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@gramos

Welcome, gramos! I, too, am new here. Aside from the facts that I'm not Spanish and that I don't much care about motorcycles, I think just about everything else in your introductory post applies to me as well... more or less. (For example, it's closer to twenty years than fifteen since I was first introduced to Linux, but eh, that's close enough...)

@design_RG

Yeah... I only realised after I'd typed that out that the timeline goes in _reverse_ chronological order, and not in chronological order. Makes a lot more sense that way, too. And thanks for the welcome!

The pinned posts point is also good. That means I can take my time, figure out the perfect introductory post, and then pin it at my leisure!

...

...I doubt I'll hurry too much about it.

Let's see... apparently, putting in a bunch of tags of things that interest me is going to make it easier for other people who share those interests to run across me.

Alright, then. Let's see...










...that should be enough for the moment, I guess.

@ElektrikChameleon

I'm no Detritus. My mental abilities are not significantly enhanced by a reduction in temperature.

@ElektrikChameleon

Oh... so it needs to be tagged with Introductions? Hmmmm. Let me think about it, then...

@ElektrikChameleon

...is there some way in which introduction posts differ from any other randomly selected post?

@starcake

Of course, *Every* first post has *something* in it that the writer later regrets.

@carlostomas @freemo

...it's certainly an approach worth looking at.

I've only looked at the overview of the proof, but there is one thing that occurs to me; and tht is that, when using the distance argument (i.e. showing that numbers have a common factor with each other, and that factor is not present in the number sought as a sum) it is *not* required that *all* the terms have the same factor.

To illustrate this, let us assume a set of N numbers, all of which are multiples of three (some may be negative). To this set of numbers, let us add the number 1, which is not a multiple of three. We now have a set of N+1 numbers, all but one of which are multiples of three; and there is no subset of these numbers which can possibly add up to 20. So we can use distance for identification of multiple subset sums even when some elements of the sum do not share a common factor.

If that common factor is 3, then we can have at most one number which does not share the common factor. However, for larger common factors, the number that we can have which do not share that factor becomes larger; with a common factor of 5, we can have at least three numbers in the set which do not share the common factor and yet still leave certain totals impossible to reach.

I'm not entirely sure how this applies to the subset problem in a general sense.

Now let's see what else is around here.

So. This will be my first mastodon post.

Whatever else I do, this will always be at the top of my timeline. However things change.

I should probably make it somehow interesting, or timelessly relevant.

But there is such a thing as overthinking things. A poor first post is very ignorable, and I don't want analysis paralysis to prevent me from ever posting anything at all.

So... overthought first post done, I guess.

Qoto Mastodon

QOTO: Question Others to Teach Ourselves
An inclusive, Academic Freedom, instance
All cultures welcome.
Hate speech and harassment strictly forbidden.