two natural integer sequences Have the integer sequences $f$ and $g$ defined below already been studied? Do they have any interesting property?
Define the maps $f,g:\mathbb Z_{ > 0}\to\mathbb Z_{ > 0}$ as follows:
Define $u,v:\mathbb Z_{ > 0}\to\mathbb Q_{ > 0}$ by 
$$
u(n):=\left(1+\frac1n\right)^n,\qquad v(n):=\sum_{k=0}^n\ \frac{1}{k!}\ ,
$$ 
let $f(n)$ be the least positive integer $k$ such that $v(k)\ge u(n)$, and let $g(n)$ be the least positive integer $k$ such that $u(k)\ge v(n)$.
I haven't found these integer sequences in the On-Line Encyclopedia of Integer Sequences. 
For the sake of completeness, let me add that the sequences $f$ and $g$ are well defined because $u$ and $v$ are increasing, bounded, and have the same limit, usually denoted by $e$. The fact that $u$ is  increasing is shown in this wonderful answer of Marc van Leeuwen. The sequences $f$ and $g$ are weakly increasing and unbounded.
We have: 
$u(1)=2,\ u(5)=2.48832,\ u(6)=2.52\dots,\ u(25)=2.665\dots,\ u(26)=2.667\dots$,
$v(1)=2,\ v(2)=2.5,\ v(3)=2.66\dots,\ v(4)=2.70833\dots$,
$f(1)=1,\ f(2)=2,\ \dots,\ f(5)=2,\dots,\ f(6)=3,\dots,\ f(25)=3,\ f(26)=4,\dots$.
The values $f(1),\ \dots, f(26)$ are: 
$1,2,2,2,2,3,3,3,3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4$, 
and the On-Line Encyclopedia of Integer Sequences returns the answer: 
Sorry, but the terms do not match anything in the table.
 A: Here is a partial answer. It gives an upper bound for the sequence $g$ in the question, but I tried to write this post in such a way that it can be read independently of the question.
As in the question, put 
$$
v(n):=\sum_{k=0}^n\ \frac{1}{k!}\ ,
$$ 
and recall that 
$(1)\quad(1+\frac1n)^n$ and $v(n)$ are increasing sequences having the same finite limit $e$. 
This answer of Marc van Leeuwen implies 
$$
\left(1+\frac1n\right)^n < v(n)
$$ 
for $n\ge2$. Let $n\ge1$ be given. By $(1)$ we have
$$
v(n) < \left(1+\frac1N\right)^N\tag2
$$ 
for $N$ large enough, and the game we are playing here is to find explicitly such an $N$. More precisely, we claim that $(2)$ holds with
$$
N:=\left\lceil\frac12\ \frac{1}{1-\log v(n)}\right\rceil.
$$ 
It suffices to prove 
$$
\log v(n) < N\ \log\left(1+\frac1N\right), 
$$ 
or, setting $x:=1/N$, 
$$
\log v(n) < \frac{\log(1+x)}{x}\ .
$$ 
Using Lagrange's expression for the remainder of the Taylor approximation of $\log(1+x)$ around $0$, we get 
$$
\frac{\log(1+x)}{x}=1-\frac{x}{2\ (1+y)^2}
$$ 
for some $y$ satisfying $0 < y < x$, and thus 
$$
\frac{\log(1+x)}{x}=1-\frac{x}{2\ (1+y)^2} > 1-\frac x2\ge 1-\frac12\ 2\ (1-\log v(n))=\log v(n).
$$
