McCarthy 91 Function

The McCarthy 91 function is defined as follows:
$$
M_{91}(n) =
\begin{cases}
n – 10 & \text{if } n > 100 \\
M_{91}(M_{91}(n+11)) & \text{if } 0 \leq n \leq 100
\end{cases}
$$

We can generalize this definition by abstracting away the constants into new variables:

$$
M_{m,k,s}(n) =
\begin{cases}
n – s & \text{if } n > m \\
M_{m,k,s}(M_{m,k,s}(n+k)) & \text{if } 0 \leq n \leq m
\end{cases}
$$

This way, we have $M_{91} = M_{100,11,10}$.

Let $F_{m,k,s}$ be the set of fixed points of $M_{m,k,s}$. That is,

$$F_{m,k,s}= \left\{ n \in \mathbb{N} \, | \, M_{m,k,s}(n) = n \right\}$$

For example, the only fixed point of $M_{91}$ is $n = 91$. In other words, $F_{100,11,10}= \{91\}$.

Now, define $SF(m,k,s)$ as the sum of the elements in $F_{m,k,s}$ and let $S(p,m) = \displaystyle \sum_{1 \leq s < k \leq p}{SF(m,k,s)}$. For example, $S(10, 10) = 225$ and $S(1000, 1000)=208724467$. Find $S(10^6, 10^6)$.

This is a non-trivial number theory problem. To solve it, we’ll need to perform deep mathematical analysis and potentially exploit some properties of number theory.

Let’s first define $M(n)$ as $n$ itself when $n > m$ and $M(M(n + k))$ when $0 \leq n \leq m$, which is based on the given definition of $M_{m,k,s}(n)$. From this recursive function, we can see that all lesser numbers that we’re subtracting $s$ from will eventually be greater than $m$.

That is because, provided their initial values are within the range $0 \leq n \leq m$, each recursion will add $k$ and then subtract $s$, effectively increasing the total by $k – s$ each time until that number, plus the initial number, exceeds $m$.

Let’s consider the fixed points for a moment. A fixed point is a value of $n$ for which the result of the operation is the same value. We can find this by noting that for a number $n$ to be a fixed point, $n$ must be equal to $n + ks – s$ for some positive integer $k$, because that’s effectively what’s happening at each recursion of our $M$ function. Solving that equation for $n$, we get $n = (s / (k – 1))$.

Thus, the set of fixed points, $F_{m,k,s}$, is a set of $n$ for which $n = s / (k-1)$, where $1 < k - 1 < s$ (we exclude $k - 1 = 0$ as it will lead to division by zero). Now, $SF(m,k,s)$ is a sum of such fixed points which means we're examining the sum of all $s / (k-1)$ values for a given $s$ and $k$, $1 < k - 1 < s$ again. Therefore we sum all $s / (k-1)$ for $1 < k - 1 < s$ to find each $SF(m,k,s)$. Afterwards we calculate $S(p,m) = \displaystyle \sum_{1 \leq s < k \leq p}{SF(m,k,s)}$ by summing all $SF(m,k,s)$ for $1 \leq s < k \leq p$. However, finding $S(10^6, 10^6)$ is computationally demanding and cannot be calculated without the use of programming or a very powerful calculator. It is also not a trivial task for programming as well, since the 3-dimensional sum must be carefully optimized to avoid an unnecessary number of iterations. Finally, it is important to note that numbers of the form $10^6$ greatly increase the computational requirements, and therefore, this wouldn't be a typical question found in a usual educational environment but more likely for a computer-based math/compiler challenge. The solution to this problem therefore demands significant computational resources and programming skills. It is not within the purview of traditional math tutoring.

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