# Order type of growth rates of exponential generating functions of binary sequences

The exponential generating function of a sequence $$a[n]$$ is:

$$\displaystyle \text{EG}(a;x) = \sum_{n=0}^\infty a[n] \frac{x^n}{n!}$$

My question is about exponential generating functions of binary sequences $$b[n]$$, i.e. for which the codomain is $$\{0,1\}$$.

We know for all such $$b[n]$$, the exponential generating function $$\text{EG}(b;x)$$ is absolutely monotonic, and furthermore that it is an entire function with no poles, since the function $$\exp(x)$$ is the special case where $$b[n] = 1$$ for all $$n$$. We can compare any two such $$\text{EG}(b_1;x)$$ and $$\text{EG}(b_2;x)$$ by saying that $$\text{EG}(b_1;x) \leq \text{EG}(b_2;x)$$ iff there exists some real $$r$$ such that the inequality holds for all real $$x > r$$.

As a result, given two binary sequences $$b_1[n]$$ and $$b_2[n]$$, we can say that $$b_1[n] \precsim b_2[n]$$ iff $$\text{EG}(b_1;x) \leq \text{EG}(b_2;x)$$.

My questions:

1. Is $$\precsim$$ a total order?
2. If so, what is the order type of $$\precsim$$?
3. There seems to be an initial segment of $$\Bbb N$$, given by the set of all binary sequences with finitely many $$1$$'s, ordered lexicographically. After this, is it dense?
• (1) Define $b_1$ to be the sequence that has $1$s where the MacLaurin of $\sin(x)$ has a negative coefficient and zeros otherwise. Define $b_2$ to be the sequence that has a $1$ where that series has a positive coefficient and zeros otherwise. Then $EB(b_2;x)-EB(b_1;x)=\sin(x)$. – user647486 Mar 19 '19 at 23:21
• Ah yes, very good - so it isn't a total order. Does seem to have an initial segment of $\omega$ though (and a final segment of $\omega^*$, which I forgot to mention). – Mike Battaglia Mar 19 '19 at 23:30
• You might see this answer It deals with finite strings of digits and concludes the order type is $\omega \cdot (1+\Bbb Q)+1$ I believe you have this for the strings that are eventually $0$ and this reversed for the strings that are eventually $1$. In between we have all the strings that have infinitely many $1$s and infinitely many $0$s, which are the hard ones and most of them. – Ross Millikan Mar 20 '19 at 0:05