this might be close: http://users.telenet.be/Wouter.Meeussen/InfiniteProducts.txt Wouter Meeussen Senior Scientist R&D dept. NV Vandemoortele Izegem tel: +32 (0)51 33 21 24 fax: +32 (0)51 33 21 75 -----Original Message----- From: math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Simon Plouffe Sent: donderdag 1 september 2011 16:31 To: math-fun Subject: [math-fun] euler's infinite product, primes of the form 4n+1, 4n+3... ?? Hello, I am working on a simple problem which is the following, consider the Euler product with primes, infinity --------' ' | | 1 | | -------- = Zeta(2) | | 1 | | 1 - ---- p = 2 2 p this is well known. Now if we take only the primes of the form 4n+1 and 4n+3 and split (avoiding the number 2), then the product is p1*p3 = Pi^2/8 , p1 is the Euler product with primes of the form 4n+1 and p3 = with the primes of the form 4n+3. Now these 2 values are approximately : p1 = 1.05618212168678504596905 and p3 = 1.16807558536624241169267 The product of p1*p3 being as expected Pi^2/8. the good question is : What is p1 made of actualy ? and p3 ? I tried all the tricks I know and came with NO answer at all. My first guess was a simple rational combination : this is wrong, my second guess was that p1 is a algebraic multiple of Pi : wrong too. And then log(p1) is related to some log(Gamma(a/b)) values ? There are no numerical evidence of that, then I tried all the tables, all the algorithms against that number and my quest was incorrect. Does anybody knows what are these numbers ? Is this a trivial question related to Dirichlet series, I don't get it. The values are correct to about 8 digits. I also tried with the exponent being, 3,4 and 8 : nothing too, a complete mystery. Any clue would help, thanks in advance. My gut feeling is that p1 should be simple (as well as p3 ). Bonne journée à tous. Simon Plouffe _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun =============================== This email is confidential and intended solely for the use of the individual to whom it is addressed. If you are not the intended recipient, be advised that you have received this email in error and that any use, dissemination, forwarding, printing, or copying of this email is strictly prohibited. You are explicitly requested to notify the sender of this email that the intended recipient was not reached.