Could Euler have conjectured the prime number theorem?
... agree “unless there is an obvious reason that they don’t.” Of course this is only a rough heuristic: after all, one person’s obvious reason may be another person’s deep result. While the Cramér model is only a heuristic, it is sometimes possible not only to make conjectures but also to prove theore ...
... agree “unless there is an obvious reason that they don’t.” Of course this is only a rough heuristic: after all, one person’s obvious reason may be another person’s deep result. While the Cramér model is only a heuristic, it is sometimes possible not only to make conjectures but also to prove theore ...
Document
... We can use either of two methods to divide one polynomial by another. These are by: ...
... We can use either of two methods to divide one polynomial by another. These are by: ...
Odd perfect numbers are greater than 10 1500
... Suppose that N is an odd perfect number, and that p is a prime factor of N . If pq k N for a q > 0, then σ(pq ) | 2N . Thus if we have a prime factor p′ > 2 of σ(pq ), we can recurse on the factor p′ . We make all suppositions for q up we get a contradiction (e.g. pq is greater than the limit we wan ...
... Suppose that N is an odd perfect number, and that p is a prime factor of N . If pq k N for a q > 0, then σ(pq ) | 2N . Thus if we have a prime factor p′ > 2 of σ(pq ), we can recurse on the factor p′ . We make all suppositions for q up we get a contradiction (e.g. pq is greater than the limit we wan ...