Warren, That's the iteration notation used by Mathematica. In particular, the iterator {n, 1, k, 8} means "for all n <= k of the form 1 + 8t, where t is a non-negative integer". I constructed the identity by beginning with the Borwein-like integral: I(k) := Integrate[Sinc[x] Product[If[PrimeQ[n], Sinc[x/n], 1], {n, 1, k, 8}], {x, 0, Infinity}] where I take the product of sinc(x/n) := sin(x/n) / (x/n) for all prime n <= k of the form 8t + 1. This integral evaluates to exactly pi/2 if and only if the sum of 1/n is bounded above by 1, which happens whenever k <= 10^2000 (relying on the doubly-logarithmic nature of the prime harmonic function). Then, I took the sum of 2^-k I(k) for all non-negative integers k, resulting in the sum: pi/2 + pi/4 + pi/8 + pi/16 + pi/32 + ... + pi/2^(10^2000) + ... + pi/2^k + (something slightly less than pi/2^(k+1)) + ... which almost equals pi. Finally, in an attempt to obfuscate the Borwein integrals slightly, I appealed to Fubini's theorem and swapped the order of integration and summation to give the identity: Integrate[f(x), {x, 0, Infinity}] ~= pi/2 Best wishes, Adam P. Goucher

Sent: Friday, October 06, 2017 at 10:49 PM From: "Warren D Smith" <warren.wds@gmail.com> To: apgoucher@gmx.com Cc: "Bill Gosper" <billgosper@gmail.com> Subject: Re: [math-fun] Borweins' high precision fraud

On 10/6/17, Adam P. Goucher <apgoucher@gmx.com> wrote:

...

Bill,

Define the function f(x) as follows:

f(x) := Sum[2^-k Sinc[x] Product[ If[PrimeQ[n], Sinc[x/n], 1], {n, 1, k, 8}], {k, 0, Infinity}]

Then the integral between 0 and infinity of f(x) agrees to pi for slightly more than 10^2000 digits.

--wow!! Now that is a fraud. But unfortunately I do not understand your notation. E.g. Product[ GoucherFunction[n], {n, 1, k, 8}] means what?

-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)