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A331235 The number of simple polygons having all points of a 3 X n grid as vertices. 0
0, 1, 8, 62, 532, 4846, 45712, 441458, 4337468, 43187630, 434602280, 4411598154, 45107210436, 464047175770, 4799184825632, 49860914628042, 520109726201420, 5444641096394926, 57176049036449464, 602125661090565914, 6357215467283967404, 67274331104623532042 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
The polygons are allowed to have flat angles (angles of exactly Pi) at some of the grid points. Empirically this sequence appears to be asymptotic to phi^(5n)/(66n), where phi is the golden ratio.
LINKS
David Eppstein, Counting grid polygonalizations, Jan 12 2020
PROG
(Python)
from math import log
memo = {}
def K(x, y, z):
"""Number of strings of length y from two sorted alphabets of lengths x, z"""
if (x, y, z) in memo:
return memo[(x, y, z)]
if y == 0:
result = 1
else:
# i = length of the last block of equal characters in the string
# xx or zz = the largest remaining character in its alphabet
result = (sum(K(xx, y-i, z) for xx in range(x) for i in range(1, y+1)) +
sum(K(x, y-i, zz) for zz in range(z) for i in range(1, y+1)))
memo[(x, y, z)] = result
return result
def GC(n):
"""Number of polygonalizations of 3xn grid"""
sum = 0
for i in range(n-1): # number of points in K(...) can be up to n-2
mid = K(n-1, i, n-1)
for left in range(n-1-i):
right = n-2-i-left
contrib = mid
if left:
contrib *= 2
if right:
contrib *= 2
sum += contrib
return sum
def exponent(p):
return p**(-4*p) * (1-p)**(-2*(1-p)) * (1-2*p)**(-1*(1-2*p))
base = ( (1+5**0.5)/2 )**5
#for n in range(2, 50):
# print(n, (base**n/(66*n))/GC(n), GC(n))
[GC(n) for n in range(1, 50)]
CROSSREFS
Sequence in context: A190975 A287815 A244831 * A053095 A356461 A099335
KEYWORD
nonn,easy
AUTHOR
David Eppstein, Jan 12 2020
STATUS
approved

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Last modified May 3 06:31 EDT 2024. Contains 372206 sequences. (Running on oeis4.)