Is $ 2.7182818281828...$ a semiconvergent of e?

Question

Euler's number $e=2.71828 18284 59045... $ can be approximated by the rational number: $$ x=\frac{271,828-27}{100,000-10}= \frac{271,801}{99,990} =2.7182818281... $$

Also, $e$ has the well-known continued fraction expansion $$ e = [2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,...]. $$

Is the fraction $x$ a semiconvergent of $e$? That is, does there exist $ a $ such that
$$ x=\frac{h_{n-1} +a h_n}{k_{n-1}+a k_n}, $$ where $h_{n-1}, h_n$ and $k_{n-1}, k_n$ are successive numerators and denominators of convergents of $e$.

Background

It is known that $y=355/113$ is a close rational approximation to $\pi$ which is also easy to remember (113 355). I want to know if there is a similar way to relate $2.718281828...$ to $e$.

Answer 1

Yes!

In Mathematica, define

Semiconvergent[a_, b_, k_: 1] := (Numerator[a] + k Numerator[b])/(Denominator[a] + k Denominator[b])

Then

es = Convergents[E, 15]
Semiconvergent[es[[13]], es[[14]], 5]

Gives

271801/99990