OFFSET
0,2
COMMENTS
A kind of two-dimensional ternary Catalan number. This sequence enumerates the decompositions of the unit square into 2n+1 rectangles obtained by the following algorithm.
(a) Start with the unit square.
(b) Perform the following operation n times:
(1) Choose a rectangle in the current decomposition.
(2) Trisect this rectangle into three rectangles horizontally or vertically.
Note that different sequences of trisections can produce the same decomposition.
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..889
Yu Hin (Gary) Au, Fatemeh Bagherzadeh, Murray R. Bremner, Enumeration and Asymptotic Formulas for Rectangular Partitions of the Hypercube, arXiv:1903.00813 [math.CO], Mar 03 2019.
FORMULA
Recurrence relation: a(n) = C(2n+1) with C(1) = 1 and C(n) = 2 Sum_{i1,i2,i3} C(i1)C(i2)C(i3) - Sum_{i1,i2,i3,i4,i5,i6,i7,i8,i9} C(i1)C(i2)C(i3)C(i4)C(i5)C(i6)C(i7)C(i8)C(i9). The first sum is over all 3-compositions of n into positive integers (i1+i2+i3=n), and the second sum is over all 9-compositions of n into positive integers (i1+i2+...+i9=n).
a(n) = [x^(2n+1)] G(x), where G(x) satisfies: G(x)^9 - 2*G(x)^3 + G(x) - x = 0.
MAPLE
a:= n-> coeff(series(RootOf(G^9-2*G^3+G-x, G), x, 2*n+2), x, 2*n+1):
seq(a(n), n=0..25); # Alois P. Heinz, Dec 14 2018
MATHEMATICA
a[n_] := SeriesCoefficient[InverseSeries[x - 2 x^3 + x^9 + O[x]^(2n+2), x], {x, 0, 2n+1}];
Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Aug 13 2019, from PARI *)
PROG
(PARI) a(n)={polcoef(serreverse(x - 2*x^3 + x^9 + O(x^(2*n+2))), 2*n+1)} \\ Andrew Howroyd, Dec 14 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Yu Hin Au, Dec 14 2018
STATUS
approved