A292688 -id:A292688 - cp's OEIS frontend

A153490 Sierpinski carpet, read by antidiagonals.

1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1

Offset: 1

Roger L. Bagula, Dec 27 2008

The Sierpinski carpet is the fractal obtained by starting with a unit square and at subsequent iterations, subdividing each square into 3 X 3 smaller squares and removing (from nonempty squares) the middle square. After the n-th iteration, the upper-left 3^n X 3^n squares will always remain the same. Therefore this sequence, which reads these by antidiagonals, is well-defined.

Row sums are {1, 2, 2, 4, 5, 4, 6, 6, 4, 8, 10, 8, ...}.

			The Sierpinski carpet matrix reads
   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 ...
   (...)
so the antidiagonals are
  {1},
  {1, 1},
  {1, 0, 1},
  {1, 1, 1, 1},
  {1, 1, 1, 1, 1},
  {1, 0, 1, 1, 0, 1},
  {1, 1, 1, 0, 1, 1, 1},
  {1, 1, 1, 0, 0, 1, 1, 1},
  {1, 0, 1, 0, 0, 0, 1, 0, 1},
  {1, 1, 1, 1, 0, 0, 1, 1, 1, 1},
  {1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1},
  {1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1},
  ...

Eric Weisstein's World of Mathematics, Sierpinski Carpet.
Wikipedia, Sierpinski carpet.

Cf. A292688 (n-th antidiagonal concatenated as binary number), A292689 (decimal representation of these binary numbers).

Cf. A293143 (number of vertex points in a Sierpinski Carpet).

<< MathWorld`Fractal`; fractal = SierpinskiCarpet;
a = fractal[4]; Table[Table[a[[m]][[n - m + 1]], {m, 1, n}], {n, 1, 12}];
Flatten[%]

A153490_row(n,A=Mat(1))={while(#AM. F. Hasler, Oct 23 2017

Edited by M. F. Hasler, Oct 20 2017

A292689 Decimal values of the antidiagonals of the Sierpinski carpet considered as binary numbers.

Original entry on oeis.org

1, 3, 5, 15, 31, 45, 119, 231, 325, 975, 2015, 2925, 8191, 16383, 23405, 61431, 118759, 166725, 499151, 1030623, 1495405, 4186623, 8372735, 11960685, 31392247, 60686823, 85197125, 255591375, 528222175, 766774125, 2147229695, 4294721535, 6135503725, 16103829495, 31132078055

Offset: 1

M. F. Hasler, Oct 23 2017

Term a(n) is the decimal value of A292688 = concatenation of the terms in row n of A153490 considered as a binary number.
The Sierpinski carpet is the fractal obtained by starting with a unit square and at subsequent iterations, subdividing each square into 3 X 3 smaller squares and removing the middle square. After the n-th iteration, the upper-left 3^n X 3^n squares will always remain the same. Therefore this sequence, which considers the antidiagonals of this infinite matrix, is well-defined.
The n-th term a(n) has n binary digits.
The Hamming weights of the terms (also row sums of A153490) are (1, 2, 2, 4, 5, 4, 6, 6, 4, 8, 10, 8, 13, 14, 10, 14, 13, 8, 14, 16, 12, 18, 18, 12, 16, ...).

			The Sierpinski carpet matrix A153490 reads
   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 ...
   ...
The concatenation of the terms in the antidiagonals yields A292688 = (1, 11, 101, 1111, 11111, 101101, 1110111, 11100111, 101000101, 1111001111, 11111011111, 101101101101, 1111111111111, 11111111111111, 101101101101101, ...).
Considered as binary numbers and converted to base 10, this yields 1, 3, 5, 15, 31, 45, 119, 231, 325, ... .

Paolo Xausa, Table of n, a(n) for n = 1..729
Eric Weisstein's World of Mathematics, Sierpinski Carpet.
Wikipedia, Sierpinski carpet.

Cf. A153490, A292688.

A292689[i_]:=With[{a=Nest[ArrayFlatten[{{#,#,#},{#,0,#},{#,#,#}}]&,{{1}},i]},Array[FromDigits[Diagonal[a,#],2]&,3^i,1-3^i]];A292689[4] (* Generates 3^4 terms *) (* Paolo Xausa, May 13 2023 *)

PARI
```
A292689(n,A=Mat(1))={while(#A
				
```

a(k+1) = 2*a(k)+1 for all k in A003462 = (1, 4, 13, 40, 121, 364, ...). (Conjectured.) - R. J. Cano, Oct 25 2017

This is true, moreover, a(k) = 2^k-1 for these k (and k' = k+1), and the neighboring antidiagonals (k-1 and k+2) have bitmaps of the form {101}*(101 repeated). - M. F. Hasler, Oct 25 2017

A293974 Row sums of antidiagonals of the Sierpinski carpet A153490.

Original entry on oeis.org

1, 2, 2, 4, 5, 4, 6, 6, 4, 8, 10, 8, 13, 14, 10, 14, 13, 8, 14, 16, 12, 18, 18, 12, 16, 14, 8, 16, 20, 16, 26, 28, 20, 28, 26, 16, 29, 34, 26, 40, 41, 28, 38, 34, 20, 34, 38, 28, 41, 40, 26, 34, 29, 16, 30, 36, 28, 44, 46, 32, 44, 40, 24, 42, 48, 36, 54, 54, 36, 48, 42, 24

Offset: 1

M. F. Hasler, Oct 24 2017

Also, sums of digits of terms of A292688, or Hamming weights of terms of A292689. See there or A153490 for definition / construction of the Sierpiski carpet.

Rémy Sigrist, Table of n, a(n) for n = 1..19683

Cf. A153490, A292688, A292689, A000120, A007953, A000217.

A293974[i_]:=With[{a=Nest[ArrayFlatten[{{#,#,#},{#,0,#},{#,#,#}}]&,{{1}},i]},Array[Total[Diagonal[a,#]]&,3^i,1-3^i]];A293974[5] (* Generates 3^5 terms *) (* Paolo Xausa, May 14 2023 *)

PARI
```
A293974(n,A=Mat(1))={while(#A
				
```

a(n) = A007953(A292688(n)) = A000120(A292689(n)) = sum(k=1..n, A153490(n,k)), considering A153490 as triangle; could also be indexed as matrix (m,n = 1,...,oo) or "flattened" (linearized) using A000217.

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.

Original entry on oeis.org

1, 111101111, 111111111101101101111111111111000111101000101111000111111111111101101101111111111

Offset: 0

M. F. Hasler, Oct 20 2017

nonn

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.

			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.

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).

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

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)}

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A153490 Sierpinski carpet, read by antidiagonals.

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A292689 Decimal values of the antidiagonals of the Sierpinski carpet considered as binary numbers.

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A293974 Row sums of antidiagonals of the Sierpinski carpet A153490.

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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.

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