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%I #35 May 13 2023 13:54:34
%S 1,111101111,
%T 111111111101101101111111111111000111101000101111000111111111111101101101111111111
%N 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.
%C The term a(n) has 9^n = A001019(n) digits.
%C 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.
%C See A292686 for a 1-dimensional variant.
%e 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.
%e 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
%e [ 1 1 1 1 1 1 1 1 1 ]
%e [ 1 0 1 1 0 1 1 0 1 ]
%e [ 1 1 1 1 1 1 1 1 1 ]
%e [ 1 1 1 0 0 0 1 1 1 ]
%e [ 1 0 1 0 0 0 1 0 1 ]
%e [ 1 1 1 0 0 0 1 1 1 ]
%e [ 1 1 1 1 1 1 1 1 1 ]
%e [ 1 0 1 1 0 1 1 0 1 ]
%e [ 1 1 1 1 1 1 1 1 1 ].
%e Concatenating all elements yields a(2) = 111111111101101101111111111111000111101000101111000111111111111101101101111111111.
%t A293973[n_]:=FromDigits[Flatten[Nest[ArrayFlatten[{{#,#,#},{#,0,#},{#,#,#}}]&,{{1}},n]]];Array[A293973,4,0] (* _Paolo Xausa_, May 12 2023 *)
%o (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)}
%Y 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).
%K nonn
%O 0,2
%A _M. F. Hasler_, Oct 20 2017
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