IIRC, the definition of "irrationality measure" M for X uses the expression | X - n/d | > 1/d^M except for a finite number of ds. So the digits n through 7.61n (of pi) can't be all 9s (or all 0s), with only a finite number of exceptional n. At a symbolic math conference in NYC ~1985, the Chudnovsky's offered a good result for pi^2, perhaps 6 or 7, and remarked that that implied a less good result for pi, perhaps 12-14. Note that the ordinary continued fraction algorithm gives an infinite stream of d when M=2, for any irrational X. So we'd need something a little stronger to address Jim's original question. And Thue-Seigel-Roth says that M<2+epsilon for (irrational, real) algebraic X. Rich ________________________________________ From: math-fun-bounces@mailman.xmission.com [math-fun-bounces@mailman.xmission.com] on behalf of Michael Kleber [michael.kleber@gmail.com] Sent: Thursday, March 20, 2014 8:20 AM To: math-fun Subject: [EXTERNAL] Re: [math-fun] Nines in Pi The irrationality measure of pi is at most 7.6063 [Salikhov, "On the Irrationality Measure of $\pi$", 2008]. I think that translates into "There are at most finitely many n for which digits n through k*n are all 9's" for some k, but I don't know what k is. But none of this answers Jim's / Dick's question (whose answer is surely "it never happens but we don't have a proof"). --Michael On Thu, Mar 20, 2014 at 9:34 AM, James Propp <jamespropp@gmail.com> wrote:
Is it possible that for some n, the n+1st through 2nth digits of pi are all 9's?
Dick Hess just gave a wonderful G4G talk which raised this issue.
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