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%I #11 Feb 15 2019 14:58:35
%S 1,2,4,7,10,14,19,24,30,36,43,49,58,66,75,85,95,105,116,128,139,152,
%T 164,178,193,206,222,236,251,268,285,302,318,338,357,377,395,416,437,
%U 457,478,501,522,547,569,591,617,641,667,691,717,746,771,799,827,856,885,914,943,974,1004,1034,1067
%N The second row of the order of square grid cells touched by a circle expanding from the middle of a cell.
%C Related to, but not the same as the case with the circle centered at the corner of a cell, see A232499.
%H Rok Cestnik, <a href="/A323621/a323621_1.gif">Visualization</a>
%o (Python)
%o N = 12
%o from math import sqrt
%o # the distance to the edge of each cell
%o edges = [[-1 for j in range(N)] for i in range(N)]
%o edges[0][0] = 0
%o for i in range(1,N):
%o edges[i][0] = i-0.5
%o edges[0][i] = i-0.5
%o for i in range(1,N):
%o for j in range(1,N):
%o edges[i][j] = sqrt((i-0.5)**2+(j-0.5)**2)
%o # the values of the distances
%o values = []
%o for i in range(N):
%o for j in range(N):
%o values.append(edges[i][j])
%o values = list(set(values))
%o values.sort()
%o # the cell order
%o board = [[-1 for j in range(N)] for i in range(N)]
%o count = 0
%o for v in values:
%o for i in range(N):
%o for j in range(N):
%o if(edges[i][j] == v):
%o board[i][j] = count
%o count += 1
%o # print out the sequence
%o for i in range(N):
%o print(str(board[i][1])+" ", end="")
%Y For the grid read by antidiagonals see A323621.
%Y For the first row of the grid see A323622.
%Y For the diagonal of the grid see A323624.
%Y For the (2,1) diagonal of the grid see A323625.
%Y Cf. A232499.
%K nonn
%O 0,2
%A _Rok Cestnik_, Jan 20 2019