A363146 Triangle T(n,k) in which the n-th row encodes the inverse of a 3n X 3n Jacobi matrix, with 1's on the lower, main, and upper diagonals in GF(2), where the encoding consists of the decimal representations for the binary rows (n >= 1, 1 <= k <= 3n).

3, 7, 6, 27, 59, 48, 3, 55, 54, 219, 475, 384, 27, 443, 432, 3, 439, 438, 1755, 3803, 3072, 219, 3547, 3456, 27, 3515, 3504, 3, 3511, 3510, 14043, 30427, 24576, 1755, 28379, 27648, 219, 28123, 28032, 27, 28091, 28080, 3, 28087, 28086, 112347, 243419, 196608, 14043, 227035, 221184, 1755, 224987, 224256, 219, 224731

Offset: 1

Author

Nei Y. Soma, May 19 2023

Keywords

• nonn
• tabf

Comments

Each term of the sequence encodes a line of the inverse of a Jacobi matrix that has 1s on its lower, main, and upper diagonals in GF(2). The associated inverse matrix column values come from the binary representation of that base-10 number, being a bit per column. These matrices have ascending and consecutive multiples of 3 sizes. If the binary number has fewer bits than the number of columns, it must be zero-padded to the left. To obtain the inverse matrices in real numbers instead of GF(2), alternate between + and - between the 1s in a row. If a row is a multiple of 3, alternate between - and +. The determinants of these 3m x 3m Jacobi matrices are 1 in GF(2), as proven by Sutner (1989), and alternate between -1 and 1 in R if m is odd or even, as proven by Melo (1987).

The recurrence, in line 3, uses the Iverson notation as presented in Graham, Knuth, and Patashnik (2002).

The proof of the correctness of that sequence of inverses is done by induction.

Examples

			For n = 1, the Jacobi 3 X 3 matrix has as rows
     1, 1, 0
     1, 1, 1
     0, 1, 1.
Its inverse has the rows
     0, 1, 1
     1, 1, 1
     1, 1, 0.
Representing these rows as decimal numbers the first three terms of the sequence are: 3, 7, and 6.
The next terms in the sequence occur for n = 2, given a sequence of six numbers. The Jacobi 6 X 6 matrix has as its rows:
      1, 1, 0, 0, 0, 0
      1, 1, 1, 0, 0, 0
      0, 1, 1, 1, 0, 0
      0, 0, 1, 1, 1, 0
      0, 0, 0, 1, 1, 1
      0, 0, 0, 0, 1, 1.
Its inverse has as rows:
      0, 1, 1, 0, 1, 1
      1, 1, 1, 0, 1, 1
      1, 1, 0, 0, 0, 0
      0, 0, 0, 0, 1, 1
      1, 1, 0, 1, 1, 1
      1, 1, 0, 1, 1, 0.
These 6 latter rows from binary to decimal give the next 6 terms of the sequence: 27, 49, 48, 3, 55, and 54.
Triangle T(n,k) begins:
     3,    7,    6;
    27,   59,   48,   3,   55,   54;
   219,  475,  384,  27,  443,  432,  3,  439,  438;
  1755, 3803, 3072, 219, 3547, 3456, 27, 3515, 3504, 3, 3511, 3510;
  ...
		

References

• R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Boston, 2nd Ed., 12th printing, 2002, pp. 24-25.
• P. Lancaster and M. Tismenetsky, The Theory of Matrices, Academic Press, Boston, 1985, p. 35.
• J. P. Melo, Reversibility of John von Neumann cellular automata, M.Sc. Thesis, Division of Computer Science, Instituto Tecnológico de Aeronáutica, 1997 (in Portuguese), p. 18.
• K. Sutner, Linear Cellular Automata and the Garden-of-Eden, The Mathematical Intelligencer, 11(2), 1989, 49-53, p. 52.

Links

• Nei Y. Soma, Rows n = 1..30, flattened

Crossrefs

Column k=1 gives A083713.

Column k=3 gives A083233.

T(n,3n) gives A125837(n+1).

T(n,3n-1) gives A083068.

T(n,3n-2) gives A010701.

Cf. A038184 one-dimensional cellular automaton (Rule 150) in a tape with 3n cells has as adjacency matrix the Jacobi matrices, 3n X 3n, with 1s on the lower, main and upper diagonals and the operations on it are in GF(2).

Programs

Maple

T:= n-> (M-> seq(add(abs(M[j, n*3-i])*2^i, i=0..n*3-1), j=1..n*3))
               (Matrix(n*3, (i, j)-> `if`(abs(i-j)<2, 1, 0))^(-1)):
seq(T(n), n=1..10);  # Alois P. Heinz, May 20 2023

Mathematica

sequence = {};
m = 6;
For[k = 1, k <= m, k++, {
  n = 3*k;
  J = ConstantArray[0, {n, n}];
  For[i = 1, i < n, i++,
   J[[i, i]] = J[[i + 1, i]] = J[[i, i + 1]] = 1];
  J[[1, 1]] = J[[n, n]] = 1;
  InvJ = Mod[Inverse[J], 2];
  For[i = 1, i <= n, i++, AppendTo[sequence, FromDigits[InvJ[[i]], 2]]]
  }
 ]
sequence

PARI

row(n)=my(m=3*n, A=lift(matrix(m, m, i, j, Mod(abs(i-j)<=1, 2))^(-1))); vector(m, i, fromdigits(A[i,], 2)) \\ Andrew Howroyd, May 20 2023

Formula

The recurrence has as its base: r(1, 1) = 3; r(2, 1) = 7 and r(3, 1) = 6;

For 2 <= k <= m, and i = 1, 2, ..., 3(k-1):

r(i, k) = 8*r(i, k-1) + r(1,1) (i != 0 (mod 3)).

And r(3k-2, k) = r(1, 1);

r(3k-1, k) = 8*r(3(k-1), k-1) + r(2,1);

r(3k, k) = 8*r(3(k-1), (k-1)) + r(3, 1).