25 2025
EXTENSIONS
a(19)-a(24) from Andrew Howroyd, Aug 31 2025
STATUS
approved