0,1

This is related to Ramanujan's surprising formula: Prod[from n = 1 to infinity] (prime(n)^2 - 1)/ (prime(n)^2 + 1) = 2/5 and we use it in finding A112407 as the semiprime analog. We also use: A090986 = Decimal expansion of Pi csch Pi = Prod[from n = 2 to infinity] (n^2 - 1)/(n^2 + 1).

Since every integer above 1 is a k-almost prime for some k, we factor the (n^2 - 1)/(n^2 + 1) infinite product, and use Ramanujan's formula, to have: Prod[from n = 1 to infinity] (prime(n)^2-1)/(prime(n)^2+1) * Prod[from n = 1 to infinity] (semiprime(n)^2 - 1)/(semiprime(n)^2 + 1) * Prod[from n = 1 to infinity] (3-almostprime(n)^2 - 1)/ (3-almostprime(n)^2 + 1) * ... * Prod[from n = 1 to infinity] (k-almostprime(n)^2 - 1)/ (k-almostprime(n)^2 + 1) * ... = Prod[from n = 2 to infinity] (n^2 - 1)/(n^2 + 1) = pi csch pi as each integer appear once and only once in numerator and once and only once in denominator.

2/5 is the first (Ramanujan, prime) term in this infinite product of infinite products. A112407 is the second (semiprime) term. A114259 is the third (3-almost prime) term. A114260 is the fourth (4-almost prime) term. All of these have slow convergence.

Robert G. Wilson v writes: <10^7 p = 0.6599827269969193328259200536514893540328393737545363456803187594 <10^8 p = 0.6599826953336500842319051496177348200173865913706990918964875494 "So you see that brute force convergence is not very good or fast. The last term took 3/2 hours. I think that we can safely say that 0.659982 is correct. Everything after that is suspect."

Borwein, J.; Bailey, D.; and Girgensohn, R. "Two Products." Section 1.2 in Experimentation in Mathematics: Computational Paths to Discovery. Natick, MA: A. K. Peters, pp. 4-7, 2004.

Eric Weisstein's World of Mathematics, Infinite Product.

Eric Weisstein's World of Mathematics, Hyperbolic Cosecant

a(n) = decimal expansion of Prod[from n = 1 to infinity] (semiprime(n)^2 - 1)/(semiprime(n)^2 + 1). a(n) = decimal expansion of Prod[from n = 1 to infinity] (A001358(n)^2 - 1)/(A001358(n)^2 + 1).

0.659982...

Robert G. Wilson v (rgwv(AT)rgwv.com): spQ[n_] := Plus @@ Last /@ FactorInteger@n == 2; p = 1; Do[If[spQ[n], p = N[p*(n^2 - 1)/(n^2 + 2), 64]], {n, 10}]; p

Cf. A001358, A090986, A114529-A114536.

Sequence in context: A011284 A073230 A134881 this_sequence A046615 A103132 A046627

Adjacent sequences: A112404 A112405 A112406 this_sequence A112408 A112409 A112410

cons,nonn,uned

Jonathan Vos Post (jvospost2(AT)yahoo.com), Dec 21 2005