4 ms·
My intuition was already screaming 'it'll be the other way around for a geometric progression' before I read that far, but I'm damned if I can understand or eve
by desc 8y ago
My intuition was already screaming 'it'll be the other way around for a geometric progression' before I read that far, but I'm damned if I can understand or even speculate why. Most likely I don't properly understand it.
I'm inclined to wonder if there's a third operator which could be tested like this, such as exponentiation, but that's not commutative over integers.
Of course, if there are similarly intriguing patterns for noncommutative operators (and their sequences) the obvious next step would be to look at complex numbers and quaternions...
- metrognome 8y agoIt's not too surprising that there are fewer unique products for a geometric progression, because subsequent terms in a geometric progression are found by multiplying by the common ratio. There will be fewer unique products when all of the numbers share that common ratio.
- thaumasiotes 8y ago> My intuition was already screaming 'it'll be the other way around for a geometric progression' before I read that far, but I'm damned if I can understand or even speculate why. It's a question of forcing the number of sums/products to be low. An arithmetic progression forces a large number of identical sums for the obvious reason: (a+b) = ((a-k)+(b+k)) = ((a-2k) + (b+2k)), and so on, and those differences of k are... the definition of an arithmetic progression. The reason a geometric progression produces a lot of identical products is exactly the same. (ab) = (a/k · bk) = (a/kk · bkk)...