[seqfan] Re: ceiling inequality

Bob Selcoe rselcoe at entouchonline.net
Wed Nov 5 10:58:19 CET 2014


Yes, it does seem likely true.

Simplifying Clark's inequality we get ceiling(n/sqrt(3)) > n^2/sqrt(3n^2-5),
n>1.  As n increases, it approaches ceiling(n/sqrt(3)) >
n/sqrt(3), which obviously is true since sqrt(3) is irrational.  So the
inequality seems more likely (though not necessarily) to hold as n
increases.

The question becomes: is ceiling(n/sqrt(3) > n^2/sqrt(3n^2-5) for n>1?  If
so, Clark's inequality holds.

As Charles points out, potential problems arise when n/sqrt(3) is "close" to
(and less than) an integer.  Charles' insight for finding integers "close to
n" is:  n = a(m) = 4a(m-1) - a(m-2); a(0)=1, a(1)=5, a(2)=19.   Let's call 
these
"Charles numbers" (C(m)), and for simplicity denote C(m) as k and C(m+1) as 
k+1.

Unless I'm mistaken, it can be shown that the inequality holds if the
following two propositions hold:

Prop 1:  k/sqrt(3) is the minimum value of ceiling(n/sqrt(3)) -
n/sqrt(3) for all n<=k, (i.e., k is "closest" to ceiling(n/sqrt(3)) compared
with all smaller n); and

Prop 2:  ceiling(k/sqrt(3)) - k^2/sqrt(3k^2-5) is positive for k>=1.

One way to show that Prop 2 holds is if the following formula is asymptotic
and positive:

sqrt(3k^2-5)*(sqrt(3(k+1)^2-5)*ceiling((k+1)/sqrt(3)) - (k+1)^2) /
sqrt(3(k+1)^2-5)*(sqrt(3k^2-5)*ceiling(k/sqrt(3)) - k^2).

This is the ratio of ceiling((k+1)/sqrt(3) - (k+1)^2/sqrt(3(k+1)^2-5) to
ceiling(k/sqrt(3)) - k^2/sqrt(3k^2-5), and it appears that it approaches 
1/(2+sqrt(3)) (~
0.26794919243) and therefore is positive.

To illustrate, let k' = ceiling(k/sqrt(3)) - k^2/sqrt(3k^2-5) and k" be the
result of Eq.1. for k (i.e., (k+1)'/k' with slight rounding improvements):

k=1 == n=5.  k' ~ 0.01102848;  k" ~ 0.44743111
k=2 == n=19.  k' ~ 0.004934485;  k" ~ 0.274151198
k=3 == n=71.  k' ~ 0.001352795;  k" ~ 0.26838176
k=4 == n=265.  k' ~ 0.000363066;  k" ~ 0.26798019
k=5 == n=989.  k' ~ 0.000097294;  k" ~ 0.26795142

(done by hand; pardon any potential typos or calculation errors).

I imagine the two propositions are provable; but if not, then perhaps
there's a way to find maximum j such that ceiling(n/sqrt(3) > 
n^2/sqrt(3n^2-j) for n>1.  If j>=5, then the
inequality holds.

Cheers,
Bob Selcoe

--------------------------------------------------
From: "Charles Greathouse" <charles.greathouse at case.edu>
Sent: Monday, November 03, 2014 9:36 AM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: ceiling inequality

> We (potentially) run into trouble where n/sqrt(3) falls just short of an
> integer. Using the continued fraction of sqrt(3) I'd check 5, 19, 71, 265,
> ... (where each term is 4 times the previous minus the one before that).
> Up
> to 15,000 digits these hold, so it seems likely that this is true. Maybe
> some effective form of Roth's theorem can be used to prove it.
>
> Charles Greathouse
> Analyst/Programmer
> Case Western Reserve University
>
> On Mon, Nov 3, 2014 at 10:03 AM, Kimberling, Clark <ck6 at evansville.edu>
> wrote:
>
>>
>> Hello SeqFans,
>>
>> It appears that  (3n^2 - 5)*ceiling(n/sqrt(3))^2 > n^4      for all n >
>> 1.
>>
>> Can someone prove this are cite something?
>>
>> Thanks.
>>
>> Clark Kimberling
>>
>>
>>
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>>
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
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