I am working on a recursive sequence and wondering if it's possible to convert it into a non-recursive form. The sequence is defined as follows:
$$ a_n = a_{n-1} \times (n + 1) + n! $$
with the initial condition $ a_1 = 3 $. The term $ n! $ represents the factorial of $ n $, and $ a_{n-1} $ is the previous term in the sequence.
So far, I've considered breaking down the sequence into sums and products, but I'm still not sure how to approach the problem effectively. The interaction between the factorial term and the multiplication by the previous term adds complexity to the sequence.
My questions are:
- Is it possible to express this sequence in a non-recursive form?
- What strategies or mathematical concepts might be helpful in approaching this problem, especially considering my thoughts on dividing it into sums and products?
Any insights or suggestions would be greatly appreciated. Thank you for your time and help!
I consider the indexing of OEIS sequence A000254 (the link provided above by @RobertIsrael), that is: $$ a_{n+1} = a_n\times (n+1) + n! \quad \mbox{with} \quad a_0\in\mathbb{C}.$$ Expressing $a_{n+2}$ depending on $a_n$ and then $a_{n+3}$ depending on $n$ leads to conjecturing the formula $$ a_n = n!\left( a_0 + \sum_{k=1}^n \frac 1 k \right).$$ The proof is straightforward by induction: denoting by $b$ the sequence corresponding to the right-hand term, $b_0=a_0$ by definition and \begin{align*} b_{n+1} &= (n+1)n! \left( a_0 + \sum_{k=1}^n \frac 1 k + \frac 1 {n+1} \right) \\ &= (n+1)b_n + n!. \end{align*}
With your indexing this corresponds to $$ a_n = (n+1)!\left( a_{-1} + \sum_{k=1}^{n+1} \frac 1 k\right).$$ Using that $a_{1}=3$, the above yields $$ 3 = 2! \left( a_{-1} + \frac 3 2 \right),$$ that is, $a_{-1} = 0$ and $$ a_n = (n+1)! \sum_{k=1}^{n+1} \frac 1 k.$$
We recognize Harmonic numbers for the sum, and thus $a_n$ can be expressed as a Stirling number of the first kind: $$ a_n = \begin{bmatrix} n+2 \\ 2\end{bmatrix}.$$