A131218 -id:A131218 - cp's OEIS frontend

A140820 Triangle read by rows: T(n,k) = 1 if and only if the Gray codes for n and k have no bits in common.

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

Offset: 0

Roger L. Bagula and Gary W. Adamson, Oct 17 2008

			Triangle begins as:
  1;
  1, 0;
  1, 0, 0;
  1, 1, 0, 0;
  1, 1, 0, 0, 0;
  1, 0, 0, 0, 0, 0;
  1, 0, 0, 1, 0, 0, 0;
  1, 1, 1, 1, 0, 0, 0, 0;
  1, 1, 1, 1, 0, 0, 0, 0, 0;
  1, 0, 0, 1, 0, 0, 0, 0, 0, 0;
  1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
  1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;

G. C. Greubel, Rows n = 0..100 of the triangle, flattened

Cf. A131218.

A140280:= func< n,k | BitwiseAnd(BitwiseXor(n, ShiftRight(n, 1)), BitwiseXor(k, ShiftRight(k, 1))) eq 0 select 1 else 0 >; // based on Kevin Ryde's code
[A140280(n, k): k in [0..n], n in [0..15]]; // G. C. Greubel, Sep 05 2025

A140820[n_, k_]:= Boole[BitAnd[BitXor[n,BitShiftRight[n,1]], BitXor[k,BitShiftRight[k, 1]]]==0]; (* based on Kevin Ryde's code *)
Table[A140280[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, May 30 2019; Sep 05 2025 *)

T(n,k) = !bitand(bitxor(n,n>>1), bitxor(k,k>>1)); \\ Kevin Ryde, Jul 13 2020

def A140820(n,k): return int( (n^^(n>>1)) & (k^^(k>>1)) ==0) # based on Kevin Ryde's code
print(flatten([[A140820(n, k) for k in range(n+1)] for n in range(16)])) # G. C. Greubel, May 30 2019; Sep 05 2025

Edited by G. C. Greubel, May 30 2019
New title from Charlie Neder, Jun 03 2019
Offset changed by G. C. Greubel, Sep 05 2025

A134265 Coefficients of the polynomials of a three level Hadamard matrix substitution set based on the game matrix set: MA={{0,1},{1,1}};MB={{1,0},{3,1}} Substitution rule is for m[n]:If[m[n - 1][[i, j]] == 0, {{0, 0}, {0, 0}}, If[m[n - 1][[i, j]] == 1, MA, MB]] Based on the Previte idea of graph substitutions as applied to matrices of graphs in the Fibonacci/ anti-Fibonacci game.

Original entry on oeis.org

1, 1, -1, 1, -2, 1, 1, 2, -1, -2, 1, 1, -2, -7, 6, 20, 6, -7, -2, 1, 1, 2, -25, -10, 225, -184, -498, 500, 610, -500, -498, 184, 225, 10, -25, -2, 1

Offset: 1

Roger L. Bagula, Jan 24 2008

m[n_] := Table[Table[If[m[n - 1][[i, j]] == 0, {{0, 0}, {0, 0}}, If[m[n - 1][[i, j]] == 1, ma, {{1, 0}, {3, 1}}]], {j, 1, 2^(n - 1)}], {i, 1, 2^(n - 1)}]

Michelle Previte and Sean Yang say Have you ever wanted to build your own fractal? This article will describe a procedure called a vertex replacement rule that can be used to construct fractals. We also show how one can easily compute the topological and box dimensions of the fractals resulting from vertex replacements.

			{1},
{1, -1},
{1, -2, 1},
{1, 2, -1, -2, 1},
{1, -2, -7, 6, 20, 6, -7, -2,1},
{1, 2, -25, -10, 225, -184, -498, 500, 610, -500, -498,184, 225, 10, -25, -2, 1}

Michelle Previte and Sean Yang, A Novel Way to Generate Fractals

Cf. A122947, A131218.

m[0] = {{1}} m[1] = {{1, 0}, {3, 1}} m[2] = {{0, 1, 0, 0}, {1, 1, 0, 0}, {1, 0, 0, 1}, {3, 1, 1, 1}} m[3] = {{0, 0, 0, 1, 0, 0, 0, 0}, {0, 0, 1, 1, 0, 0, 0, 0}, {0, 1, 0, 1, 0, 0, 0, 0}, {1, 1, 1, 1, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0, 0, 1}, {1, 1, 0, 0, 0, 0, 1, 1}, {1, 0, 0, 1, 0, 1, 0, 1}, {3, 1, 1, 1, 1, 1, 1, 1}} m[4] = {{0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1}, {0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1}, {1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1}, {0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1}, {1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1}, {1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1}, {3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}}; Table[CharacteristicPolynomial[m[i], x], {i, 0, 4}]; a = Join[{{1}}, Table[CoefficientList[CharacteristicPolynomial[m[i], x], x], {i, 0, 4}]]; Flatten[a] (* visualization*) Table[ListDensityPlot[m[i]], {i, 0, 4}]

m[n] = If[m[n - 1][[i, j]] == 0, {{0, 0}, {0, 0}}, If[m[n - 1][[i, j]] == 1, MA, MB]] m[0] = {{1}} m[1] = {{1, 0}, {3, 1}} m[2] = {{0, 1, 0, 0}, {1, 1, 0, 0}, {1, 0, 0, 1}, {3, 1, 1, 1}} m[3] = {{0, 0, 0, 1, 0, 0, 0, 0}, {0, 0, 1, 1, 0, 0, 0, 0}, {0, 1, 0, 1, 0, 0, 0, 0}, {1, 1, 1, 1, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0, 0, 1}, {1, 1, 0, 0, 0, 0, 1, 1}, {1, 0, 0, 1, 0, 1, 0, 1}, {3, 1, 1, 1, 1, 1, 1, 1}} m[4] = {{0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1}, {0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1}, {0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1}, {1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1}, {0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1}, {1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1}, {1, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1}, {3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1}}

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A140820 Triangle read by rows: T(n,k) = 1 if and only if the Gray codes for n and k have no bits in common.

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