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From: rancid moth on 27 Jun 2010 03:43 product(k=1..m) Binomial(2k,k)^(8*(2k+1)) = ?
From: rancid moth on 27 Jun 2010 04:28 "rancid moth" <rancidmoth(a)yahoo.com> wrote in message news:i06vf1$2mg$1(a)speranza.aioe.org... > product(k=1..m) Binomial(2k,k)^(8*(2k+1)) = ? Actually i think i have it. i believe i can express it in terms of a finite product of Barnes-G functions. perhaps there is an easier way to obtain it though?
From: Robert Israel on 27 Jun 2010 19:01 "rancid moth" <rancidmoth(a)yahoo.com> writes: > product(k=1..m) Binomial(2k,k)^(8*(2k+1)) = ? Well, if p is a prime, by a theorem of Kummer the highest power of p in binomial(2k,k) is the number of times you "carry 1" in adding k + k in base p. If you call that C(k,p), your answer is product_p p^(8 sum_{k=1}^m (2k+1) C(k,p)) the product being over all primes <= 2m. -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
From: rancid moth on 27 Jun 2010 21:51 "Robert Israel" <israel(a)math.MyUniversitysInitials.ca> wrote in message news:rbisrael.20100627222655$02d7(a)news.acm.uiuc.edu... > "rancid moth" <rancidmoth(a)yahoo.com> writes: > >> product(k=1..m) Binomial(2k,k)^(8*(2k+1)) = ? > > Well, if p is a prime, by a theorem of Kummer the highest power of p in > binomial(2k,k) is the number of times you "carry 1" in adding k + k in > base p. > If you call that C(k,p), your answer is > > product_p p^(8 sum_{k=1}^m (2k+1) C(k,p)) > > the product being over all primes <= 2m. > -- > Robert Israel israel(a)math.MyUniversitysInitials.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada Thank you...that's the kind of thing i was after. Now I've just got to get my head around it. cheers
From: Rob Johnson on 27 Jun 2010 22:31
In article <rbisrael.20100627222655$02d7(a)news.acm.uiuc.edu>, Robert Israel <israel(a)math.MyUniversitysInitials.ca> wrote: >"rancid moth" <rancidmoth(a)yahoo.com> writes: > >> product(k=1..m) Binomial(2k,k)^(8*(2k+1)) = ? > >Well, if p is a prime, by a theorem of Kummer the highest power of p in >binomial(2k,k) is the number of times you "carry 1" in adding k + k in >base p. >If you call that C(k,p), your answer is > >product_p p^(8 sum_{k=1}^m (2k+1) C(k,p)) > >the product being over all primes <= 2m. The number of base p carries in adding k+k is (2 S_p(k) - S_p(2k))/p where S_p(k) is the sum of the base p digits of k. Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font |