Three days ago I started publicly investigating a deterministic sequence of binary numbers created by prepending 1 followed by as few 0s as possible so that each succeeding number is relatively prime to all of its predecessors.
42.
If you are choosing large numbers sequentially so that each new one must be relatively prime to all of its predecessors, what is a good algorithm for determining each succeeding term's suitability for this?
43.
Just as the quadratic reciprocity law for the Legendre symbol is also true for the Jacobi symbol, the requirement that the numbers be prime is not needed; it suffices that they be odd relatively prime nonunits.
44.
As stated this doesn't really make sense, but is an informal way of saying that as n approaches infinity the probability that two random numbers from 1 to n are relatively prime approaches 6 / ? 2.
45.
Moreover, as long as the polynomial factors at each stage are relatively prime ( which for polynomials means that they have no common roots ), one can construct a dual algorithm by reversing the process with the Chinese Remainder Theorem.
46.
The elementary divisors can be obtained from the list of invariant factors of the module by decomposing each of them as far as possible into pairwise relatively prime ( non-unit ) factors, which will be powers of irreducible elements.
47.
The four linear factors p, q, p + q, and p-q are relatively prime and therefore must themselves be squares; let p + q = r ^ 2 and p-q = s ^ 2.
48.
It is not possible to produce a definite test of primality based on whether a number is an Euler pseudoprime because there exist " absolute Euler pseudoprimes ", numbers which are Euler pseudoprimes to every base relatively prime to themselves.
49.
The factors 1, 2, 4, 5, 8, 10 may start to make the pattern clear : they are those integers less than that are relatively prime to ( or have no prime factors in common with ) 21.
50.
The total number of positive divisors of n is a multiplicative function d ( n ), meaning that when two numbers m and n are relatively prime, then d ( mn ) = d ( m ) \ times d ( n ).
