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The mean of X(2X-n)^2 + Y(2Y-n)^2 is 2n, or A005843.
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The mean of X^2 + Y^2 is 2n, or A005843.
So the squared-distance is 2 with probability 36/64, 10 with probability 24/16, 64, and 18 with probability 4/64; the median squared-distance is therefore 2.
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For n=3 the probabilities of ending up at the lattice points in [-3,3]x[-3,3] are 1/64 of:1 0 3 0 3 0 10 0 0 0 0 0 03 0 9 0 9 0 30 0 0 0 0 0 03 0 9 0 9 0 30 0 0 0 0 0 01 0 3 0 3 0 1So the squared-distance is 2 with probability 36/64, 10 with probability 24/16, and 18 with probability 4/64; the median squared-distance is therefore 2.
For n=3 the probabilities of ending up at the lattice points in [-3,3]x[-3,3] are 1/64 of:
1 0 3 0 3 0 1
0 0 0 0 0 0 0
3 0 9 0 9 0 3
0 0 0 0 0 0 0
3 0 9 0 9 0 3
0 0 0 0 0 0 0
1 0 3 0 3 0 1
So the squared-distance is 2 with probability 36/64, 10 with probability 24/16, and 18 with probability 4/64; the median squared-distance is therefore 2.
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A continuous analog draws each move from N(0,1) rather than from {+1,-1}, so the final x- and y- coordinates are distributed as N(0,Sqrtsqrt(n)). Then the final point has probability 1 - exp(-r^2/2n) of being within r of the origin, and the median squared-distance for this continuous analog is n lnlog(4). We also observe empirically that for this discrete sequence, a(n)/n approaches lnlog(4).
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