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Triangle read by rows: T(n,k) = number of nonequivalent dissections of an n-gon into k polygons by nonintersecting diagonals up to rotation and reflection.
+10
8
1, 1, 1, 1, 1, 1, 1, 2, 3, 3, 1, 2, 6, 7, 4, 1, 3, 11, 24, 24, 12, 1, 3, 17, 51, 89, 74, 27, 1, 4, 26, 109, 265, 371, 259, 82, 1, 4, 36, 194, 660, 1291, 1478, 891, 228, 1, 5, 50, 345, 1477, 3891, 6249, 6044, 3176, 733, 1, 5, 65, 550, 3000, 10061, 21524, 29133, 24302, 11326, 2282
EXAMPLE
Triangle begins: (n >= 3, k >= 1)
1;
1, 1;
1, 1, 1;
1, 2, 3, 3;
1, 2, 6, 7, 4;
1, 3, 11, 24, 24, 12;
1, 3, 17, 51, 89, 74, 27;
1, 4, 26, 109, 265, 371, 259, 82;
1, 4, 36, 194, 660, 1291, 1478, 891, 228;
...
PROG
(PARI) \\ See A295419 for DissectionsModDihedral() T=DissectionsModDihedral(apply(i->y, [1..12]));
for(n=3, #T, for(k=1, n-2, print1(polcoeff(T[n], k), ", ")); print)
Number of nonequivalent dissections of an n-gon into n-4 polygons by nonintersecting diagonals up to rotation and reflection.
(Formerly M1673)
+10
6
1, 2, 6, 24, 89, 371, 1478, 6044, 24302, 98000, 392528, 1570490, 6264309, 24954223, 99253318, 394409402, 1565986466, 6214173156, 24647935156, 97732340680, 387428854374, 1535588541762, 6085702368796, 24116801236744, 95569050564444, 378718095630676
COMMENTS
In other words, the number of (n - 5)-dissections of an n-gon modulo the dihedral action.
Equivalently, the number of two-dimensional faces of the (n-3)-dimensional associahedron modulo the dihedral action.
The dissection will always be composed of either 1 pentagon and n-5 triangles or 2 quadrilaterals and n-6 triangles. - Andrew Howroyd, Nov 24 2017
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
MAPLE
C:=n->binomial(2*n, n)/(n+1);
T32:=proc(n) local t1; global C;
if n mod 2 = 0 then
t1 := (n-3)^2*(n-4)*C(n-2)/(8*n*(2*n-5));
if n mod 5 = 0 then t1:=t1+(2/5)*C(n/5-1) fi;
if n mod 2 = 0 then t1:=t1+((3*(n-4)*(n-1))/(16*(n-3)))*C(n/2-1) fi;
else
t1 := (n-3)^2*(n-4)*C(n-2)/(8*n*(2*n-5));
if n mod 5 = 0 then t1:=t1+(2/5)*C(n/5-1) fi;
if n mod 2 = 1 then t1:=t1+((n^2-2*n-11)/(8*(n-4)))*C((n-3)/2) fi;
fi;
t1; end;
[seq(T32(n), n=5..40)];
MATHEMATICA
c = CatalanNumber;
T32[n_] := Module[{t1}, If[EvenQ[n], t1 = (n-3)^2*(n-4)*c[n-2]/(8*n*(2*n - 5)); If[Mod[n, 5] == 0, t1 = t1 + (2/5)*c[n/5-1]]; If[EvenQ[n], t1 = t1 + ((3*(n-4)*(n-1))/(16*(n-3)))*c[n/2-1]], t1 = (n-3)^2*(n-4)*c[n-2]/(8*n *(2*n - 5)); If[Mod[n, 5] == 0, t1 = t1 + (2/5) * c[n/5-1]]; If[OddQ[n], t1 = t1 + ((n^2 - 2*n - 11)/(8*(n-4)))*c[(n-3)/2]]]; t1];
PROG
(PARI) \\ See A295419 for DissectionsModDihedral() { my(v=DissectionsModDihedral(apply(i->if(i>=3&&i<=5, y^(i-3) + O(y^3)), [1..30]))); apply(p->polcoeff(p, 2), v[5..#v]) } \\ Andrew Howroyd, Nov 24 2017
Number of nonequivalent dissections of an n-gon into n-3 polygons by nonintersecting diagonals up to rotation.
+10
6
1, 1, 4, 12, 43, 143, 504, 1768, 6310, 22610, 81752, 297160, 1086601, 3991995, 14732720, 54587280, 202997670, 757398510, 2834510744, 10637507400, 40023636310, 150946230006, 570534578704, 2160865067312, 8199711378716, 31170212479588, 118686578956272
COMMENTS
This is almost identical to A003444, but has a different offset and a more precise definition. In other words, the number of almost-triangulations of an n-gon modulo the cyclic action.
Equivalently, the number of edges of the (n-3)-dimensional associahedron modulo the cyclic action.
The dissection will always be composed of one quadrilateral and n-4 triangles. - Andrew Howroyd, Nov 25 2017 Also number of necklaces of 2 colors with 2n-4 beads and n black ones. - Wouter Meeussen, Aug 03 2002
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
a(n) = (1/(2n-4)) Sum_{d |(2n-4, n)} phi(d)*binomial((2n-4)/d, n/d) for n >= 4. - Wouter Meeussen, Aug 03 2002
MAPLE
C:=n->binomial(2*n, n)/(n+1);
T2:= proc(n) local t1; global C;
t1 := (n-3)*C(n-2)/(2*n);
if n mod 4 = 0 then t1:=t1+C(n/4-1)/2 fi;
if n mod 2 = 0 then t1:=t1+C(n/2-1)/4 fi;
t1; end;
[seq(T2(n), n=4..40)];
MATHEMATICA
c[n_] := Binomial[2*n, n]/(n+1);
T2[n_] := Module[{t1}, t1 = (n-3)*c[n-2]/(2*n); If[Mod[n, 4] == 0, t1 = t1 + c[n/4-1]/2]; If[Mod[n, 2] == 0, t1 = t1 + c[n/2-1]/4]; t1];
a[n_] := Sum[EulerPhi[d]*Binomial[(2n-4)/d, n/d], {d, Divisors[GCD[2n-4, n] ]}]/(2n-4);
PROG
(PARI)
a(n) = if(n>=4, sumdiv(gcd(2*n-4, n), d, eulerphi(d)*binomial((2*n-4)/d, n/d))/(2*n-4)) \\ Andrew Howroyd, Nov 25 2017
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