A245094 Total squares count in n-th generation of Pythagoras tree variation which is rhombitrihexagonal tiling.

1, 2, 4, 8, 13, 20, 24, 27, 33, 36, 42, 45, 51, 54, 60, 63, 69, 72, 78, 81, 87, 90, 96, 99, 105, 108, 114, 117, 123, 126, 132, 135, 141, 144, 150, 153, 159, 162, 168, 171, 177, 180, 186, 189, 195, 198, 204, 207, 213, 216, 222, 225, 231, 234, 240, 243, 249, 252, 258, 261, 267

Offset: 0

Author

Kival Ngaokrajang, Nov 12 2014

Keywords

• nonn

Comments

Refer to Pythagoras tree (fractal) in the link. In the article "Varying the angle", the construction rule is changed from the standard Pythagoras tree by changing the base angle from 90 degrees to 60 degrees. It is easily seen that the size of the unit squares remains constant and equal to sin(30 degrees)/(1/2) = 1. The first overlap occurs at the fifth generation (n=4). The general pattern produced is the rhombitrihexagonal tiling, an array of hexagons bordered by the constructing squares. a(n) gives total count of squares in n-th generation which excluding the overlap into (n-1)-th generation and count only 1 for the overlap among current one. See illustration.

Conjecture: In the limit n -> infinity this construction produces one of the eight planar semiregular tessellations (one of the 11 Archimedean tessellations, the other three being regular). This is the tessellation (3,4,6,4) because of the sequence of regular 3-, 4- and 6-gons around each vertex. See the Eric Weisstein link. - Wolfdieter Lang, Nov 23 2014

Links

• Kival Ngaokrajang, Illustration of initial terms
• John Riordan and N. J. A. Sloane, Correspondence, 1974
• Eric Weisstein's World of Mathematics, Semiregular Tessellation
• Wikipedia, Pythagoras tree
• Wikipedia, Rhombitrihexagonal tiling
• Index entries for linear recurrences with constant coefficients, signature (1, 1, -1).

Crossrefs

Cf. A226454, A227298.

Programs

Mathematica

a[n_] := a[n] = If[n <= 6, {1, 2, 4, 8, 13, 20, 24}[[n+1]], a[n-1] + 6 - 3 Mod[n, 2]]; Table[a[n], {n, 0, 60}] (* Jean-François Alcover, Nov 24 2016, adapted from PARI *)

PARI

{a=24;print1("1, 2, 4, 8, 13, 20, ",a,", ");
for (n=7,100,if (Mod(n,2)==1,d1=3,d1=6);a=a+d1;print1(a,", "))}

Formula

Conjectures from Colin Barker, Nov 12 2014: (Start)

a(n) = 3*((-1)^n + 6*n-5)/4 for n > 5.

a(n) = a(n-1) + a(n-2) - a(n-3) for n > 8.

G.f.: (2*x^8 - 4*x^7 - x^6 + 3*x^5 + 3*x^4 + 3*x^3 + x^2 + x + 1) / ((x-1)^2*(x+1)).

(End)

It follows from the above conjecture that this sequence consists of interlaced polynomials for n > 5: a(2n) = 3*(3n-1) and a(2n+1) = 9*n. - Avi Friedlich, May 09 2015