tag:blogger.com,1999:blog-264226589944705290.post4437095302922870452..comments2023-09-29T02:57:42.471-07:00Comments on God Plays Dice: A little number theory problemMichael Lugohttp://www.blogger.com/profile/15671307315028242949noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-264226589944705290.post-18363299298130378412010-02-15T13:35:03.554-08:002010-02-15T13:35:03.554-08:00buy bactrim without prescription buy bactrim onlin...buy bactrim without prescription buy bactrim online buy bactrim buy bactrim es online without prescription buy bactrim without a prescription overnight buy bactrim f <br />[url=http://bactrim.eventbrite.com/]buy bactrim without prescription [/url]<br />glucophage side effects glucophage metformin when to take glucophage doses glucophage side effect glucophage kidney pain glucophage recall how to take glucophage <br />[url=http://takeglucophage.eventbrite.com/]glucophage odor cause [/url]<br />proscar and propecia prostate cancer resistance to proscar generic for proscar proscar finasteride proscar pi proscar without a prescription proscar for hair loss <br />[url=http://proscar.eventbrite.com/]proscar and psa levels [/url]<br />impotence solutions levitra users how does levitra work impotence drug name acquisto levitra levitra buy generic levitra <br />[url=http://virb.com/yalevi]levitra website [/url]<br />http://proscar.eventbrite.com/Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-83512769442088563562008-05-26T21:25:00.000-07:002008-05-26T21:25:00.000-07:00The missing link appears to be the von on Staudt-C...The missing link appears to be the von on Staudt-Clausen theorem, namely that the denominator of the a-th Bernoulli number is equal to the product of the primes dividing a-1.<BR/><BR/>If we work one prime at a time in David's argument: For q an odd prime, then q^b divides the gcd if and only if (q-1)*q^(b-1) divides a. This is since we know explicitly (q-1)*q^(b-1)=lambda(q^b).<BR/><BR/>Using the above quoted theorem, this is equivalent to q^b dividing the denominator of B_a/(2a). To complete the desired result, we have to do a similar analysis at the prime 2, which is slightly different since lambda(2^m)=2^(m-2) for m>2, but similar.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-44997520023636381192008-05-26T16:01:00.000-07:002008-05-26T16:01:00.000-07:00Peter J. Cameron's comment, on Sloane's table, alm...Peter J. Cameron's comment, on Sloane's table, almost proves the desired result. Let me spell this out a little. Let G be the GCD of p^n-1, where p runs over all sufficiently large primes.<BR/><BR/>Now, the condition<BR/><BR/>(0) m divides G <BR/><BR/>is equivalent to<BR/><BR/>(1) m divides p^n-1, for all sufficiently large p<BR/><BR/>is equivalent to<BR/><BR/>(2) for every unit u in the ring Z/m, we have u^n=1.<BR/><BR/>This latter is because, by Dirichlet's theorem, every unit in Z/m occurs as the image of some prime in Z. Carmichael defines lambda(m) to be the exponent of (Z/m)^*. In other words, (2) is equivalent to <BR/><BR/>(3) lambda(m) divides n.<BR/><BR/>Now, Cameron's comment is that (3) is equivalent to <BR/><BR/>(4) m divides lambda^*(n), where lambda^*(n) is the sequence defined from denominators of Bernouli numbers.<BR/><BR/>So m divides G if and only if m divides lambda^*(n) and thus G=lambda^*(n). In short, the GCD of p^n-1, where p ranges over sufficiently large primes, is lambda^*(n).<BR/><BR/>The first thing that worries me is that it is not clear to me whether just taking p to be larger than n is sufficiently large. I also don't know why Cameron's comment is true.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-264226589944705290.post-79236261032451686162008-05-19T02:47:00.000-07:002008-05-19T02:47:00.000-07:00You should ask Tanya Khovanova, she likes puzzles ...You should ask <A HREF="http://www.tanyakhovanova.com/" REL="nofollow">Tanya Khovanova</A>, she likes puzzles like this and may know the answer, or figure it out.mishahttps://www.blogger.com/profile/01166708933155105921noreply@blogger.com