[go: up one dir, main page]

login
A245094
Total squares count in n-th generation of Pythagoras tree variation which is rhombitrihexagonal tiling.
5
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
0,2
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
John Riordan and N. J. A. Sloane, Correspondence, 1974
Eric Weisstein's World of Mathematics, Semiregular Tessellation
Wikipedia, Pythagoras tree
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
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 *)
PROG
(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, ", "))}
CROSSREFS
Sequence in context: A186752 A360512 A030503 * A164486 A084684 A011907
KEYWORD
nonn
AUTHOR
Kival Ngaokrajang, Nov 12 2014
STATUS
approved