A292686 -id:A292686 - cp's OEIS frontend

A292687 a(n) = Product_{k=0..n-1} (4^(3^k) + 1) = decimal value of the Sierpinski-type iteration result A292686(n) (replace 0 with 000 and 1 with 101) considered as a binary number.

1, 5, 325, 85197125, 1534774961612150361293125, 8972304477322525702813810177861539421333393918862058319149818714344653125

Offset: 0

M. F. Hasler, Oct 20 2017

nonn

The next term, a(6), has 202 digits and does not fit on one line.
This is the decimal representation of the terms of A292686 considered as binary numbers.
To get a(n+1) from a(n), write a(n) in binary, replace digits 0 with 000 and 1 with 101, and convert back to decimal. Equivalently, consider the binary expansion of a(n) as base-8 expansion, multiply it by 5, and convert back from octal to decimal.

			a(0) = 1 is already written in binary; multiplied by 5 it yields 5, read in octal is the same as in decimal, a(1) = 5.
a(1) = 5  = 101[2] in binary; consider 101 in base 8 (or base 10), multiply by 5 to get 505, convert from octal to decimal to get a(2) = 5*8^2 + 5 = 325.
a(2) = 325  = 101000101[2] in binary; consider this in base 8 (or base 10), multiply by 5 to get 505000505, convert from octal to decimal to get a(2) = 325*8^6 + 325 = 85197125.

Cf. A292686 for the binary representation of a(n), and for more links, references and motivation.

A292687[nmax_]:=FoldList[Times,1,4^(3^Range[0,nmax-1])+1];A292687[6] (* Paolo Xausa, May 13 2023 *)

PARI
```
A292687(n)=prod(k=0,n-1,4^3^k+1)
```

a(n+1) = (4^(3^n)+1)*a(n).

a(n) = Product_{k=0..n-1} (4^(3^k)+1).

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)}

cp's OEIS Frontend

A292687 a(n) = Product_{k=0..n-1} (4^(3^k) + 1) = decimal value of the Sierpinski-type iteration result A292686(n) (replace 0 with 000 and 1 with 101) considered as a binary number.

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