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Number of planar hexagon trees with n hexagons.
(Formerly M2936)
12

%I M2936 #38 Aug 01 2024 21:50:32

%S 1,1,3,12,68,483,3946,34485,315810,2984570,28907970,285601251,

%T 2868869733,29227904840,301430074416,3141985563575,33059739636198,

%U 350763452126835,3749420616902637,40348040718155170,436827335493148600

%N Number of planar hexagon trees with n hexagons.

%C Number of nonequivalent dissections of a polygon into n hexagons by nonintersecting diagonals up to rotation and reflection. - _Andrew Howroyd_, Nov 20 2017

%C Number of unoriented polyominoes composed of n hexagonal cells of the hyperbolic regular tiling with Schläfli symbol {6,oo}. A stereographic projection of this tiling on the Poincaré disk can be obtained via the Christensson link. For unoriented polyominoes, chiral pairs are counted as one. - _Robert A. Russell_, Jan 23 2024

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H G. C. Greubel, <a href="/A004127/b004127.txt">Table of n, a(n) for n = 1..925</a>

%H Malin Christensson, <a href="http://malinc.se/m/ImageTiling.php">Make hyperbolic tilings of images</a>, web page, 2019.

%H L. W. Beineke and R. E. Pippert, <a href="https://doi.org/10.1017/S0017089500002305">On the enumeration of planar trees of hexagons</a>, Glasgow Math. J., 15 (1974), 131-147.

%H L. W. Beineke and R. E. Pippert, <a href="/A004127/a004127.pdf">On the enumeration of planar trees of hexagons</a>, Glasgow Math. J., 15 (1974), 131-147. [Annotated scanned copy]

%H <a href="/index/Tra#trees">Index entries for sequences related to trees</a>

%F See Theorem 3 on p. 142 in the Beineke-Pippert paper; also the Maple and Mathematica codes here.

%F a(n) ~ 5^(5*n + 1/2) / (sqrt(Pi) * n^(5/2) * 2^(8*n + 13/2)). - _Vaclav Kotesovec_, Mar 13 2016

%F a(n) = A221184(n-1) - A369473(n) = (A221184(n-1) + A143546(n)) / 2 = A369473(n) + A143546(n). - _Robert A. Russell_, Jan 23 2024

%p T := proc(n) if floor(n)=n then binomial(5*n+1,n)/(5*n+1) else 0 fi end: U := proc(n) if n mod 2 = 0 then binomial(5*n/2+1, n/2)/(5*n/2+1) else 6*binomial((5*n+1)/2,(n-1)/2)/(5*n+1) fi end: S := n->T(n)/4/(2*n+1)+T(n/2)/6+(5*n-2)*T((n-1)/3)/6/(2*n+1)+T((n-1)/6)/6+7*U(n)/12: seq(S(n),n=1..25); (Emeric Deutsch)

%t p=6; Table[(Binomial[(p-1)n, n]/(((p-2)n+1)((p-2)n+2)) + If[OddQ[n], If[OddQ[p], Binomial[(p-1)n/2, (n-1)/2]/n, (p+1)Binomial[((p-1)n-1)/2, (n-1)/2]/((p-2)n+2)], 3Binomial[(p-1)n/2, n/2]/((p-2)n+2)]+Plus @@ Map[EulerPhi[ # ]Binomial[((p-1)n+1)/#, (n-1)/# ]/((p-1)n+1)&, Complement[Divisors[GCD[p, n-1]], {1, 2}]])/2, {n, 1, 20}] (* _Robert A. Russell_, Dec 11 2004 *)

%Y Column k=6 of A295260.

%Y Cf. A002294.

%Y Polyominoes: A221184{n-1} (oriented), A369473 (chiral), A143546 (achiral), A005040 {5,oo}, A005419 {7,oo}.

%K nonn

%O 1,3

%A _N. J. A. Sloane_

%E More terms from _Emeric Deutsch_, Jan 22 2004