tag:blogger.com,1999:blog-264226589944705290.post8746880986570482531..comments2023-11-05T03:45:25.001-08:00Comments on God Plays Dice: Playing around with sums of squaresMichael Lugohttp://www.blogger.com/profile/15671307315028242949noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-264226589944705290.post-87561510464043148432010-01-25T23:06:21.282-08:002010-01-25T23:06:21.282-08:00ftlus insults eliminated tours mozambique equipmen...ftlus insults eliminated tours mozambique equipment receipts families kinship joplin studios <br />servimundos melifermulyAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-42749844071651523392008-02-02T22:03:00.000-08:002008-02-02T22:03:00.000-08:00Google "Leech Lattice", a very neat lattice in 24 ...Google "Leech Lattice", a very neat lattice in 24 dimensions that has some amazing properties.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-37361359357896447572007-12-18T02:43:00.000-08:002007-12-18T02:43:00.000-08:00Proof sketch:The idea is to check whether n(n+1)(2...Proof sketch:<BR/>The idea is to check whether n(n+1)(2n+1)/6 is 0 modulu p^2 for every prime p. This is similar to the analysis of the probability of an integer to be squarefree, which is \prod_p (1-1/p^2) = 6/pi^2<BR/><BR/>So, for all prime p>3 we get (1-3/p^2) because there are exactly 3 values of n mod p^2 yielding 0, which are 0, -1 and -1/2. For p=3 we need to work modulo 27 instead because we lose a factor of 3 when dividing by 6, so we get (1-3/27). For 2 we work modulo 8 so there's no -1/2 so we get (1-2/8).<BR/><BR/>I guess that this number is indeed transcendental, though I didn't actually bother to prove it.Ori Gurel-Gurevichhttps://www.blogger.com/profile/03570318768726906432noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-90209025238581968992007-12-16T16:13:00.000-08:002007-12-16T16:13:00.000-08:00Hey ori,Any chance of a proof?Hey ori,<BR/>Any chance of a proof?I. J. Kennedyhttps://www.blogger.com/profile/04805435564360543720noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-67200459326619351932007-12-16T14:42:00.000-08:002007-12-16T14:42:00.000-08:00Actually, I believe that c=(2/3)*(\prod_{p>3, p pr...Actually, I believe that c=(2/3)*(\prod_{p>3, p prime} (1-3/p^2)) =~0.501948Ori Gurel-Gurevichhttps://www.blogger.com/profile/03570318768726906432noreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-58976004527103764132007-12-15T13:45:00.000-08:002007-12-15T13:45:00.000-08:00It looks like there are no other solutions : http:...It looks like there are no other solutions : <BR/><A HREF="http://www.daviddarling.info/encyclopedia/C/Cannonball_Problem.html" REL="nofollow">http://www.daviddarling.info/encyclopedia/C/Cannonball_Problem.html</A>Anonymousnoreply@blogger.com