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A374391
Number of ways to tile a hexagonal strip made up of n equilateral triangles, using diamonds and trapezoids.
0
0, 1, 0, 1, 1, 1, 2, 5, 4, 4, 6, 22, 17, 17, 25, 89, 70, 73, 104, 369, 291, 302, 434, 1534, 1209, 1253, 1802, 6375, 5023, 5205, 7485, 26483, 20867, 21626, 31095, 110018, 86689, 89843, 129182, 457055, 360138, 373239, 536670, 1898777, 1496146, 1550570, 2229523
OFFSET
-1,7
COMMENTS
Here is the hexagonal strip:
________________ ____
/\ /\ /\ /\ / \ /\
/__\/__\/__\/__\/ ... \/__\
\ /\ /\ /\ /\ /\ /
\/__\/__\/__\/__\ /__\/
The two types of tiles are diamonds and trapezoids (each of which can be rotated). Here are the two types of tiles:
____ _______
\ \ \ /
\___\ and \___/.
This is related to A356622 and A356623 (tilings with triangles and diamonds), and to A365352 (tilings with triangles and trapezoids).
LINKS
FORMULA
a(4*n) = a(4*n-4) + 2*a(4*n-5) + a(4*n-6).
a(4*n+1) = a(4*n-1) + a(4*n-4) + a(4*n-5).
a(4*n+2) = a(4*n) + 2*a(4*n-1) + a(4*n-2) + 2*a(4*n-3) + a(4*n-4).
a(4*n+3) = a(4*n+1) + a(4*n) + a(4*n-2) + a(4*n-3).
a(n) = 4*a(n-4) + 2*a(n-12) + 3*a(n-16) - a(n-20).
a(4*n+2) = p(2*n+1)^2 + Sum_{i=0..n} (a(4*i) + 2*a(4*i-1))*p(2*n-2*i)^2 + 2*a(4*i)*p(2*n-2*i)*p(2*n-2*i-1), where p(n) = the Padovan number A000931(n+3) and where a(-1)=0.
EXAMPLE
For n=13, here is one of the a(13)=25 ways to tile this strip (of 13 triangles) using diamonds and trapezoids.
____________
/\ / /
/ \____/___/___
\ / \ /
\/______\____/.
MATHEMATICA
T[0] = 1; T[1] = 0; T[2] = 1; T[3] = 1; T[4] = 1; T[5] = 2;
T[n_] := T[n] = Switch[Mod[n, 4],
0, T[n - 4] + 2 T[n - 5] + T[n - 6],
1, T[n - 2] + T[n - 5] + T[n - 6],
2, T[n - 2] + 2 T[n - 3] + T[n - 4] + 2 T[n - 5] + T[n - 6],
3, T[n - 2] + T[n - 3] + T[n - 5] + T[n - 6]
]; Table[T[n], {n, 0, 50}]
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Greg Dresden and Haoran Xu, Jul 07 2024
STATUS
approved