Exact Sums of Powers
Now:  23 Sep 2020 15:37 
Description  Submit An Entry  Progress 
Introduction
This event is not a contest. There is no standings page and there will be no prizes at the end. In fact, there will be no end.
This event is an ongoing collaborative effort to find good upper bounds for yettobediscovered terms of Sequence A030052 in the OnLine Encyclopedia of Integer Sequences.
Everything you need to know to participate is explained on this web page. In particular, you don't need to be familiar with Sequence A030052 or even with the OEIS.
The Goal
We want to find, for various values of n, the smallest integer s whose n^{th} power is a sum of two or more distinct smaller positive n^{th} powers. For example, for n = 7 the smallest such integer s is 40, and 40^{7} can be written as
1^{7} + 3^{7} + 5^{7} + 9^{7} + 12^{7} + 14^{7} + 16^{7} + 17^{7} + 18^{7} + 20^{7} + 21^{7} + 22^{7} + 25^{7} + 28^{7} + 39^{7}.
How to Submit
Paste your solutions into the appropriate box on the Submit page and click Submit Entry.
You can submit solutions in either of two formats: Verbose or Compact.
Verbose
The verbose format for submitting solutions is easily explained by example. If your solution is
15^{4} = 14^{4} + 9^{4} + 8^{4} + 6^{4} + 4^{4}
then submit
15^4 => {14,9,8,6,4}
Include spaces and line breaks anywhere you like (except within a number or within the arrow) to improve readability.
Compact
If your solution is
s^{n} = a_{1}^{n} + a_{2}^{n} + a_{3}^{n} + ... + a_{k}^{n}
then calculate
K = (2^{a1} + 2^{a2} + 2^{a3} + ... + 2^{ak}) / 2
and then, substituting the appropriate values for s, n and K, submit
s^n => K
.
So, for example, if your solution is
15^{4} = 14^{4} + 9^{4} + 8^{4} + 6^{4} + 4^{4}
then calculate
K = (2^{14} + 2^{9} + 2^{8} + 2^{6} + 2^{4}) / 2 = 2^{13} + 2^{8} + 2^{7} + 2^{5} + 2^{3} = 8192 + 256 + 128 + 32 + 8 = 8616
and submit
15^4 => 8616
As with the verbose format, include spaces and line breaks anywhere you like (except within a number or within the arrow) to improve readability.
Getting Your Questions Answered
First, check the FAQ section below. If you can't find the information you need there, send your question to the discussion group. If your question is of a personal nature, and not of general interest, send an email directly to Al Zimmermann.
The Discussion Group
If you think you might participate in the collaboration, you should join the
contest discussion group. You can join either by sending a blank email to
AZsPCs+subscribe@groups.io
or by visiting the AZsPCs group at
http://groups.io/g/AZsPCs.
The discussion group serves two purposes. First,
it allows contestants to ask for clarifications to the goal. Be aware that
these requests can result in changes to the goal, and the first place
those changes are announced is in the discussion group. Second, the discussion
group allows contestants to interact with each other regarding algorithms, programming
techniques, and anything else related to the collaboration.
Once you've joined the discussion group, you can post messages by emailing your messages to
AZsPCs@groups.io.
You can leave the group at any time by sending a blank email to
AZsPCs+unsubscribe@groups.io.
Frequently Asked Questions

Can teams participate?
Everyone who participates is a member of the same team. This is a collaboration, not a contest. Subteams would be meaningless.

May I write a program that submits entries by bypassing my browser?
Yes.

What topics are appropriate for the discussion group?
If it's related to this collaboration, then it's fine to talk about it in the discussion group. In particular, you are encouraged to share your results and discuss algorithms.