I have watched some videos on YouTube on finding an explicit formula from a recursive sequence but none of the tricks they have shown seem to work in my case. I have the recursive relation defined as $a_n = (1-a_{n-1})f(n-1)$ where $f(n)=\tfrac{1555+3n}{1612535}$ and $a_0=0$. (I'm not replacing $f(n-1)$ by $\tfrac{1552+3n}{1612535}$) because it gets messy). I want to find the general formula for the n-th term. I think i found a solution but no calculator is able to calculate it and im not even sure if it's true or not. Here is my thought process:
$$a_n = (1-a_{n-1})f(n-1)$$ $$a_n = f(n-1)-(a_{n-1})f(n-1)$$ $$a_n = f(n-1)-(f(n-1))(f(n-2))+(a_{\mathbf{n-2}})(f(n-1))(f(n-2))$$ $$a_n = f(n-1)-(f(n-1))(f(n-2))+(f(n-1))(f(n-2))(f(n-3))-(a_{\mathbf{n-3}})(f(n-1))(f(n-2))(f(n-3))$$
Seeing the term with $a_n$ only appears in the last term and $a_0=0$, $a_0$ will cancel it. So i have come to the conclusion that:
$$a_n = f(n-1)-(f(n-1))(f(n-2))+(f(n-1))(f(n-2))(f(n-3))\cdots \pm f(n-1) \cdots f(1)$$
After that, i cleaned things up a bit: $$a_n =\prod_{i=1}^{1}f(n-i)-\prod_{i=1}^{2}f(n-i)+\prod_{i=1}^{3}f(n-i) \cdots \pm\prod_{i=1}^{n-1}f(n-i)$$ $$a_{n}=\sum_{k=1}^{n-1}((-1)^{k+1}\prod_{i=1}^{n}(f(n-i))))$$
The $(-1)^{k+1}$ part is to determine the $\pm$.
So far, i am unsure if i made any mistakes while clearing things up a bit, since the product and sum notations are sometimes difficult to implement.
Even if i didn't made any mistakes, no calculator is able to calculate this expression.
I wonder if there is an easier way to find the formula for the n-th term of the series without using the product and sum notations.
Thanks in advance!
Looking at the sequence you properly wrote, it is almost sure that the incomplete gamma function would appear somewhere.
Making the problem more general $$a_n = (1-a_{n-1})\,f_{n-1}\qquad \text{with} \qquad f_n=\alpha+\beta\, n$$ with $\alpha>0$ and $\beta>0$.
I shall skip the intermediate steps which are quite tedious. The result is : $$\color{blue}{a_n=\frac{e^{-1/\beta }\,\, (-\beta )^{\frac{\alpha }{\beta }+n-2}}{\Gamma \left(\frac{\alpha }{\beta }\right)}\,\,\, \Delta_n}$$ where $$\color{blue}{\Delta_n=(\beta -\alpha )\,\, \Gamma \left(\frac{\alpha-\beta }{\beta },-\frac{1}{\beta }\right)\,\, \Gamma \left(n+\frac{\alpha }{\beta }\right)+}$$ $$\color{blue}{ (\alpha +\beta (n-1))\,\,\Gamma \left(\frac{\alpha }{\beta }\right)\,\, \Gamma \left(n+\frac{\alpha -\beta}{\beta},-\frac{1}{\beta }\right)}$$
For your specific case, make $\alpha=\frac{1}{1037}$ and $\beta=\frac{3}{1612535}$ to have an almost linear function with an asymptotic slope $\sim 1.79\times 10^{-6}$.
The problem is not more difficult for $a_0 \neq 0$.
We also have an explicit solution if $f_n=\alpha+\beta\, n+\gamma\,n^2$ but in terms of the $\, _1F_2(.)$ generalized hypergeometric function.