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A255250
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Array T(n, m) of numbers of points of a square lattice in the first octant covered by a circular disk of radius n (centered at any lattice point taken as origin) with ordinate y = m.
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2
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1, 2, 3, 1, 4, 2, 1, 5, 3, 2, 6, 4, 3, 2, 7, 5, 4, 3, 1, 8, 6, 5, 4, 2, 9, 7, 6, 5, 3, 2, 10, 8, 7, 6, 5, 3, 1, 11, 9, 8, 7, 6, 4, 3, 1, 12, 10, 9, 8, 7, 5, 4, 2, 13, 11, 10, 9, 8, 6, 5, 3, 1, 14, 12, 11, 10, 9, 8, 6, 4, 3, 1, 15, 13, 12, 11, 10, 9, 7, 6, 4, 2, 16, 14, 13, 12, 11, 10, 8, 7, 5, 4, 2
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OFFSET
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0,2
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COMMENTS
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The row length of this array (irregular triangle) is 1 + flpoor(n/sqrt(2)) = 1 + A049472(n) = 1, 1, 2, 3, 3, 4, 5, 5, 6, 7, 8, 8, 9, 10, 10, 11, ...
This entry is motivated by the proposal A255195 by Mats Granvik, who gave the first differences of this array.
See the MathWorld link on Gauss's circle problem.
The first octant of a square lattice (x, y) with n = x >= y = m >= 0 is considered. The number of lattice points in this octant covered by a circular disk of radius R = n around the origin having ordinate value y = m are denoted by T(n, m), for n >= 0 and m= 0, 1,..., floor(n/sqrt(2)).
The row sums give RS(n) = A036702(n), n >= 0. This is the total number of square lattice points in the first octant covered by a circular disk of radius R = n.
The alternating row sums give A256094(n), n >= 0.
The total number of square lattice points in the first quadrant covered by a circular disk of radius R = n is therefore 2*RS(n) - (1 + floor(n/sqrt(2))) = A000603(n).
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LINKS
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FORMULA
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T(n, m) = floor(sqrt(n^2 - m^2)) - (m-1), n >= 0, m = 0,1, ..., floor(n/sqrt(2)).
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EXAMPLE
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The array (irregular triangle) T(n, m) begins:
n\m 0 1 2 3 4 5 6 7 8 9 10 ....
0: 1
1: 2
2: 3 1
3: 4 2 1
4: 5 3 2
5: 6 4 3 2
6: 7 5 4 3 1
7: 8 6 5 4 2
8: 9 7 6 5 3 2
9: 10 8 7 6 5 3 1
10: 11 9 8 7 6 4 3 1
11: 12 10 9 8 7 5 4 2
12: 13 11 10 9 8 6 5 3 1
13: 14 12 11 10 9 8 6 4 3 1
14: 15 13 12 11 10 9 7 6 4 2
15: 16 14 13 12 11 10 8 7 5 4 2
...
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CROSSREFS
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KEYWORD
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nonn,easy,tabf
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AUTHOR
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STATUS
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approved
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