Maybe I don't understand the full implications of the problem, but it seems obvious that a multiplicative table will result in more 'numerical diversity' than a summation table. Both multiplication and addition yield points on lines described by the usual mx+b equation. Addition constrains x to 1, so all of the results will be clustered near the origin, with more coincidences as an unavoidable result. Conversely, if x is allowed to exceed 1, the resulting 'lines' will spread out on the graph and leave more room for unique points.
”but it seems obvious that a multiplicative table will result in more 'numerical diversity' than a summation table”
That isn’t true. If you pick the set of numbers {1, 2, 4, 8}, you get 10 different sums (2, 3, 4, 5, 6, 8, 9, 10, 12, and 16) but only 7 different products (1, 2, 4, 8, 16, 32, and 64)
Sure, you can always find special-case exceptions, but I still don't see why the result is surprising in the general case. There are a lot more arithmetic progressions than geometric ones.
Ah but the conjecture isn't about what happens “on average” or for a “random set” of integers. It's about what is forced to happen in every possible case. (In the “random” case there are probably very few or no duplicates anyway, as both a+b=c+d and ab=cd will be unlikely for completely random integers — note that they say d is exactly a+b-c or ab/c respectively — the conjecture is about limiting what you could possibly arrange to happen, non-randomly.)
Basically, when you take integers like {1, 2, 3, 4} or {1, 3, 5, 7} (so that there are additive relations between them) (e.g. small integers), there tends to be a lot of duplicates in the addition table. But if you take integers like {1, 2, 4, 8} or {1, 3, 9, 27} (so that there are multiplicative relations between them) (e.g. very similar prime factorizations), there tend to be a lot of duplicates in the multiplication table. In other words, it's very easy to create sets with either lots of duplicates in the addition table or lots of duplicates in the multiplication table.
The conjecture is not about which kind of set is more common (so it doesn't matter even if it's true that there are a lot “more” arithmetic progressions than geometric ones, in whatever sense), but rather says there is no way you can create a set with both sorts of collisions — it's saying that if a set has many additive coincidences it can't have many multiplicative coincidences and vice-versa.
In some deep sense, what you are saying is true : for example, if an infinite sequence of numbers is sufficiently dense, then it has arbitrarily long arithmetic progressions.[1] On the other hand, Rankin in 1990 constructed a set with positive density which does not have long geometric progressions.
If you consider all finite sequences of positive integers with all values below some limit M, it might be the case that there are more arithmetic sequences than geometric sequences. I don't think that's relevant to the article, but might be where your head was when you made that post.
Did you read the article? The numbers doesn't have to be consecutive, for example what you said is only true for arithmetic progresions, but for geometric progressions it happens the opposite that multiplications has less diversity. Read the full article.
dunno, but it's neat to think about. multiplication can cover more ground in integer space than addition, so it can reach more unique places than addition for a given set of numbers. it's kinda like comparing how many squares knights and pawns can reach on a chessboard for a given number of moves.
If you give well-spaced source numbers, they can reach the same number of spaces.
For products it's easy: distinct prime numbers.
For addition you can pick numbers so that each one is at least [previous number] x [number of moves] + 1. To keep it extra simple go with [number of moves + 1]^n.
So this is very easy if the number of moves is known upfront, and impossible if it's not.
And in the source puzzle the number of moves is always 2.
What am I missing?