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

Mon Oct 19 18:37:15 CEST 2020

```I think it's probably true. To try to prove it, I would want to start by looking at the continued fraction for e.

-----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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