t's been weeks (months?) since the 2048 game--by Gabriele Cirulli--took Internet by storm. I have an explicit integer $X$ which is greater or equal than any score of this game. Possibly my $X$ is the actual maximal score, I don't know, it may go either way. Thus my question: what is the maximal possible score for game 2048? (I'll provide $X$ in an answer below, as a SPOILER; it was quite straightforward to get $X$, and similar upper bounds for a way larger family of similar games). If $X$ is not sharp than sharper bounds are welcome too.
BTW, have anybody published an upper bound for 2048 on Internet or elsewere? (If there is a respective link then I'd like to see the article).
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**REMARK:** *While the earlier nice post:*
http://mathoverflow.net/questions/160703/expected-halting-time-for-the-2n-game-aka-2048-with-random-moves
*is related to the game 2048, this post and my question and my answer below are otherwise **not** related. It's nice to know about both posts, but that's all.*
*The other two posts mentioned provided a construction but not a **proof** that it provides maximum, and there is no other **proof** of a specific lower bound there. Their style seems to be of the type: I cannot do it better (without even attempting a mathematical proof of a bound). Please, show me in those posts that it's otherwise (a feeling that a construction is the best is not a proof).*
**PS:** I'd like to see a quote of an essential moment of a **rigorous** proof of an upper bound for a maximal **tile**.
I'll clarify the simple issue of an upper bound for the *total sum*:
Since I have my bound for the maximal tile, let me **follow it** with an upper bound for the **total** sum. Actually, it is obtained as corollary to a maximal configuration (even if such a configuration can be fictitious--nevertheless, all tiles have values not exceeding the respective tiles of the *fictitious* configuration).
Once again, it will be easier to consider the general *templates* (see the answer below) and the logarithmic notation.
Consider $\ A=A_0=A_1=\ldots$ and $\ \forall_{n=0\ 1\ \ldots}\ |A_n|=b\ $ be finite (the playing board is finite so-to-speak). Consider an arbitrary non-negative integer $n$. For each $\ x\in A_n\ $ consider one of its oldest ancestors. By permuting sets $\ A_n\ $ we may assume that $\ x\ $ itself is always its ancestor whenever there is an ancestor for $\ x.\ $ Now let $\ \xi\ $ be the set of all $\ y\in A_n\,\ $ different from $\ x,\ $ for which there is an ancestor at least as old as the one for $\ x.\ $. Then $\ x\ $ has all its ancestors in $\ A(x) := A\setminus\xi.\ $ Observe that functions (*moves*) preserve $\ A(x).\ $ Thus we can consider the induced template for $\ A(x)\ $ and the respective bound:
$$h_n(x)\ =\ (b-|\xi|) + S - 1$$
That's all.
**REMARK** The permutation argument is obvious in the context of templates while it feels messy in the special case of the *game 2048*.
**REMARK** Let me repeat what I said from the start, that the whole thing is straightforward. A simple proof when it appears in a *result+proof* combination can still have a value (when the theorem is significant). @Pietro's above comment is too sketchy. The @David's post math.stackexchange.com/a/902535/448 still doesn't mention some points, as simple as the whole thing is (see my comment over there at stackexchange). But anyway, a complete proof, with all essential moments was already provided by me below (and also above for the **total** sum), and in an extended generality. David's post simply repeats my obvious main idea--anyway, there is perhaps but one sensible way, but for details.