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%I #11 Feb 15 2019 14:58:44
%S 0,4,11,21,34,49,68,90,112,137,167,198,230,265,305,345,388,432,480,
%T 529,583,635,692,752,812,876,941,1010,1079,1151,1225,1305,1383,1462,
%U 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
%N The (2,1) diagonal 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 = 24
%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(int(round(N/2))):
%o print(str(board[2*i][i])+" ", end="")
%Y For the grid read by antidiagonals see A323621.
%Y For the first row of the grid see A323622.
%Y For the second row of the grid see A323623.
%Y For the (1,1) diagonal of the grid see A323624.
%Y Cf. A232499.
%K nonn
%O 0,2
%A _Rok Cestnik_, Jan 20 2019
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