[seqfan] Re: A214686
Vladimir Shevelev
shevelev at bgu.ac.il
Fri Jul 27 16:17:46 CEST 2012
Consider the sequence: b(n) is the greatest integer x<=(n-1)^2 such that gcd(x,n!)=1. It is clear that b(n) is the maximal prime <=(n-1)^2-2. If n!(1-sum{j=2,...,n-1}a(j)/j!)<=(n-1)^2, then, evidently,
a(n)<=b(n)<=(n-1)^2-2. If there exists an exceptional set E={n: n!(1-sum{j=2,...,n-1}a(j)/j!)>=
(n-1)^2+1}, then we always have a(n)<=(n-1)^2-2 iff in every interval [(n-1)^2+1, n!(1-sum{j=2,...,n-1}a(j)/j!)] all numbers from E are not respectively prime to n! I think that the natural conjecture is that E is empty for n>=3.
Regards,
Vladimir
----- Original Message -----
From: israel at math.ubc.ca
Date: Friday, July 27, 2012 4:10
Subject: [seqfan] A214686
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> I recently submitted a new sequence, A214686. This is defined so
> that, for
> n >= 2, a(n) is the greatest integer x such that gcd(x,n!) = 1
> and x/n! < 1
> - sum_{j=2}^{n-1} a(j)/j!. It is not hard to show that
> sum_{j=2}^infinity
> a(j)/j! = 1: my proof uses the fact that for any epsilon > 0, if
> $n$ is
> large enough there is a prime p with n < epsilon n!/2 < p
> < epsilon n!.
> This came up in a discussion on MathOverflow.net
> http://mathoverflow.net/questions/103129/irrationality-proof-
> technique-no-factorial-in-the-denominator/103136#103136
> as a counterexample to the conjectured rationality of sums of
> convergent
> series with terms of the form c(n)/n! in lowest terms.
>
> Since gcd(a(n),n!) = 1, for each n we have either a(n) = 1 or
> a(n) > n, and
> indeed both of these cases are well represented in the data. An
> interesting
> pattern came to light when I plotted the terms. Most of those
> a(n) > n are
> not too much bigger than n, maybe up to 40000 when n <= 1000.
> But for those
> n >= 6 for which a(n-1) = 1 (of which there are 166 up to
> n=1000), a(n)
> seems to be very close to (n-1)^2. Can anyone explain this pattern?
>
> Robert Israel
> University of British Columbia
>
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>
Shevelev Vladimir
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