A323625 The (2,1) diagonal of the order of square grid cells touched by a circle expanding from the middle of a cell.

0, 4, 11, 21, 34, 49, 68, 90, 112, 137, 167, 198, 230, 265, 305, 345, 388, 432, 480, 529, 583, 635, 692, 752, 812, 876, 941, 1010, 1079, 1151, 1225, 1305, 1383, 1462, 1545, 1630, 1720, 1811, 1901, 1995, 2092, 2190, 2287, 2391, 2499, 2606, 2715, 2827, 2941, 3056, 3174, 3295, 3421, 3541, 3668, 3792, 3923, 4058, 4193, 4333, 4466, 4609, 4754, 4899, 5042, 5194, 5344, 5498, 5654, 5813, 5972, 6133

Offset: 0

Author

Rok Cestnik, Jan 20 2019

Keywords

• nonn

Comments

Related to, but not the same as the case with the circle centered at the corner of a cell, see A232499.

Links

• Rok Cestnik, Visualization

Crossrefs

For the grid read by antidiagonals see A323621.

For the first row of the grid see A323622.

For the second row of the grid see A323623.

For the (1,1) diagonal of the grid see A323624.

Cf. A232499.

Programs

Python

N = 24
from math import sqrt
# the distance to the edge of each cell
edges = [[-1 for j in range(N)] for i in range(N)]
edges[0][0] = 0
for i in range(1,N):
    edges[i][0] = i-0.5
    edges[0][i] = i-0.5
for i in range(1,N):
    for j in range(1,N):
        edges[i][j] = sqrt((i-0.5)**2+(j-0.5)**2)
# the values of the distances
values = []
for i in range(N):
    for j in range(N):
        values.append(edges[i][j])
values = list(set(values))
values.sort()
# the cell order
board = [[-1 for j in range(N)] for i in range(N)]
count = 0
for v in values:
    for i in range(N):
        for j in range(N):
            if(edges[i][j] == v):
                board[i][j] = count
    count += 1
# print out the sequence
for i in range(int(round(N/2))):
    print(str(board[2*i][i])+" ", end="")