On Fri, Jun 22, 2018 at 8:34 PM Dan Asimov <dasimov@earthlink.net> wrote:
These things form a very interesting set.
Conjecture (either vague enough to not be controversial, or dead wrong): ---------- If we think of these bit strings as vectors in the vector space V that is a countably infinite direct product of the 2-element field:
∞ V = ∏ Z/2 1
then every sequence
b : Z+ —> {0,1}
of bits is the sum of elements {E_(n_j) | j in Z+} on Jim's list, i.e., where each element occurs at most once. (Since only finitely many elements have any 1's among the first n bits, any such countable sum will make sense.)
It's not true that only finitely many elements have any 1's among the first n bits. For example, the vectors that repeat the pattern "1 followed by n zeros" are all periodic, and all have a 1 in the first position. But you can still make sense of the questions below, by saying that the sum is undefined if there are infinitely many terms with a non-zero value in the same position, and requiring all the sums to be defined.
Now I'm wondering if Jim's list of pure frequencies among the periodic bit strings can also be seen as generating — *in some sense* — all sequences of integers, or of positive integers.
Questions: ----------
I) What are the sets of sequences of arbitrary integers
x : Z+ —> Z
of the form
{x = Sum N_j E(j), N_j in Z}
This is all sequences of integers. Choose E(j) with E(j)[i] = 0 for i < j (this ensures that the sum is defined), and choose N_j to be x(j) - sum_{i<j} N_i E(i) (this ensures that x(j) has the desired value).
and
II) What are the sets of sequences of positive integers
of the form
{x = Sum N_j E(j), N_j in Z+}
This seems much more difficult to characterize. I can't even manage a simple statement of the constraints on the sequence imposed by the value of x(1). All periodic series are representable, but so are lots of aperiodic ones. For example, adding together all the sequences where the periodic part is all zeros except for a final 1 gives x(n) = tau(n), the number of divisors of n. Andy
—Dan
Jim Propp wrote: ----- There are only countably many sets of natural numbers that form (the nonnegative part of) a congruence classes. Here's a natural way to list them as bit strings:
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ... 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 ... 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 ... 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 ... 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 0 ... 0 0 1 0 0 1 0 0 1 0 0 1 0 0 1 0 ... 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 ... 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 ... 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 ... 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 ... 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 ... 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 ... 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 ... 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 ... 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 ... 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 ... ... ... ... -----
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