A few days ago, Terence Tao proposed a challenge to decompose the 300K! as the product of 300K factors larger than 100K. A smaller example is decomposing 10! as the product of 10 factor greater or equal than 3. 10! = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 10! = 3 * 3 * 4 * 4 * 4 * 5 * 5 * 6 * 6 * 7He was able to show a decomposition of 300K! as the product of 300K factors larger than only 90K, and proposed a method to try to use larger numbers. This is part of an attempt to prove a conjecture, so the complete problem is more general, but let´s try to solve this case.We will fix N=300K that is the case analyzed by Tao in the post. The idea by Tao is to start with a odd number B that is a product of 300K odds numbers larger than 100K, and attempt to fix the difference. One important point is that 300! is even, so the idea is to get try to use the 2 to help in this task.He defines the B-heavy primes that are the primes that appear more time in B than in N!=300!, and the N!-heavy primes that are the primes that appear more time in N!=300! than in B. The idea is to replace each B-heavy prime with bigger a N!-heavy prime n multiplied by a power of 2, or by a pure power of 2. This will increase the factor of B and we will get the requested decomposition.It's important to use as few 2 as possible, so we don't run out of them. All of this is better explained in Tao's post, so it's worth reading it.He was able to apply this method to get a decomposition with number that are larger than 90K bu the mapping and use of 2 was not optimal, so it was not good enough for 100K. We are going to find the optimal way to do this mapping and get a decomposition with number that are larger than 100K.A preambleWe will use Racket. First we will try to reproduce the result with 90K.We will fix N=300K and write a memoized version of factorize, because factorization is hard and slow, and we will reuse the same number a lot.#lang racket(require math/number-theory)(define N 300000)(define secret-f...
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