Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: $Revision: 1.6.2.16 $; site ada-uts.UUCP Path: utzoo!watmath!clyde!burl!ulysses!ucbvax!decvax!yale!ada-uts!az From: az@ada-uts.UUCP Newsgroups: net.math Subject: Re: Re: resistor cube and N dim cubes Message-ID: <16100004@ada-uts.UUCP> Date: Wed, 30-Oct-85 10:58:00 EST Article-I.D.: ada-uts.16100004 Posted: Wed Oct 30 10:58:00 1985 Date-Received: Sat, 2-Nov-85 07:24:40 EST References: <40@birtch.UUCP> Lines: 25 Nf-ID: #R:birtch:-4000:ada-uts:16100004:000:1268 Nf-From: ada-uts!az Oct 30 10:58:00 1985 The only formula for n-dimensional cube I could find is: the sum for k from 0 to (n-1) of k!(n-k-1)!/n! (I could not simplify it any further.) A sketchy proof: for any given k (0<=k<=n) the cube has n!/(k!(n-k)!) of vertices which are k edges from the "leftmost" vertex. Let's call this vetices k-vertices. There are (n-k) edges going from any given k-vertex to k+1-vertex. After we merge equipotential vertices, corresponding "bunch" has resistence (k!(n-k-1)!/n! Ohm. Just in case anybody is interested, first 40 values follow: n=1, r=1 n=11, r=.206782 n=21, r=.100616 n=31, r=.066834 n=2, r=1 n=12, r=.186724 n=22, r=.095762 n=32, r=.064667 n=3, r=.833333 n=13, r=.170285 n=23, r=.091359 n=33, r=.062636 n=4, r=.666667 n=14, r=.156571 n=24, r=.087346 n=34, r=.06073 n=5, r=.533333 n=15, r=.144952 n=25, r=.083673 n=35, r=.058936 n=6, r=.433333 n=16, r=.134976 n=26, r=.080298 n=36, r=.057246 n=7, r=.359524 n=17, r=.126311 n=27, r=.077186 n=37, r=.05565 n=8, r=.304762 n=18, r=.118711 n=28, r=.074307 n=38, r=.054141 n=9, r=.263492 n=19, r=.111987 n=29, r=.071636 n=39, r=.052711 n=10, r=.231746 n=20, r=.105993 n=30, r=.069151 n=40, r=.051356 Alex Zatsman, Intermetrics,Inc., Cambridge, Mass.