[seqfan] Re: Maximizing product for a fixed sum n

Brendan McKay Brendan.McKay at anu.edu.au
Tue Oct 20 07:20:29 CEST 2020

For the case of a sum of real numbers, if there is an integer m
in the interval ( n/e+1/2, n/e+1/2+e/(24n)+e^3/(1920n^3) )
then (n/m)^m > (n/(m-1))^(m-1) even though m is further from n/e.

However I can't find any n with an integer in that range.  I tried
up to 10^11 but some rounding error could have caused me to
miss one.

As Franklin says, the continued fraction of e is relevant.


On 20/10/20 3:37 am, Frank Adams-watters via SeqFan wrote:
> I think it's probably true. To try to prove it, I would want to start by looking at the continued fraction for e.
> Franklin T. Adams-Watters
> -----Original Message-----
> From: Jack Grahl <jack.grahl at gmail.com>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Sent: Sun, Oct 18, 2020 4:31 am
> Subject: [seqfan] Maximizing product for a fixed sum n
> Dear all,
> Suppose I wish to find some set of positive numbers a1,a2...,ak such that
> their sum is n, and their product is as large as possible. Clearly, if I
> fix the cardinality k, the optimal solution is k copies of n/k. What is the
> best choice of k? In other words, what is the natural number k which
> maximises (n/k)^k?
> Calculus shows that, if we replace integer k by real number x, x = n/e is
> optimal, and hence than either k = floor(n/e) or k = ceil(n/e) is the
> optimal integer solution, with the choice of floor and ceiling possibly
> depending on n.
> Experimentation shows that in fact the best k is the closest integer to n/e
> for all 2 <= n <= 1000. (n = 4 is slightly exceptional, as the closest
> integer to 4/e is 1, but there is a tie between 1 and 2 for the optimal
> value of k, using the well known result that 2*2 = 4.)
> 1. Can anyone show that the closest integer to n/e must always be optimal?
> 2. If so, we could add the sequence 'closest integer to n/e' with a remark
> that it solves the above problem. If not, we could add both sequences, with
> a remark that they are conjectured to be the same.
> Best wishes,
> Jack Grahl
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