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May 8

What is the term for...

What is a proper term for an ordered set of values {a, b, c}, where each value can be independently selected from an allowed set for that value. Each allowed set may be different (but need not be distinct). For example:

a from {a₁, a₂, a₃}
b from {b₁, b₂}
c from {c₁, c₂, c₃, c₄}

The set {a, b, c} could then have any of 24 possible values.

A more concrete example would be all upper/lower/mixed case variations of the word "dog" - "dog", "doG", "dOg", "dOG", "Dog", ....

I saw this called a permutation, which I'm pretty sure is not correct. I was thinking combination, but this doesn't quite seem to match the definition. Is there some standard term that is a good fit for this? Tom N talk/contrib 20:21, 8 May 2024 (UTC)[reply]

See Cartesian product. --Trovatore (talk) 20:33, 8 May 2024 (UTC)[reply]

Come to think of it, though, the direct answer to your question would not be "Cartesian product" but rather "element of the Cartesian product". I don't know that there's a snappy one-word term for this. --Trovatore (talk) 20:42, 8 May 2024 (UTC)[reply]

I agree with Trovatore's answer. If we call A = {a₁, a₂, a₃}, B = {b₁, b₂}, and C = {c₁, c₂, c₃, c₄} the three sets in your example, then the ordered sets {a, b, c} that you are describing in your question are the elements of the set A×B×C : the cartesian product of A, B, and C. This set is precisely the set of such ordered sets.

On the other hand, a permutation of a set is, intuitively, a "shuffling" of this set. If you consider the set of cards in a 52 cards deck, this gives a good intuition. A combination of those cards would just be a subset of the whole deck, i.e. any number of cards drawn from the deck, without any regard for the order. So the whole diamonds suit is a combination of cards from the deck. So would the unordered set {3 of clubs, 10 of hearts, queen of spades}. Tommpouce (talk) 17:23, 18 May 2024 (UTC)[reply]

Collatz Conjecture

From time to time I noodle around the Collatz conjecture, in an attempt to bring my "special insight" into an apparently simple question that has stumped the best and the brightest. Most of the work I've seen seems to be based on trying to find a counter-example, which would render the conjecture null and void. They've tested every number up to some ridiculous number of trillions, so far without any luck, but they take the view, quite rightly, that a giga-zillion examples do not prove the general proposition, and the elusive counter-example could be just round the corner, so they keep searching, and trying new ways to attack the problem.

I've never seen any work that started at the opposite end: the number 1. That is, take 1, and ask "What could produce 1?": Answer: only 2. Repeat. This results in the series 1, 2, 4, 8, 16. Then it starts getting interesting, because 16 could be derived from either 5 or 32. (Being odd, 5 produces 16, because 3*5+1 = 16; and being even, 32 produces 16, because 32/2=16.) Then 5 and 32 can be investigated separately, and so on. The tree quickly sprouts new branches and it just gets more bushy the further we go.

The question in my mind is: Can it be shown that every integer must belong to this tree? If so, would that not prove the Conjecture? Or, if it could be shown that not all numbers are captured, even if we could not identify any specific examples, would that not disprove it?

Yours simplistically, Jack of Oz ^{[pleasantries]} 22:51, 8 May 2024 (UTC)[reply]

Nice way of re-thinking the problem! I agree that if you could prove that the tree hits every positive integer, or that it doesn't, then you've solved the problem. I seriously doubt that it hasn't been tried, but it's a good example of things to try. --Trovatore (talk) 23:07, 8 May 2024 (UTC)[reply]

The specific approach is mentioned in Collatz conjecture § In reverse, which also has a diagram of the first 21 layers of the tree. A more parsimonious representation is achieved by considering that it suffices to show that all odd positive integers are reached. The onset of the odd tree is shown in the very first image in the article; a more extensive one, not shown in the article, is found here --Lambiam 05:41, 9 May 2024 (UTC)[reply]

Well, there you go. Thanks, Lambiam. (At least I should be given credit as the independent co-discoverer of this idea.) -- Jack of Oz ^{[pleasantries]} 07:46, 9 May 2024 (UTC)[reply]

Here's another exploration of the Collatz tree. Double sharp (talk) 11:27, 11 May 2024 (UTC)[reply]

Problem: Give positive integer n, how many natural density of positive integers reach n in their Collatz (3x+1) sequence? Of course, for n = 1, 2, 4, 8, 16, the natural density is 100%, but for n = 32 and 5, what will be the answers? For n divisible by 3, the answer is 0%, since only numbers of the form n*2^k reach n, and I think that the answer for 13 and 80 (which are the two numbers before 40) should be equal. 2402:7500:942:8E8F:A4D8:9B73:8E52:1E7B (talk) 07:47, 10 May 2024 (UTC)[reply]

You can eliminate this by calculate (for example) all positive integers <= 2^16 = 65536, how many positive integers reach 32 (or 5, or 13, or 80)? 2402:7500:900:DEEB:B513:C07E:8EF3:8275 (talk) 04:15, 11 May 2024 (UTC)[reply]