A038184 -id:A038184 - cp's OEIS frontend

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

Nei Y. Soma, May 19 2023

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.

			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;
  ...

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.

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

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).

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

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

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

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).

A177960 Numbers of the form A001317(t), excluding those at places of the form t=m*(2^k-1), m>=0, k>=2.

Original entry on oeis.org

3, 5, 17, 51, 257, 1285, 3855, 13107, 65537, 196611, 983055, 1114129, 5570645, 16711935, 50529027, 84215045, 858993459, 4294967297, 21474836485, 219043332147, 365072220245, 1103806595329, 3311419785987

Offset: 1

Vladimir Shevelev, Dec 24 2010

nonn

m-nomial (m>=2) coefficients are coefficients of the polynomial (1+x+...+x^(m-1))^n (n>=0), see A007318 (m=2), A027907 (m=3), A008287 (m=4), A035343 (m=5) etc. For k>=1, consider the triangle of 2^k-nomial coefficients, each entry reduced mod 2, and convert each row of the reduced triangle to a single number by interpreting the sequence of bits as binary representation of a number. This defines sequences A001317 (k=1), A177882 (k=2), A177897 (k=3), etc. The current sequence lists terms of A001317 which are not derived from any of the sequences for k >=2, not from 4-nomial, not from 8-nomial, not from 16-nomial etc.

Conjecture: If for every m>=2, to consider triangle of m-nomial coefficients mod 2 converted to decimal, then the sequence lists terms of A001317 which are not in the union of other sequences for m=3 (A038184), 4 (A177882), 5, 6,...

Cf. A001317, A177882, A177897, A027907, A008287.

Denote by B(n) the number of terms of the sequence among the first n terms of A001317. Then lim_{n->infinity} B(n)/ = Product_{prime p>=2} (1 - 1/(2^p-1)) = A184085.

A363099 Triangle T(n,k) in which the n-th row encodes the inverse of a 3n+1 X 3n+1 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+1).

Original entry on oeis.org

11, 3, 12, 13, 91, 27, 96, 107, 3, 108, 109, 731, 219, 768, 859, 27, 864, 875, 3, 876, 877, 5851, 1755, 6144, 6875, 219, 6912, 7003, 27, 7008, 7019, 3, 7020, 7021, 46811, 14043, 49152, 55003, 1755, 55296, 56027, 219, 56064, 56155, 27, 56160, 56171, 3, 56172, 56173, 374491, 112347, 393216, 440027, 14043, 442368

Offset: 1

Nei Y. Soma, May 20 2023

Each term in 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 start with a 4 X 4 matrix and the consecutive terms came by adding ascending and consecutive multiples of 3. 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+1 X 3m+1 Jacobi matrices are 1 in GF(2), and alternate between -1 and 1 in R if m is odd or even. These properties were proven by Sutner (1989) and Melo (1997), respectively.

			For m = 1, the Jacobi 4 X 4 matrix has as rows
     1, 1, 0, 0
     1, 1, 1, 0
     0, 1, 1, 1
     0, 0, 1, 1
Its inverse has the rows
     1, 0, 1, 1
     0, 0, 1, 1
     1, 1, 0, 0
     1, 1, 0, 1
Representing these rows as binary numbers in base 10 the first three terms of the sequence are: 11, 3, 12, 13.
The next numbers in the sequence occur for m = 2, given a sequence of six numbers. The Jacobi 7 X 7 matrix has as its rows:
     1, 1, 0, 0, 0, 0, 0
     1, 1, 1, 0, 0, 0, 0
     0, 1, 1, 1, 0, 0, 0
     0, 0, 1, 1, 1, 0, 0
     0, 0, 0, 1, 1, 1, 0
     0, 0, 0, 0, 1, 1, 1
     0, 0, 0, 0, 0, 1, 1
Its inverse has as rows:
     1, 0, 1, 1, 0, 1, 1
     0, 0, 1, 1, 0, 1, 1
     1, 1, 0, 0, 0, 0, 0
     1, 1, 0, 1, 0, 1, 1
     0, 0, 0, 0, 0, 1, 1
     1, 1, 0, 1, 1, 0, 0
     1, 1, 0, 1, 1, 0, 1
These 7 latter rows from binary to base 10 give the next 7 terms of the sequence: 91, 27, 96, 107, 3, 108, 109.
Triangle T(n,k) begins:
    11,    3,   12,   13;
    91,   27,   96,  107,   3,  108,  109;
   731,  219,  768,  859,  27,  864,  875,  3,  876,  877;
  5851, 1755, 6144, 6875, 219, 6912, 7003, 27, 7008, 7019, 3, 7020, 7021;
  ...

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.

Nei Y. Soma, Rows n = 1..30, flattened
K. Sutner, Linear Cellular Automata and the Garden-of-Eden, The Mathematical Intelligencer, 11(2), 1989, pp. 49-53; see p. 52

Column k=1 gives A245599(n+1).
Column k=2 gives A083713.
Column k=3 gives A204623.
T(n,3n-1) gives A010701.
Cf. A038184 One-dimensional cellular automaton (Rule 150) in a tape with 3m cells has as adjacency matrix the Jacobi matrices, 3m X 3m, with 1s on the lower, main and upper diagonals and the operations on it are in GF(2). And A363146 for the inverse of Jacobi matrices 3m X 3m, with 1s on the lower, main, and upper diagonals in GF(2).

T:= n-> (M-> seq(add(abs(M[j, n*3+1-i])*2^i, i=0..n*3), j=1..n*3+1))
               (Matrix(n*3+1, (i, j)-> `if`(abs(i-j)<2, 1, 0))^(-1)):
seq(T(n), n=1..6);  # Alois P. Heinz, May 20 2023

sequence = {};
For[k = 1, k <= 50, k++, {
  n = 3*k + 1;
  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

row(n)=my(m=3*n+1, 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

The recurrence has as its base:
     r(1, 1) = 11;
     r(2, 1) =  3;
     r(3, 1) = 12;
     r(4, 1) = 13.
For 2 <= k <= m, and i = 1, 2, 3, ..., 3k - 2:
    r(i, k) = 8*r(i, k-1) + r(2, 1) (i != 0 (mod 3)).
And r(3k-1, k) = r(2, 1);
    r(3k,   k) = 8*r(3(k-1), k-1) + r(3,1);
    r(3k+1, k) = 8*r(3(k-1), k-1) + r(4,1).

cp's OEIS Frontend

A048712 2nd column of Family 1 "90 X 150 array": generations 0 .. n of Rule 150 starting from seed pattern 5.

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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).

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A177960 Numbers of the form A001317(t), excluding those at places of the form t=m*(2^k-1), m>=0, k>=2.

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A363099 Triangle T(n,k) in which the n-th row encodes the inverse of a 3n+1 X 3n+1 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+1).

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Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

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