A375874 Number of distinct n X n patterns in the squiral tiling.

1, 2, 14, 70, 126, 270, 438, 630, 790, 958, 1134, 1542, 1974, 2430, 2910, 3414, 3942, 4494, 5070, 5670, 6142, 6622, 7110, 7606, 8110, 8622, 9142, 9670, 10206, 11406, 12630, 13878, 15150, 16446, 17766, 19110, 20478, 21870, 23286, 24726, 26190, 27678, 29190

Offset: 0

Author

Johan Nilsson, Sep 01 2024

Keywords

• nonn

Comments

The squiral tiling, can be obtained as the limit pattern under the binary block substitution 0 -> [[1,0,1],[0,0,0][1,0,1]] and 1 -> [[0,1,0],[1,1,1][0,1,0]], when starting with the seed 0.

Examples

			a(1) = 2, since there are 2 different 1X1 patterns in the squiral tiling; namely 0 and 1.
a(2) = 14, since there are 14 different 2X2 patterns in the squiral tiling; namely all 16 2X2 binary matrices except [[0,0],[0,0]] and [[1,1],[1,1]].
		

References

• M. Baake, and U. Grimm, Aperiodic Order. Volume 1: A Mathematical Invitation, Encyclopedia of Mathematics and its Applications No. 149 Cambridge University Press, Cambridge (2013).
• B. Grünbaum and F. C. Shephard, Tilings and Patterns, W.H. Freeman 1987, MR0857454.

Links

• Johan Nilsson, Table of n, a(n) for n = 0..10000
• Tilings Encyclopedia, Squiral
• Johan Nilsson The Pattern Complexity of the Squiral Tiling

Programs

Maple

a:= n-> `if`(n<3, [1, 2, 14][n+1], ((A, B)-> (4+8*A-8*B)*(n-1)^2+
    (12*3^A+24*3^B)*(n-1)-18*9^A)(ilog[3](n-2), ilog[3]((n-2)/2))):
seq(a(n), n=0..42);  # Alois P. Heinz, Sep 18 2024

Mathematica

a[n_] := If[n<3, {1, 2, 14}[[n+1]], With[{A = Floor@ Log[3, n-2], B = Floor@ Log[3, (n-2)/2]}, (4+8*A-8*B)*(n-1)^2+(12*3^A+24*3^B)*(n-1)-18*9^A]];
Table[a[n], {n, 0, 42}] (* Jean-François Alcover, Mar 27 2025, after Alois P. Heinz *)

PARI

a(n)=if(n<4, [1,2,14,70][n+1], my(A=logint(n-2,3), B=logint((n-2)\2,3)); (4 + 8*A - 8*B)*(n - 1)^2 + (12 * 3^A + 24 * 3^B) * (n - 1) - 18 * 9^A) \\ Andrew Howroyd, Sep 18 2024

Python

from sympy import integer_log
def A375874(n):
    if n<4: return (1,2,14,70)[n]
    a, b = integer_log(n-2,3)[0]+1, integer_log((n>>1)-1,3)[0]+1
    return (n-1)*((1+(a-b<<1))*(n-1)+((c:=3**a)+(3**b<<1))<<1)-c**2<<1 # Chai Wah Wu, Sep 18 2024

Formula

a(n) = (4 + 8*A - 8*B)*(n - 1)^2 + (12 * 3^A + 24 * 3^B) * (n - 1) - 18 * 9^A, for n>=4 where A = floor(log3(n-2)), B = floor(log3((n-2)/2)), and log3 is the logarithm in base 3.

For n>=2;

a(3*n-2) = 9*a(n),

a(9*n-7) = 5*a(3*n+1) - 16*a(3*n) + 20*a(3*n-1),

a(9*n-4) = - a(3*n+1) + 5*a(3*n) + 5*a(3*n-1),

a(9*n-1) = 2*a(3*n+1) + 8*a(3*n) - a(3*n-1),

a(3*n) = a(3*n-1) + 3*a(n+1) - 3*a(n).