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Number of triadic partitions of the unit square into (2n+1) subrectangles.
2

%I #29 Sep 21 2022 21:53:09

%S 1,2,12,96,879,8712,90972,985728,10979577,124937892,1446119664,

%T 16972881120,201526230555,2416309004872,29215072931136,

%U 355800894005760,4360705642282569,53744080256387478,665667989498682936,8281518339078928800,103441301833577854041,1296713265300164761632

%N Number of triadic partitions of the unit square into (2n+1) subrectangles.

%C 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.

%C (a) Start with the unit square.

%C (b) Perform the following operation n times:

%C (1) Choose a rectangle in the current decomposition.

%C (2) Trisect this rectangle into three rectangles horizontally or vertically.

%C Note that different sequences of trisections can produce the same decomposition.

%H Alois P. Heinz, <a href="/A322543/b322543.txt">Table of n, a(n) for n = 0..889</a>

%H Yu Hin (Gary) Au, Fatemeh Bagherzadeh, Murray R. Bremner, <a href="https://arxiv.org/abs/1903.00813">Enumeration and Asymptotic Formulas for Rectangular Partitions of the Hypercube</a>, arXiv:1903.00813 [math.CO], Mar 03 2019.

%F 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).

%F a(n) = [x^(2n+1)] G(x), where G(x) satisfies: G(x)^9 - 2*G(x)^3 + G(x) - x = 0.

%p a:= n-> coeff(series(RootOf(G^9-2*G^3+G-x, G), x, 2*n+2), x, 2*n+1):

%p seq(a(n), n=0..25); # _Alois P. Heinz_, Dec 14 2018

%t a[n_] := SeriesCoefficient[InverseSeries[x - 2 x^3 + x^9 + O[x]^(2n+2), x], {x, 0, 2n+1}];

%t Table[a[n], {n, 0, 21}] (* _Jean-François Alcover_, Aug 13 2019, from PARI *)

%o (PARI) a(n)={polcoef(serreverse(x - 2*x^3 + x^9 + O(x^(2*n+2))), 2*n+1)} \\ _Andrew Howroyd_, Dec 14 2018

%Y Cf. A000108 (Catalan numbers), A005408, A236339 (decompositions of unit square using bisections).

%K nonn

%O 0,2

%A _Yu Hin Au_, Dec 14 2018