The problem states:
Two players play a game with a single pile of stones of initial size $n$. They take stones from the pile in turn, according to the following rules which depend on a fixed real number $r > 0$:
Let $L(r)$ be the set of initial pile sizes n for which the second player has a winning strategy. For example, $L(0.5) = \{1\}$, $L(1) = \{1, 2, 4, 8, 16, \dots\}$, $L(2) = \{1, 2, 3, 5, 8, \dots\}$.
A real number $q > 0$ is a "transition value" if $L(s)$ is different from $L(t)$ for all $s \lt q \lt t$.
Let $T(i)$ be the $i$-th transition value. For example, $T(1) = 1$, $T(2) = 2$, $T(22) \approx 6.3043478261$.
Clarification Question 1: How are the transition values ordered, and are they indexed starting at 1?- In the first turn, the first player may take $k$ stones with $1 \leq k \lt n$.
- If a player takes $m$ stones in a turn, then in the next turn the opponent may take $k$ stones with $1 \leq k \leq \lfloor r\cdot m\rfloor$.
Let $L(r)$ be the set of initial pile sizes n for which the second player has a winning strategy. For example, $L(0.5) = \{1\}$, $L(1) = \{1, 2, 4, 8, 16, \dots\}$, $L(2) = \{1, 2, 3, 5, 8, \dots\}$.
A real number $q > 0$ is a "transition value" if $L(s)$ is different from $L(t)$ for all $s \lt q \lt t$.
Let $T(i)$ be the $i$-th transition value. For example, $T(1) = 1$, $T(2) = 2$, $T(22) \approx 6.3043478261$.
Clarification Question 2: Is $\frac{4}{3}$ a transition value?
My initial reading of the problem thought the answers to the above questions were:
(1) normal $\lt$ on the real numbers, ascending, starting indexing at $1$
(2) yes, based on analysis shown below
which leads to a contradiction, since $\frac{4}{3} < 2$ but $T(1) = 1$ and $T(2) = 2$.
110 people have solved the problem, so clearly I'm misreading/misunderstanding/miscalculating something.
Analysis to show that $\frac{4}{3}$ is a transition value (minimized to try to avoid spoilers):
Assumption: $0 \lt s \lt \frac{4}{3} \lt t$
Show: $L(s) \neq L(t)$
When $s \lt 1$:
$n = 1$ is losing for the first player, while $n = 2$ is winning for the first player (select $k = 1$)
When $1 \leq s \lt \frac{4}{3}$:
$n = 1, 2, 4$ are losing for the first player, while $n = 3, 5, 6, 7$ are winning for the first player (select $k = 1, 1, 2, 3$ respectively)
When $\frac{4}{3} \lt t \lt 2$:
$n = 1, 2, 4, 7$ are losing for the first player, while $n = 3, 5, 6$ are winning for the first player (select $k = 1, 1, 2$ respectively)
When $2 \leq t$:
$n = 1, 2, 3$ are losing for the first player
All combinations of $s$ and $t$ above produce $L(s) \neq L(t)$.
