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

0,2

Comments

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

Links

Table of n, a(n) for n=0..71.

Rok Cestnik, Visualization

Prog

(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="")

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.

Sequence in context: A008052 A016438 A038427 * A301096 A115067 A298787

Adjacent sequences: A323622 A323623 A323624 * A323626 A323627 A323628

Keyword

nonn

Author

Rok Cestnik, Jan 20 2019

Status

approved