So, what are the practical, real world applications/uses of being able to determine the number of partitions of large number?
It helps you pick up chicks?
Beats me. What are the practical, real world applications/uses of the Laplace transform?
“So, what are the practical, real world applications/uses of being able to determine the number of partitions of large number?”
It gives us ADD kids another way to count numbers and now I don’t have to remember up to 65536 as I double numbers from 2.
Doesn’t everybody know this?
I think someone asked a similar question to Einstein when he came up with his Theory of Relativity.
Maybe there is a pattern to the number of partitins of prime numbers, which would help in the search for larger prime numers?
I'm glad I'm not the only one to ask that question. Somehow I doubt it has any real world use.
Off-hand, I don't know any. On the other hand, that's not really a relevant question. The mathematics underpinning RSA encryption was about 300 years old when it was finally put to practical use. The mathematics underlying general relativity had been done without any application in sight over the 30 years or so before Einstein used it (and w/o understanding general relativity, neither atomic energy nor the GPS system would be possible). The Radon transform uses in medical imaging was discovered in the 1930's w/o any practical application in mind.
Mathematical results (which oddly, as some fundamental level are all tautologies -- the weird thing is that there are non-obvious tautologies that have to be discovered) have an infinite shelf-life, and often turn out to be surprisingly useful years later.
Actually, we've been able to determine the number of partitions of a large number for years, the curious thing Ono claims to have done is give an ordinary formula for the number. (The term of art is a "closed formula", which basically is a formula made out of the operations on a calculator, with nothing like a sum whose number of terms depends on the value n, or a place where one uses one formula of something is true, and a different formula if it's not.)
It turns out p(n) is the coefficient of x^n when one computes the product of (1 - x^k)^(-1) for k=1,. . . n by writing each factor as a geometric series (1 + x^k + x^(2k) + . . .) then multiplying them and collecting terms. (If one is willing to allow infinitely many factors with k ranging over all the counting numbers one gets a product whose value is the series (1 + p(1)x + p(2)x^2 + p(3)x^3 +. . . .) .
For practical purposes, one is left with the question of whether evaluating Ono's formula has a lower computational complexity than the procedure I just described. Now that we have computers, a closed formula may not even be useful if it's slower to compute than some other algorithm that computes the same value.