1, 1, 1, 1, 3, 2, 1, 5, 8, 3, 1, 8, 22, 20, 6, 1, 11, 46, 73, 49, 11, 1, 15, 87, 206, 233, 119, 23, 1, 19, 147, 485, 807, 689, 288, 46, 1, 24, 236, 1021, 2320, 2891, 1988, 696, 98, 1, 29, 356, 1960, 5795, 9800, 9737, 5561, 1681, 207, 1, 35, 520, 3525, 13088, 28586, 38216, 31350, 15322, 4062, 451, 1, 41, 730, 5989, 27224, 74280, 127465, 139901, 97552, 41558, 9821, 983
Offset: 2
From _Petros Hadjicostas_, Jan 20 2018: (Start)
According to Devadoss and Read (2015), triangle T(n,k) begins as follows:
(n=2) 1,
(n=3) 1, 1,
(n=4) 1, 3, 2,
(n=5) 1, 5, 8, 3,
(n=6) 1, 8, 22, 20, 6,
(n=7) 1, 11, 46, 73, 49, 11,
(n=8) 1, 15, 87, 206, 233, 119, 23,
(n=9) 1, 19, 147, 485, 807, 689, 288, 46,
(n=10) 1, 24, 236, 1021, 2320, 2891, 1988, 696, 98,
(n=11) 1, 29, 356, 1960, 5795, 9800, 9737, 5561, 1681, 207,
...
Geometrical example for n=5: If no twisting is allowed, the number of regular (n+1)-gons (= hexagons) with a rooted edge and dissected into k regions by non-intersecting diagonals is given by A033282(n+1, k-1) = A033282(6, k-1): 1, 9, 21, and 14 for k = 1, 2, 3, 4, respectively.
Recall that, according to Devadoss and Read (2001), two regular (unrooted) polygons G_1 and G_2 are of the same "class" if they are identified under the actions of D_n (= dihedral group of order 2*n) and twisting along the diagonals. To avoid confusion, we call two such unrooted equivalent polygons as being of the same DT-class (D for "dihedral" and T for "twisting"). According to Table 5 (p. 92) in Devadoss and Read (2001), the numbers of DT-classes for regular hexagons dissected into k regions (by k-1 non-intersecting diagonals) are 1, 2, 3, 2 for k = 1, 2, 3, 4, respectively. When an edge is rooted, each one of these DT-classes is subdivided into subclasses.
Call the regular hexagons ABCDEF with FA being the rooted edge (base).
For k=1, we have T(5,1) = 1 rooted regular hexagon with no intersecting diagonals.
For k=2, we have T(5,2) = 5 equivalent rooted regular hexagons with 1 diagonal. The equivalence classes are as follows according to the single dissecting diagonal:
Class 1a: AE, FB; Class 1b: AC, FD; Class 1c: BD, EC; (DT-class 1 has 6 hexagons)
Class 2a: AD, FC; Class 2b: BE; (DT-class 2 has 3 hexagons).
Note that 6 + 3 = 9 = A033282(5+1, 2-1).
For k=3, we have T(5,3) = 8 equivalent rooted regular hexagons with 2 non-intersecting diagonals. The equivalence classes are as follows according to the two diagonals:
Class 1a: (AC, AE), (FB, FD), (BD, BF), (EA, EC);
Class 1b: (DB, DF), (CE, CA); (DT-class 1 has 6 hexagons)
Class 2a: (DB, EA), (CE, BF);
Class 2b: (DF, CA); (DT-class 2 has 3 hexagons)
Class 3a: (DA, EA), (FC, FB), (EA, EB), (BE, BF);
Class 3b: (AC, AD), (FC, FD), (DA, DB), (CE, CF);
Class 3c: (BD, BE), (EC, EB);
Class 3d: (CF, CA), (DA, DF); (DT-class 3 has 12 hexagons)
Note that 6 + 3 + 12 = 21 = A033282(5+1, 3-1).
For k=4, we have T(5,4) = 3 equivalent rooted regular hexagons with 3 non-intersecting diagonals. The equivalence classes are as follows according to the three diagonals:
Class 1: (EA, AC, CE), (BD, DF, FB); (DT-class 1 has 2 hexagons)
Class 2a: (DF, DA, AC), (DF, FC, CA), (CE, CF, CA), (DF, DA, DB);
Class 2b: (CE, BE, BF), (BD, BE, BF), (EC, EB, EA), (DB, EB, EA), (EC, CF, FB), (DB, DA, AE), (FD, FC, FB), (AE, AD, AC); (DT class 2 has 12 hexagons)
Note that 2 + 12 = 14 = A033282(5+1, 4-1).
Recall that two rooted hexagons are equivalent iff they are a reflection of each other along the rooted edge or one can be obtained from the other by twisting a diagonal as long as the twisting does not affect the rooted edge.
The case k = 4 = n-1 above is related to the triangulation of a convex polygon and the Wedderburn-Etherington commutative bracketing problem that appear in Comtet (1974, pp. 54-55). Devadoss and Read (2001, p. 89) claim that T(n,k) is the number of possible ways of using k-1 pairs of brackets on n commutative variables, but it is not clear how each one of the above hexagons (from k=1 to k=3) can be transformed into some kind of a generalized commutative bracketing problem.
(End)
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