A232206 Triangle read by rows: T(n,k) is the number of non-equivalent regular polygons with n+1 edges, one of which is rooted, which are dissected by non-intersecting diagonals into k regions, such that two such polygons are identified up to reflection along the rooted edge and twisting along the diagonals that does not affect the root edge (for 1 <= k <= n-1 and n >= 2).
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
Examples
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)
References
- L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 54-55 and 74-75.
Links
- Andrew Howroyd, Table of n, a(n) for n = 2..1276
- S. L. Devadoss and R. C. Read, Cellular structures determined by polygons and trees, Ann. Combin., 5 (2001), 71-98.
- Ronald C. Read, On general dissections of a polygon, Preprint (1974).
- Ronald C. Read, On general dissections of a polygon, Aequat. Math., 18 (1978), 370-388.
- K. Schöbel and A. Veselov, Separation coordinates, moduli spaces and Stasheff polytopes, arXiv:1307.6132 [math.DG], 2013-2014.
- K. Schöbel and A. Veselov, Separation coordinates, moduli spaces and Stasheff polytopes, Commun. Math. Phys., 337 (2015), 1255-1274.
- Wikipedia, Associahedron.
Crossrefs
Programs
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Mathematica
rows = 12; A[, ] = 0; Do[A[s_, t_] = Series[s+(t/2)(A[s, t]^2/(1-A[s, t])+(1+A[s, t]) A[s^2, t^2] / (1-A[s^2, t^2])), {s, 0, rows+1}, {t, 0, rows+1}] // Normal, {rows+1}]; Rest[CoefficientList[#, t]]& /@ Drop[CoefficientList[A[s, t], s], 2] // Flatten (* Jean-François Alcover, Sep 02 2018 *)
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PARI
BIKxy(p)={(1/(1-p) + (1+p)/subst(subst(1-p, x, x^2), y, y^2))/2} A(n)={my(p=x); for(i=2, n+1, p+=x^i*y*polcoeff(BIKxy(p + O(x*x^i)), i)); [Vecrev(q/y) | q<-Vecrev((p-x)/x^2)]} { my(T=A(10)); for(n=1, #T, print(T[n])) } \\ Andrew Howroyd, Aug 31 2018
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SageMath
#This is a crude Sage program on how to derive the g.f. of any column in the triangle given that you have correctly derived the g.f.s of the previous columns. Suppose you have the expansion for C(s,t) w.r.t. t up to the power of 3 and you want to get the coefficient of t^4. s,t=var('s,t') C(s,t) = s + t*s^2/(1-s) + t^2*s^3*(1+s-s^2)/((1+s)*(1-s)^3) + t^3*s^4*(2+2*s-2*s^3+s^4)/((1+s)^2*(1-s)^5) B(s,t)=s+(t/2)*(C(s,t)^2/(1-C(s,t))+(1+C(s,t))*C(s^2,t^2)/(1-C(s^2,t^2)))-C(s,t) factor(taylor(B(s,t),t,0,4)/t^4) # Petros Hadjicostas, Jan 21 2018
Formula
From Petros Hadjicostas, Jan 20 2018: (Start)
G.f.: If C(s,t) = Sum_{n>=0, k>=0} T(n,k)*s^n*t^k (with T(n=1,k=0) = 1), then C(s,t) = s + (t/2)*(C(s,t)^2/(1-C(s,t)) + (1 + C(s,t))*C(s^2, t^2)/(1-C(s^2, t^2))).
Note that t*C(s,t) = Sum_{k>=1} T(k,k-1)*(s*t)^k + Sum_{k>=2} (s*t)^k*Sum_{n >= k+1} T(n,k-1)*s^{n-k}. If we let w = s*t and D(s,w) = (w/s)*C(s,w/s), then the above functional equation implies that E(w): = lim_{s -> 0} D(s,w) = Sum_{k>=1} T(k,k-1)*w^k satisfies E(w) = w + (1/2)*(E(w)^2 + E(w^2)), which is the g.f. for the Wedderburn-Etherington numbers in sequence A001190. Thus, T(k,k-1) = A001190(k) for k>=1.
Expansion w.r.t to t: C(s,t) = s + t*s^2/(1-s) + t^2*s^3*(1+s-s^2)/((1+s)*(1-s)^3) + t^3*s^4*(2+2*s-2*s^3+s^4)/((1+s)^2*(1-s)^5) + t^4*s^5*(3+8*s+2*s^2-2*s^3-2*s^4+3*s^5-s^6)/((1+s)^3*(1-s)^7) + t^5*s^6*(6+19*s+30*s^2+15*s^3+23*s^4-4*s^5+2*s^6-4*s^7+6*s^8-4*s^9+s^10)/((1+s^2)*(1+s)^4*(1-s)^9) + ... (It can be proven using a symbolic computation package by manipulating the above functional equation by Devadoss and Read.)
This means that, in the above expansion for C(s,t), the coefficient of t is the g.f. of the first column of the triangle below, the coefficient of t^2 is the g.f. of the second column of the triangle below, and so on.
(End)
Extensions
Name edited by Petros Hadjicostas, Jan 20 2018
Offset corrected by Andrew Howroyd, Aug 31 2018
Comments