Thursday, June 24, 2004
Conjecture
Here'a a conjecture (my first):
Let S{n - x} represent the set of all numbers in the set {2^n .. 2^(n + 1) - 1} with n - x factors. If n = 3, S{n - 1} therefore represents the set of all numbers in the set {2^3..(2^4 - 1)} with n - 1, or 2, factors. Let us progress through the sets from n = 1 to n = infinity. As we progress from set to set, each subset S{n - x} will eventually reach a maximum "size". For instance, in any set {2^n .. 2^n - 1}, the maximum number of members with n - 1 factors is 5. Each subset S{n - x} "begins" with a set of prime numbers (these occur when n - x = 1), and "end" in some other set. Let us define the "length" of a set as the number of sets between the "beginning" set and the "ending" set, _not_ including those two sets.
Conjecture: The "length" of a set S(n - x) = the xth member of the Lower Wythoff sequence.
Or, in other words, the "length" of S(n - x); x = 1 is 1, x = 2 is 3, x = 3 is 4, x = 4 is 6, x = 5 is 8, and so on. If we add 2 to each member of the Lower Wythoff sequence, we get the absolute length (i.e., the length including the first and last set).
Let S{n - x} represent the set of all numbers in the set {2^n .. 2^(n + 1) - 1} with n - x factors. If n = 3, S{n - 1} therefore represents the set of all numbers in the set {2^3..(2^4 - 1)} with n - 1, or 2, factors. Let us progress through the sets from n = 1 to n = infinity. As we progress from set to set, each subset S{n - x} will eventually reach a maximum "size". For instance, in any set {2^n .. 2^n - 1}, the maximum number of members with n - 1 factors is 5. Each subset S{n - x} "begins" with a set of prime numbers (these occur when n - x = 1), and "end" in some other set. Let us define the "length" of a set as the number of sets between the "beginning" set and the "ending" set, _not_ including those two sets.
Conjecture: The "length" of a set S(n - x) = the xth member of the Lower Wythoff sequence.
Or, in other words, the "length" of S(n - x); x = 1 is 1, x = 2 is 3, x = 3 is 4, x = 4 is 6, x = 5 is 8, and so on. If we add 2 to each member of the Lower Wythoff sequence, we get the absolute length (i.e., the length including the first and last set).
