1+2*(1+2+3+...+44)=1981

How did you get it?Could you please show us more clearly?

FUSK wrote: How did you get it?Could you please show us more clearly? Consider the 'first' number deleted in each process. In fact f(n) should be one first deleted number in some process. Then every first number in each process should be: 1, 2, 3, 5, 7, ... The diff should be: 1, 1, 2, 2, 3, 3, ...

yamamaya wrote: FUSK wrote: How did you get it?Could you please show us more clearly? Consider the 'first' number deleted in each process. In fact f(n) should be one first deleted number in some process. Then every first number in each process should be: 1, 2, 3, 5, 7, ... The diff should be: 1, 1, 2, 2, 3, 3, ... Could you please give a rigorous proof?

yamamaya wrote: FUSK wrote: How did you get it?Could you please show us more clearly? Consider the 'first' number deleted in each process. In fact f(n) should be one first deleted number in some process. Then every first number in each process should be: 1, 2, 3, 5, 7, ... The diff should be: 1, 1, 2, 2, 3, 3, ... can you give a complete solution?

Define $a_n$ as the greatest square less than $n$ and let $b_n = n - a_n^2$. \[ f(n) = \begin{cases} a_n^2 +1 &\text{if $b_n \leq a_n$} \\ a_n^2 + a_n + 1 &\text{if $a_n< b_n \leq 2a_n + 1$.} \end{cases} \]The proof proceeds by strong induction on $n$. The base case $n=1$ is checked by hand. Now we move on to the inductive step. If $n$ is a perfect square, then it is clear that $f(n) = f(n-1)$ and the conclusion is obvious. Suppose, then, that $n$ is not a perfect square. In the first instance, there are $a_n$ terms removed. Enumerate the remaining terms in increasing order as $x_1, x_2, \dots, x_{n-a_n}$. Then \[ f(n) = x_{f(n-a_n)} . \]If $b_n \leq a_n$, then $a_{n-a_n} = a_n -1$. It's a simple calculation to check that $b_{n-a} > a_n$, so $$f(n-a_n) = a_{n-1}^2 + a_{n-1} + 1 = a_n^2 - a_n +1$$by hypothesis. Since this is more than $a_n^2 - a_n$, it follows that \[ f(n) = x_{f(n-a_n)} = f(n - a_n) + a_n = (a_n^2 - a_n + 1) + a_n = a_n^2 + 1. \] If $b_n > a_n$, the logic is identical, except that $a_{n - a_n} = a_n$. This implies that $f(n-a_n) = a_n^2 + 1$. Thus the final step of our calculation turns to \[ f(n) = (a_n^2 + 1) +a_n , \]as desired. Substituting $n = 2019$ gives $a_n = 44$ and $b_n = 83 > 44$, so \[ f(2019) = 44^2 + 44 + 11 = 1981. \]

It's not difficult to could discover that \[f(n) = \begin{cases} k^2-k+1 &\text{if $k^2-k+1 \leq n \leq k^2, k \ge 1$} \\ k^2+1 &\text{if $k^2+1 \leq n \leq k^2+k, k\ge 1$} \end{cases}\]It's easy to prove by induction.