%I #37 May 02 2013 13:35:12
%S 0,1,2,3,6,9,12,16,21,26,32,38,45,52,61,69,78,88,99,110,121,133,146,
%T 159,173,188,203,218,234,251,268,286,305,324,343,364,384,406,428,450,
%U 473,497,521,546,571,597,624,651,679,707,736,765,795,826,857
%N Floor of the Euclidean distance of a point on the (1, 2, 3; 4, 5, 6) 3D walk.
%C Consider a standard 3-dimensional Euclidean lattice. We take 1 step along the positive x-axis, 2 along the positive y-axis, 3 along the positive z-axis, 4 along the positive x-axis, and so on. This sequence gives the floor of the Euclidean distance to the origin after n steps.
%C The (x,y,z) coordinates are (1,0,0), (1,2,0), (1,2,3), (5,2,3), (5,7,3), (5,7,9), (12,7,9) etc, where the x values run through A000326, the y-values through A005449, and the z-values through A045943. The squared Euclidean distances are s(n) = 1, 5, 14, 38, 83, 155, 274, 450,..., which obey the recurrence s(n) = 3*s(n-1) -3*s(n-2) +3*s(n-3) -6*s(n-4) +6*s(n-5) -3*s(n-6) +3*s(n-7) -3*s(n-8) +s(n-9), s(n) = (3*n^2+9*n+10)^2/108 +4*A099837(n+3)/27 -2*(-1)^n*A165202(n)/9, with a = floor(sqrt(s(n))). - _R. J. Mathar_, May 02 2013
%F a(n) ~ n^2 sqrt(3)/6. - _Charles R Greathouse IV_, May 02 2013
%e For a(4) we are at [5,2,3], so a(n) = floor(sqrt(25+4+9)) = 6.
%o (JavaScript)
%o p=new Array(0,0,0);
%o for (a=0;a<100;a++) {
%o p[a%3]+=a;
%o document.write(Math.floor(Math.sqrt(p[0]*p[0]+p[1]*p[1]+p[2]*p[2]))+", ");
%o }
%Y Cf. A054925, A224985, A225215.
%K nonn
%O 0,3
%A _Jon Perry_, Apr 14 2013