A293973 Sierpinski carpet iterations: start with a(0) = 1; read a(n) as a 3^n X 3^n binary matrix, replace 1 with [111;101;111] and 0 with [000;000;000], concatenate the 3^(n+1) rows of the new matrix.

1, 111101111, 111111111101101101111111111111000111101000101111000111111111111101101101111111111

Offset

0,2

Comments

The term a(n) has 9^n = A001019(n) digits.

See A153490 for the Sierpinski carpet seen as an infinite matrix read by antidiagonals, and A292688 for a variant where the digits on the antidiagonals are concatenated.

See A292686 for a 1-dimensional variant.

Links

Table of n, a(n) for n=0..2.

Example

Consider a(0) = 1 as a 1 X 1 matrix, replace the 1 by the 3 X 3 matrix E = [1,1,1; 1,0,1; 1,1,1], then this matrix is the result. Concatenating all elements yields a(1) = concat(111,101,111) = 111101111.
Now reconsider a(1) as the previously given 3 X 3 matrix E. Replace every 1 by that same matrix E. This yields the 9 X 9 matrix
   [ 1 1 1  1 1 1  1 1 1 ]
   [ 1 0 1  1 0 1  1 0 1 ]
   [ 1 1 1  1 1 1  1 1 1 ]
   [ 1 1 1  0 0 0  1 1 1 ]
   [ 1 0 1  0 0 0  1 0 1 ]
   [ 1 1 1  0 0 0  1 1 1 ]
   [ 1 1 1  1 1 1  1 1 1 ]
   [ 1 0 1  1 0 1  1 0 1 ]
   [ 1 1 1  1 1 1  1 1 1 ].
Concatenating all elements yields a(2) = 111111111101101101111111111111000111101000101111000111111111111101101101111111111.

Mathematica

A293973[n_]:=FromDigits[Flatten[Nest[ArrayFlatten[{{#, #, #}, {#, 0, #}, {#, #, #}}]&, {{1}}, n]]]; Array[A293973, 4, 0] (* Paolo Xausa, May 12 2023 *)

Prog

(PARI) a(n, A=Mat(1), E=2^9-1-2^4)={for(k=1, n, A=matrix(3^k, 3^k, i, j, A[(i+2)\3, (j+2)\3]&&bittest(E, (i-1)%3*3+(j-1)%3))); fromdigits(apply(t->fromdigits(t~, 10), Vec(A)), 10^3^n)}

Crossrefs

Cf. A153490, A293834, A001019, A292688 (antidiagonals 1..3^n of the term a(n) seen as 3^n X 3^n matrix), A292686 (1-dim. variant).

Sequence in context: A345631 A346286 A158755 * A094327 A180513 A202174

Adjacent sequences: A293970 A293971 A293972 * A293974 A293975 A293976

Keyword

nonn

Author

M. F. Hasler, Oct 20 2017

Status

approved