A376004 Limiting matrix {m_n}, where m_0 = 1 and m_{i+1} = [[m_i, A(m_i)], [B(m_i), C(m_i)]], read by antidiagonals, and A adds the corresponding x-coords to every element, B subtracts it, and C adds the corresponding y-coords.

1, 1, 1, 1, 1, 1, 2, 1, 0, 1, 1, 2, 1, 0, 1, 2, 1, 1, 2, 0, 1, 3, 2, 1, 2, -1, 0, 1, 5, 3, 1, 1, -1, -1, -1, 1, 1, 5, 3, 1, 1, -1, -1, -1, 1, 2, 1, 4, 4, 1, 2, -2, 0, 0, 1, 3, 2, 1, 5, 1, 2, 3, -1, -1, 0, 1, 5, 3, 1, 1, 2, 2, 2, 4, -1, -1, -1, 1, 5, 5, 3, 1, 1, 3, 3, 3, -3, -1, -1, -1, 1

Offset: 1

Author

Bryle Morga, Sep 05 2024

Keywords

• sign
• look
• tabl

Comments

Start with M_0 = [[1]]. M_n is a 2^n X 2^n matrix. M_{n+1} is constructed from M_n as follows:

| M_n A(M_n) |

M_{n+1} = | |

| B(M_n) C(M_n) |

A - adds the corresponding x-coords to the elements of the matrix

B - subtracts the corresponding x-coords to the elements of the matrix

C - adds the corresponding y-coords to the elements of the matrix

The coordinate system sets the top-left corner to be (0, 0).

Then we take the limiting matrix {M_n}, and turn it into an integer sequence by reading it by antidiagonals.

Examples

			M_0 = [1] by definition. Constructing M_1 goes as follows:
A(M_0) = M_0 + [0] = [1]
B(M_0) = M_0 - [0] = [1]
C(M_0) = M_0 + [0] = [1]
So we have:
            |  1    1 |
M_1 =       |  1    1 |
From this M_2 can be constructed:
A(M_2) = M_1+[[0, 1],[0, 1]] = [[1, 2], [1, 2]]
B(M_2) = M_1-[[0, 1],[0, 1]] = [[1, 0], [1, 0]]
C(M_2) = M_1+[[0, 0],[1, 1]] = [[1, 1], [2, 2]]
      | 1 1 1 2 |
      | 1 1 1 2 |
M_2 = | 1 0 1 1 |
      | 1 0 2 2 |
		

Links

• Bryle Morga, Table of n, a(n) for n = 1..10000
• Bryle Morga, Visualization of the first 2 million terms.

Programs

Python

def expand(m):
  i = len(m)
  res = [[0 for  in range(2*i)] for  in range(2*i)]
  for x in range(i):
    for y in range(i):
      res[y][x] = m[y][x]
      res[y][x+i] = m[y][x] + x
      res[y+i][x] = m[y][x] - x
      res[y+i][x+i] = m[y][x] + y
  return res
a = []
m = [[1]]
for _ in range(11):
  m = expand(m)
for i in range(len(m)):
  for j in range(i+1):
    a.append(m[j][i-j])