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%I #31 Feb 26 2024 11:24:01
%S 8,20,20,40,75,40,70,210,210,70,112,490,784,490,112,168,1008,2352,
%T 2352,1008,168,240,1890,6048,8820,6048,1890,240,330,3300,13860,27720,
%U 27720,13860,3300,330,440,5445,29040,76230,104544,76230,29040,5445,440
%N Table read by antidiagonals in which entry T(n,k) in row n and column k gives the number of possible rhombus tilings of an octagon with interior angles of 135 degrees and sequences of side lengths {n, k, 1, 1, n, k, 1, 1} (as the octagon is traversed), n,k in {1,2,3,...}.
%C Proof of the formula for T(n,k) is given in [Elnitsky].
%C So-called "generalized Narayana numbers" (see A145596), linking rhombus tilings of polygons to certain walks or paths through the square lattice.
%H Michael De Vlieger, <a href="/A214457/b214457.txt">Table of n, a(n) for n = 1..11325</a> (rows n = 1..150, flattened)
%H Serge Elnitsky, <a href="http://www.elnitsky.com/tilings.pdf">Rhombic tilings of polygons and classes of reduced words in Coxeter groups (preprint)</a>, J. Combin. Theory Ser. A, Vol. 77, Issue 2, 193-221 (1997).
%H L. E. Jeffery, <a href="/A214457/a214457.pdf">Worked out example for A214457(1,1)=8</a>
%H Tad White, <a href="https://arxiv.org/abs/2401.01462">Quota Trees</a>, arXiv:2401.01462 [math.CO], 2024. See p. 20.
%F T(n,k) = 2*(n+k+1)!*(n+k+2)!/[n!*k!*(n+2)!*(k+2)!].
%e See [Jeffery]. T(1,1) = 8 because there are eight ways to tile the proposed octagon with rhombuses.
%e Table begins as
%e 8 20 40 70 112 ...
%e 20 75 210 490 1008 ...
%e 40 210 784 2352 6048 ...
%e 70 490 2352 8820 27720 ...
%e 112 1008 6048 27720 76230 ...
%e ...
%t Table[2*(# + k + 1)!*(# + k + 2)!/(#!*k!*(# + 2)!*(k + 2)!) &[n - k + 1], {n, 10}, {k, n}] // Flatten (* _Michael De Vlieger_, Feb 26 2024 *)
%Y Empirical: T(1,n) = T(n,1) = 2*A000292(n+1); T(2,n) = T(n,2) = A006411(n+1); T(n,n) = A145600(n+1).
%K nonn,tabl
%O 1,1
%A _L. Edson Jeffery_, Jul 18 2012
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