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A053479
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Circle numbers (version 6): a(n) = number of points (i+j/2,j*sqrt(3)/2), i,j integers (triangular grid) contained in a circle of diameter n, centered at (1/2, 1/(2*sqrt(3))).
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6
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0, 0, 3, 6, 12, 21, 30, 42, 54, 69, 90, 102, 129, 150, 174, 198, 225, 258, 288, 327, 354, 396, 435, 471, 522, 558, 609, 654, 702, 759, 807, 864, 924, 981, 1038, 1104, 1173, 1230, 1308, 1368, 1443, 1512, 1590, 1671, 1746, 1830, 1908, 2001, 2076, 2166, 2265
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OFFSET
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0,3
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COMMENTS
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In other words, number of points in a hexagonal lattice covered by a circular disk of diameter n if the center of the circle is chosen at the deep hole. - Hugo Pfoertner, Jan 07 2007
Also number of integer coordinate pairs (s,t) satisfying s^2+t^2+st-s-t <= n^2/4-1/3. The a(2)=3 coordinate pairs are (s,t)=(0,0), (0,1) and (1,0). The a(3)=6 coordinate pairs are (-1,1),(0,0),(0,1),(1,-1),(1,0) and (1,1). - R. J. Mathar, Feb 23 2007
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LINKS
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FORMULA
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a(n)/(n/2)^2 -> Pi*2/sqrt(3).
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MAPLE
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A053479 := proc(n) local res, a, b ; res :=0 ; for a from -n to n do for b from -n to n do if a^2+b^2+a*b-a-b <= n^2/4-1/3 then res := res+1 ; fi ; od ; od ; RETURN(res) ; end : for n from 1 to 40 do printf("%d ", A053479(n)) ; od ; # R. J. Mathar, Feb 23 2007
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MATHEMATICA
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cx = 1/2; cy = 1/(2*Sqrt[3]); a[n_] := Sum[ dj = (1/2)* Sqrt[Abs[-3*cx^2 + 2*Sqrt[3]*cx*cy - cy^2 + 6*cx*i - 2*Sqrt[3]*cy*i - 3*i^2 + n^2]]; j1 = cx/2 + (Sqrt[3]*cy)/2 - i/2 - dj // Floor ; j2 = cx/2 + (Sqrt[3]*cy)/2 - i/2 + dj // Ceiling; Sum[Boole[(i + j/2 - cx)^2 + (j*(Sqrt[3]/2) - cy)^2 <= n^2/4], {j, j1, j2}], {i, -(n + 1)/2 - 2 // Floor, (n + 1)/2 + 3 // Ceiling}]; Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Jun 06 2013 *)
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jan 14 2000
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EXTENSIONS
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STATUS
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approved
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