Scatterstone Nim

Alice and Bob are playing a modified game of Nim called Scatterstone Nim, with Alice going first, alternating turns with Bob. The game begins with an arbitrary set of stone piles with a total number of stones equal to $n$.
During a player’s turn, he/she must pick a pile having at least $2$ stones and perform a split operation, dividing the pile into an arbitrary set of $p$ non-empty, arbitrarily-sized piles where $2 \leq p \leq k$ for some fixed constant $k$. For example, a pile of size $4$ can be split into $\{1, 3\}$ or $\{2, 2\}$, or $\{1, 1, 2\}$ if $k = 3$ and in addition $\{1, 1, 1, 1\}$ if $k = 4$.
If no valid move is possible on a given turn, then the other player wins the game.
A winning position is defined as a set of stone piles where a player can ultimately ensure victory no matter what the other player does.
Let $f(n,k)$ be the number of winning positions for Alice on her first turn, given parameters $n$ and $k$. For example, $f(5, 2) = 3$ with winning positions $\{1, 1, 1, 2\}, \{1, 4\}, \{2, 3\}$. In contrast, $f(5, 3) = 5$ with winning positions $\{1, 1, 1, 2\}, \{1, 1, 3\}, \{1, 4\}, \{2, 3\}, \{5\}$.
Let $g(n)$ be the sum of $f(n,k)$ over all $2 \leq k \leq n$. For example, $g(7)=66$ and $g(10)=291$.
Find $g(200) \bmod (10^9 + 7)$.

The problem belongs to the realm of combinatorial game theory and since the game ends when the largest pile has size 1, this problem can be represented by generating functions.

Let’s denote by `G(x)` the generating function of all possible moves for a certain range of `k`:
`G(x) = x^2/2 + x^3/6 + … + x^k/k` when 2 ≤ k ≤ p
`G(x) = x^2/2 + x^3/6 + … + x^p/p` when p+1 ≤ k ≤ n

The function `F(x)` with `F(x) = 1 + x*G(F(x))` gives us the generating function of the complete game, where `(x^n)/n!` gives the number of ending configurations with `n` stones that was reached by Alice who started the game and cannot move.

It gives us winning positions for Bob and to get ones for Alice we must subtract from the total number of all configurations for `n` stones – `H(n)`, which is given by Bell’s numbers `b(n)` minus ones for Bob:

`f(n,k) = H(n) – (x^n)/n!`, where `H(x)` is the exponential generating function for Bell’s numbers `H(x) = e^(e^x – 1)` and `b(n) = H^(n)(1)`, `H^(n)(x)` – n-th derivative.

Thus, `g(n) = Σ(f(n,k))`, sum over all `k, 2 ≤ k ≤ n`.

We calculate `g(n)` modulo `10^9 + 7` by implementing all steps in code remembering about function `F(x)` and its relevance for calculation `(x^n)/n!`. All steps must be done carefully because of recursion and the need of memoization to save values already computed for `F(x)`.

All the steps above require knowledge of generating functions, combinatorial game theory, calculus (derivative), and the concept of modulo operation.

Implementing this approach in real code can be challenging and would require an understanding of programming in languages such as Python or Java to execute recursive computation and memoization efficiently.

I should note that this solution involves multiple complex mathematical concepts and partial equations, meaning that it is more suitable for people who have reached at least a college-level understanding of mathematics. Therefore, I would not recommend this problem for beginners since it might be overwhelming and discourage them from continuing their study of mathematics.

More Answers: