$$a\_0 = 1, \ \ a\_{n+1} = \ 1+\frac{n \*a\_n}{n+a\_n} , \ \ n=0,1,2,3,4,...$$


I have built the above recursive formula. Some terms of sequence are: 
1, 1, 3/2, 13/7, 73/34, 501/209, 4051/1546, 37633/13327, 394353/130922, 4596553/1441729, 58941091/17572114, 824073141/234662231,... Τhe numerators of the fractions are identical to the sequence A000262 in OEIS encyclopedia and the denominators to the A002720. If we take n = {inf,.......,5,4,3,2,1}, ie to start from quite high and finish at 1. Let a (about 1.6768...) the last term. Then 1/(a\*exp(1)) converges to: 0.21938393439552027367716377546012164903 ... that is the decimal expansion of -Ei(-1), A099285 in OEIS.


**My question is:**
How it is explained, and how does the formula relate to the above sequences? 



A recursive formula
For $n\geq 1$, let $p\_n$ be the $(n+1)$-th term of [A000262](https://oeis.org/A000262), and let $q\_n$ be $n$-th term of [A002720](https://oeis.org/A002720). Then, according to the description of these two sequences (more precisely by the contributions of Dennis P. Walsh and Paul Berry)
$$p\_n=\sum\_{k=0}^{n-1}\binom{n}{k}\binom{n-1}{k}k!\qquad\text{and}\qquad
q\_n=\sum\_{k=0}^{n-1}\binom{n-1}{k}^2k!\ .$$
We derive some identities. First,
\begin{align}p\_{n+1}-q\_{n+1}&=\sum\_{k=0}^{n}\binom{n+1}{k}\binom{n}{k}k!-\sum\_{k=0}^{n}\binom{n}{k}^2k!\\[6pt]
&=\sum\_{k=0}^n\left(\binom{n+1}{k}-\binom{n}{k}\right)\binom{n}{k}k!\\[6pt]
&=\sum\_{k=1}^n\binom{n}{k-1}\binom{n}{k}k!\\[6pt]
&=n\sum\_{k=1}^n\binom{n}{k-1}\binom{n-1}{k-1}(k-1)!\\[6pt]
&=n\sum\_{k=0}^{n-1}\binom{n}{k}\binom{n-1}{k}k!\\[6pt]
&=np\_n.
\end{align}
That is,
$$p\_{n+1}=np\_n+q\_{n+1}.\tag{1}$$
Second,
\begin{align}q\_{n+1}-p\_n&=\sum\_{k=0}^{n}\binom{n}{k}^2k!-\sum\_{k=0}^{n-1}\binom{n}{k}\binom{n-1}{k}k!\\[6pt]
&=n!+\sum\_{k=0}^{n-1}\binom{n}{k}\left(\binom{n}{k}-\binom{n-1}{k}\right)k!\\[6pt]
&=n!+\sum\_{k=1}^{n-1}\binom{n}{k}\binom{n-1}{k-1}k!\\[6pt]
&=n!+n\sum\_{k=1}^{n-1}\binom{n-1}{k-1}^2(k-1)!\\[6pt]
&=n!+n\sum\_{k=0}^{n-2}\binom{n-1}{k}^2k!\\[6pt]
&=n\sum\_{k=0}^{n-1}\binom{n-1}{k}^2k!\\[6pt]
&=nq\_n.\end{align}
That is,
$$q\_{n+1}=p\_n+nq\_n.\tag{2}$$
From $(1)$ and $(2)$, it follows by induction that 
$$a\_n=\frac{p\_n}{q\_n}.\tag{3}$$ Indeed, $(3)$ holds for $n=1$. Assuming $(3)$ holds for a given $n$, it also holds for $n+1$ in place of $n$:
$$a\_{n+1}=1+\frac{na\_n}{n+a\_n}=1+\frac{np\_n}{p\_n+nq\_n}=1+\frac{p\_{n+1}-q\_{n+1}}{q\_{n+1}}=\frac{p\_{n+1}}{q\_{n+1}}.$$
This answers the second part of the OP's question. More precisely, it would also be desirable to prove that $\gcd(p\_n,q\_n)=1$, but I have not verified this.


**Added 1.** I can now prove that $\gcd(p\_n,q\_n)=1$, i.e., the fraction $(3)$ is in lowest terms. Let us proceed by induction. For $n=1$, the statement is clear. We assume therefore that $\gcd(p\_n,q\_n)=1$ for a given $n$, and we prove that $\gcd(p\_{n+1},q\_{n+1})=1$. If this is not the case, then there is a prime $\ell$ that divides both $p\_{n+1}$ and $q\_{n+1}$. From $(1)$ we get $\ell\mid np\_n$, hence from $(2)$ we get $\ell\mid n^2 q\_n$. As $p\_n$ and $q\_n$ are relatively prime by the induction hypothesis, this forces $\ell\mid n$, and then from $(2)$ we get $\ell\mid p\_n$. However,
$$p\_n=1+\sum\_{k=1}^{n-1}\binom{n}{k}\binom{n-1}{k}k!=1+n\sum\_{k=1}^{n-1}\binom{n-1}{k-1}\binom{n-1}{k}(k-1)!\ ,$$
whence $p\_n\equiv 1\pmod{n}$, and therefore $p\_n\equiv 1\pmod{\ell}$. This contradiction shows that $\gcd(p\_{n+1},q\_{n+1})=1$, and we are done.


**Added 2.** I can now answer the remaining first part of the question. Briefly, as pointed out by მამუკა ჯიბლაძე in a comment, the observed "property of $a$" is equivalent to the identity
$$e E\_1(1)=[1/1,1,1/2,1,1/3,1,1/4,1,1/5,1,\dots],$$
where (cf. [Wikipedia](https://en.wikipedia.org/wiki/Exponential_integral))
$$E\_1(x):=\int\_x^\infty\frac{e^{-t}}{t}\,dt,\qquad x>0.$$
More generally, it is known that (cf. [Wikipedia](https://en.wikipedia.org/wiki/Exponential_integral))
$$e^x E\_1(x)=[x/1,1,x/2,1,x/3,1,x/4,1,x/5,1,\dots].$$


