Sequences from Group Theory

[@yatyayat]

10:43 You made a mistake here: 9 × 13 = 117 ≠ 121.

[@TheGrayCuber]

Agh I sure did, thank you for noting this

[@yatyayat]

@TheGrayCuber Thanks. 🙂

[@elendiastarman]

Congrats on contributing these sequences to the OEIS! That's a great accomplishment.

[@yfidalv]

This and the previous were two of the most interesting/novel group theory videos i’ve seen, thanks for sharing your findings :)

[@d1scocubes]

your voice is so calming

[@djsmeguk]

Lovely stuff. The patterns are fascinating. Glad you got them up onto oeis!

[@1.4142]

31^(1/3)=3.14138065 nice

[@RuleAndLine]

This is incredible

[@ShenghuiYang]

Clean and concise!

[@mostly_mental]

These are some really interesting patterns you've found. Your approach to the maximal representations (and the variance) looks a lot like the 0-1 knapsack problem, which is a very well explored topic. Have you looked at applying any of those approaches here?

[@TheGrayCuber]

I hadn't heard of the knapsack problem, thanks for pointing me that way! I'm working now to optimize the logic for this sequence, with the hopes of finding a million or so terms. It looks like the knapsack approach applies and is similar to what I've been building, but this list of factors is a little nicer than the general knapsack setup, particularly because for n>1, the 'cheapest' factor with weight n is cheaper than that of n+. I may make a follow-up video on this optimized logic

[@TheGrayCuber]

I've been working more on this, and found a good algorithm to find terms of my sequence. This is essentially the 0-1 Knapsack problem, but with infinite items. Are you aware of any materials on this? I've found a lot of resources about the 0-1 knapsack with some N amount of items to pick from, but I can't find any discussion on an unlimited number of items. The 'unbounded' version of the problem allows for many copies of the same item but still uses finite items

[@mostly_mental]

@@TheGrayCuber If you can efficiently find an upper bound for factors with a given number of cycles (or precompute the cycles and costs for a sufficiently large list of primes), you can treat it as the finite problem. Otherwise, I don't think you're going to have much luck with this approach.

[@TriangularCosmos]

Cool!

[@marigold2257]

Here’s a fun sequence, 1,2,3,1,5,6,7,2,3,10,11,3,13,14,15,1,17,6,19,5,21,22,23,6,5,26,1,7,29,30,31,2,33,34,35,3,37,38,39,10,41,42,43,11,15,46,47,3,49,10. Can you find the pattern

[@TheGrayCuber]

It seems like for any prime power p^k that divides n, if p divides k then remove the p factor, else use just p. The only exception I can see is 49 which would go to 7 under my rule.

[@marigold2257]

@@TheGrayCuber this is completely correct and I made a mistake on 49😅

[@TheGrayCuber]

Cool, thanks for the puzzle!!

[@Kyoz]

🤍