[seqfan] Re: Interesting sequence

Frank Adams-Watters franktaw at netscape.net
Tue Nov 22 03:30:41 CET 2016

If we ignore the floors, the result will be (n+1)!/(n-1)! = n*(n+1), so this is an upper bound for a(n).

Franklin T. Adams-Watters

-----Original Message-----
From: David Wilson <davidwwilson at comcast.net>
To: 'Sequence Fanatics Discussion list' <seqfan at list.seqfan.eu>
Sent: Mon, Nov 21, 2016 7:40 pm
Subject: [seqfan] Interesting sequence

Starting with integer n,  multiply by ((n+1)/n), take the floor, multiply by(n/(n-1), take the floor, all the way down to 2/1, call the result f(n).For example, starting with n = 5floor(5*(6/5)) = 6,floor(6*(5/4)) = 7,floor(7*(4/3)) = 9,floor(9*(3/2)) = 13,floor(14*(2/1)) = 26.so f(5) = 26.Starting at n = 1, we havef = (2, 6,  12, 18, 26, 38, 48, 62, 78, 90, ...)It's trivial that all elements are even, given the final multiplier 2/1.It looks to me as if f(n) ~ pi*n^2/4, but I couldn't begin to prove this.--Seqfan Mailing list - http://list.seqfan.eu/

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