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EXAMPLE
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In a two-dimensional vector format the optimal circles are given as the first two terms are the coordinates of the center, the third is the squared radius. If you don't see a particular n value then consider a larger one: for example, as a(9)=a(10), it is enough to give the circles only for n=10. Short Pari code to check these solutions: f(v,n)={a=matrix(n,n,i,j,0);L=length(v);for(i=1,L,x0=v[i][1];y0=v[i][2];r2=v[i][3];for(x=0,n-1,for(y=0,n-1,d2=(x-x0)^2+(y-y0)^2-r2;if(d2==0,a[x+1,y+1]=1)))); s=sum(i=1,n,sum(j=1,n,a[i,j]));print("Using ",L," circles, covered "s," points on the [0,",n-1"]^2 grid.")} Call for example f(v,8), where v is the vector listed for n=8.
n=2: 1 circle: v=[[1/2, 1/2, 1/2]];
n=4: 3 circles: v=[[3/2, 1/2, 5/2], [3/2, 3/2, 5/2], [3/2, 5/2, 5/2]];
n=5: 5 circles: v=[[3/2, 1/2, 5/2], [3/2, 3/2, 5/2], [1/2, 1/2, 25/2], [5/2, 5/2, 5/2], [13/6, 17/6, 85/18]];
n=6; 6 circles: v=[[3/2, 1/2, 5/2], [5/2, 2, 25/4], [5/2, 3, 25/4], [3/2, 9/2, 5/2], [7/2, 1/2, 5/2], [7/2, 9/2, 5/2]];
n=8: 8 circles: v=[[7/2, 7/2, 25/2], [5/2, 5/2, 25/2], [7/2, 7/2, 37/2], [5/2, 9/2, 25/2], [9/2, 5/2, 25/2], [9/2, 9/2, 25/2], [7/2, 7/2, 5/2], [7/2, 7/2, 1/2]];
n=10: 11 circles: v=[[9/2, 9/2, 25/2], [11/2, 11/2, 25/2], [7/2, 7/2, 25/2], [7/2, 11/2, 25/2], [11/2, 7/2, 25/2], [9/2, 9/2, 65/2], [9/2, 9/2, 53/2], [9/2, 9/2, 37/2], [9/2, 9/2, 5/2], [9/2, 1/2, 41/2], [9/2, 17/2, 41/2]];
n=11: 14 circles: v=[[13/2, 13/2, 25/2], [13/2, 9/2, 25/2], [11/2, 11/2, 25/2], [9/2, 13/2, 25/2], [9/2, 9/2, 25/2], [11/2, 11/2, 65/2], [11/2, 11/2, 5/2], [11/2, 11/2, 37/2], [9/2, 11/2, 65/2], [11/2, 9/2, 65/2], [15/2, 15/2, 125/2], [9/2, 9/2, 65/2], [11/2, -1/2, 61/2], [11/2, 21/2, 41/2]];
n=12: 15 circles: v=[[11/2, 11/2, 65/2], [13/2, 11/2, 65/2], [9/2, 11/2, 65/2], [15/2, 11/2, 25/2], [9/2, 11/2, 25/2], [15/2, 11/2, 65/2], [13/2, 11/2, 25/2], [11/2, 15/2, 13/2], [11/2, 7/2, 13/2], [7/2, 11/2, 65/2], [7/2, 11/2, 25/2], [11/2, 19/2, 65/2], [11/2, 3/2, 65/2], [103/10, 11/2, 1517/50], [7/10, 11/2, 1517/50]];
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