Hello, On Thu, Aug 01, 2019 at 11:17:30PM +1200, Richard O'Keefe wrote:
2(N-1)/N = 2 - 2/N. So one way to get exactly that mean is to make all the numbers 2 except for two of them which are 1.N < 2 : can't be done. N = 2 : only [1,1] does the job. N = 3 : the sum of the three numbers must be 4, so none of them can be 3, so [1,1,2] [1,2,1] [2,1,1] are the only possibilities. N = 4 : [1,1,1,3] [1,1,2,2] [1,1,3,1] [1,2,1,2] [1,2,2,1] [1,3,1,1] [2,1,1,2] [2,1,2,1] [2,2,1,1] [3,1,1,1] Is there a pattern here? Yes. There must be an integer k such that 0 <= 2k <= N-2 and then you have a rearrangement of (k+2) 1s, (N-2-2k) 2s, and k 3s.
This is what I did by solving a set of equations n1+n2+n3=N (i.e. the number of 1, 2 and 3 equals N, the length of my sample) and n2+2*n2+3*n3=2*(N-1) without delving into the details, you are composing (fractal) structures of N spheres in 3D, with an algorithm which establishes that every sphere has on average (2*N-1)/N neighbours. One solution is a simple chain, without bifurcations, where every sphere has two neighbours apart from the two terminal spheres, each of which has just one neighbour. Whenever a bifurcation arises, one sphere will have three neighbors. L. ______________________________________________ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
