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A333574
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Number of Hamiltonian paths in the n X 2 grid graph which start at any of the n vertices on left side of the graph and terminate at any of the n vertices on the right side.
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2
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1, 2, 4, 6, 10, 14, 20, 26, 34, 42, 52, 62, 74, 86, 100, 114, 130, 146, 164, 182, 202, 222, 244, 266, 290, 314, 340, 366, 394, 422, 452, 482, 514, 546, 580, 614, 650, 686, 724, 762, 802, 842, 884, 926, 970, 1014, 1060, 1106, 1154, 1202, 1252, 1302, 1354, 1406, 1460
(list;
graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,2
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COMMENTS
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LINKS
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FORMULA
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G.f.: x*(1+2*x*(1-x^2+x^3)/((1+x)*(1-x)^3)).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n>5.
a(n) = (9 + (-1)^(1+n) - 4*n + 2*n^2) / 4 for n>1. (End)
E.g.f.: ((4 - x + x^2)*cosh(x) + (5 - x + x^2)*sinh(x) - 2*(2 + x))/2. - Stefano Spezia, Jun 14 2023
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EXAMPLE
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a(1) = 1;
+--+
a(2) = 2;
+ + *--*
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*--* + +
a(3) = 4;
+ + +--* *--+ *--*
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* * *--* *--* * *
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*--* *--+ +--* + +
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PROG
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(PARI) N=66; x='x+O('x^N); Vec(x*(1+2*x*(1-x^2+x^3)/((1+x)*(1-x)^3)))
(Python)
# Using graphillion
from graphillion import GraphSet
import graphillion.tutorial as tl
def A(start, goal, n, k):
universe = tl.grid(n - 1, k - 1)
GraphSet.set_universe(universe)
paths = GraphSet.paths(start, goal, is_hamilton=True)
return paths.len()
if n == 1: return 1
s = 0
for i in range(1, n + 1):
for j in range(k * n - n + 1, k * n + 1):
s += A(i, j, k, n)
return s
print([A333574(n) for n in range(1, 25)])
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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