$$S(n)=S(n-1)+\operatorname{lcm}[n,S(n-1)]-1$$
Given the initial $S(1)=1$.
Where $n\ge2$ and $\operatorname{lcm}$ is the Least Common Multiple
As an example the $\operatorname{lcm}(6,8)=24$.
Let $n:=2,3,4$ and $5$ as an example to see the sequence terms $$S(n)=S(n-1)+\operatorname{lcm}[n,S(n-1)]-1$$ $$S(1)=1$$
$$S(2)=s(1)+\operatorname{lcm}[2,S(1)]-1=1+\operatorname{lcm}(2,1)-1=1+2-1=2$$ $S(2)=2$
$$S(3)=s(2)+\operatorname{lcm}[3,S(2)]-1=2+\operatorname{lcm}(3,2)-1=2+6-1=7$$ $S(3)=7$
$$S(4)=S(3)+\operatorname{lcm}[4,S(3)]-1=7+\operatorname{lcm}(4,7)-1=7+28-1=34$$ $S(4)=34$
$$S(5)=s(4)+\operatorname{lcm}[5,S(4)]-1=34+\operatorname{lcm}(5,34)-1=34+170-1=203$$ $S(5)=203$
We think the limit will hold, but how to shows it, is a hard part we can't do. $$\lim_{n \to \infty}{s(n)\over (n+1)!}=3-e$$
$${2\over 3!}\approx. 0.333$$
$${7\over 4!}\approx. 0.2916$$
$${34\over 5!}\approx. 0.283$$
$${203\over 6!}\approx. 0.2819$$
$$3-e=0.28171817...$$
Does this limit hold as $n \to\infty?$