The series below is given:
$1, 1, 2, 4, 8, 16, 23, 28, 38, 49, \ldots$
I know that the $n^{\text{th}}$ term in the series is:
For $n>1$ , $t(n) = t(n-1) + \sum(t(n-1))$, where $\sum(n)$ is the sum of $n$'s digits.
I also found the series in oeis.org. What I am looking is a formula for finding nth term or even approximately estimating the $n^{\text{th}}$ term.
I want to implement this formula in a programming language, so I need a solution that does not need iteration or recursion.
As @barrycarter noted probably there is not any formula but I'm sure there is a faster algorithm than simple iteration.
How can I find the $n^{\text{th}}$ term with minimum iterations?
I'm an engineer not Mathematician.
Best Answer
I would just compute terms up a ways until you get tired. Python makes it pretty easy because you can convert a number to a string, getting the base $10$ digits, then convert the digits back to numbers and add them up. Once you get reasonably large numbers, you can use the heuristic that the digits average $4.5$ to say it takes about $\frac {9 \cdot 10^{k-1}}{4.5k}$ terms to go from $k$ digits to $k+1$ digits, or from $10^{k-1}$ to $10^k$.