Formula for finding the nth number in this sequence: $[0, 1, 3, 11, 50, 274...]$?

Question

The sequence here is what I have discovered to be this: https://oeis.org/A000254, which references Stirling numbers of the First Kind. The formulas provided in that link are very ambiguous with their notation, and I have no idea what to take from them. Attempting to come up with a formula that will give me the nth number in the sequence, I've had to delve into research of Stirling numbers of the first kind and its definitions, which have proved largely ineffective in helping me derive this formula. I'm sure a formula for this already exists and is on the internet (or in someone's memory), I just haven't found it, and if anyone could provide this formula, or lead me toward deriving this formula, that would be fantastic. Thanks!

Answer 1

The link you provide has the following formula: $$a(n)=n!\cdot\sum_{k=0}^{n-1}\frac{1}{n-k}$$ where $a(n)$ is the $n$th number in the series and $n\ge 1$. Thus,

$a(0)=0$
$a(1)=1!\cdot\frac{1}{1-0}=1$
$a(2)=2!\cdot\left(\frac{1}{2-0}+\frac{1}{2-1}\right)=3$
$a(3)=3!\cdot\left(\frac{1}{3-0}+\frac{1}{3-1}+\frac{1}{3-2}\right)=11$
and so on.

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There are a lot of other formulas on the link you provide, for example, $a(n)=a(n−1)\cdot(2n−1)−a(n−2)\cdot(n−1)^2$, if $n>1$.

$a(0)=0$
$a(1)=1$
$a(2)=a(2-1)\cdot(2\cdot 2-1)-a(2-2)\cdot(2-1)^2=a(1)\cdot 3-a(0)\cdot 1=1\cdot 3-0\cdot 1=3$
$a(3)=a(3-1)\cdot(2\cdot 3-1)-a(3-2)\cdot(3-1)^2=a(2)\cdot 5-a(1)\cdot 4=3\cdot 5-1\cdot 4=11$
and so on.

Another formula is $a(n+1)=(n+1)*a(n)+n!$, where $n\ge 0$

$a(0)=0$
$a(1)=1\cdot a(0)+0!=1\cdot 0+1=1$
$a(2)=2\cdot a(1)+1!=2\cdot 1+1=3$
$a(3)=3\cdot a(2)+2!=3\cdot 3+2\cdot 1=11$
and so on.