Mastering Recursive Sequences: Back-Tracking & Tree-Pruning Techniques for Advanced Math Competitions
Cracking Piecewise Recursive Sequences: The Power of "Back-Tracking" and "Tree-Pruning" (Complete 10-Problem Guide)
In competitive high school math and elite entrance exams, brute-forcing recurrence relations from the initial term $a_1$ forward is a guaranteed trap. When a downstream term like $a_5$ or $a_6$ is fixed, true mastery lies in inverting the relation, traversing backwards, and pruning invalid branches before they explode.
1. Anatomy of Piecewise Recurrence Relations
High-level test questions deliberately introduce multi-case piecewise definitions that branch at each step:
① Parity-Dependent Branching: Different arithmetic operations applied depending on whether $a_n$ is odd or even (reminiscent of the Collatz 3x+1 problem).
② Threshold / Sign-Dependent Branching: Rules split across a zero threshold ($a_n \le 0$ vs $a_n > 0$) or dynamic indices ($a_n \ge n$).
③ Modular / Residue State Transitions: Branching determined by $a_n \pmod 3$, where residues dictate growth or factor-division.
π‘ Yul's Golden Rules: 3 Steps to Efficient Tree-Pruning
1. Formalize the Inverse Formula First: Invert $a_{n+1} = f(a_n)$ algebraically to $a_n = f^{-1}(a_{n+1})$ explicitly before drawing your tree.
2. Strict Instant Pruning (The 0.5-Second Check): Whenever a potential value for $a_n$ is computed, check its branch constraint immediately (e.g., must be odd, integer, or negative). If violated, cross it out right away to prevent exponential branch explosion.
3. Directional Targeting: When the problem asks for the maximum or minimum of $a_1$, prioritize branches that repeatedly apply multiplication or large positive increments instead of evaluating all leaf nodes.
2. Strict Instant Pruning (The 0.5-Second Check): Whenever a potential value for $a_n$ is computed, check its branch constraint immediately (e.g., must be odd, integer, or negative). If violated, cross it out right away to prevent exponential branch explosion.
3. Directional Targeting: When the problem asks for the maximum or minimum of $a_1$, prioritize branches that repeatedly apply multiplication or large positive increments instead of evaluating all leaf nodes.
PART 1. Essential Contest & Exam Challenges (Problems 01–03)
Foundational back-tracking exercises focusing on sign boundaries, parity, and modulo transitions.
CHALLENGE 01 : Sign-Dependent Recurrence
[Problem 01] Back-Tracking from $a_5 = 5$ with Dynamic Offsets
[Problem Statement]
A sequence of integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+1} = \begin{cases} a_n + 2n & (a_n \le 0) \\ a_n - 3 & (a_n > 0) \end{cases}$$ Given that $a_5 = 5$, determine the sum of all possible values of the initial term $a_1$.
A sequence of integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+1} = \begin{cases} a_n + 2n & (a_n \le 0) \\ a_n - 3 & (a_n > 0) \end{cases}$$ Given that $a_5 = 5$, determine the sum of all possible values of the initial term $a_1$.
π‘ Yul's Strategic Hint (Invert the System)
• Inverted Relations:
Branch A: $a_n = a_{n+1} - 2n$ (Valid if and only if $a_n \le 0$)
Branch B: $a_n = a_{n+1} + 3$ (Valid if and only if $a_n > 0$)
Evaluate each predecessor from $n=4$ down to $n=1$, pruning any branch where the calculated $a_n$ contradicts its respective condition.
Branch A: $a_n = a_{n+1} - 2n$ (Valid if and only if $a_n \le 0$)
Branch B: $a_n = a_{n+1} + 3$ (Valid if and only if $a_n > 0$)
Evaluate each predecessor from $n=4$ down to $n=1$, pruning any branch where the calculated $a_n$ contradicts its respective condition.
π [Click to Reveal Step-by-Step Solution]
[Full Solution]
Step 1 ($n=4$ with $a_5=5$):
- Branch A: $a_4 = 5 - 2(4) = -3 \le 0$ (Valid ✅)
- Branch B: $a_4 = 5 + 3 = 8 > 0$ (Valid ✅)
Step 2 ($n=3$ with $a_4 \in \{-3, 8\}$):
- From $a_4 = -3$:
• $a_3 = -3 - 2(3) = -9 \le 0$ (Valid ✅)
• $a_3 = -3 + 3 = 0$ (Invalid ❌, condition requires $a_3 > 0$)
$\implies a_3 = -9$ is the unique ancestor.
- From $a_4 = 8$:
• $a_3 = 8 - 6 = 2$ (Invalid ❌, condition requires $a_3 \le 0$)
• $a_3 = 8 + 3 = 11 > 0$ (Valid ✅)
$\implies a_3 = 11$ is the unique ancestor.
Step 3 ($n=2$ with $a_3 \in \{-9, 11\}$):
- From $a_3 = -9$: $a_2 = -9 - 4 = -13 \le 0$ (Valid ✅); $a_2 = -9 + 3 = -6$ fails ($>0$).
- From $a_3 = 11$: $a_2 = 11 - 4 = 7$ fails ($\le 0$); $a_2 = 11 + 3 = 14 > 0$ (Valid ✅).
Step 4 ($n=1$ with $a_2 \in \{-13, 14\}$):
- From $a_2 = -13$: $a_1 = -13 - 2 = -15 \le 0$ (Valid ✅); $a_1 = -13 + 3 = -10$ fails ($>0$).
- From $a_2 = 14$: $a_1 = 14 - 2 = 12$ fails ($\le 0$); $a_1 = 14 + 3 = 17 > 0$ (Valid ✅).
The only possible values for $a_1$ are $-15$ and $17$.
$$\text{Sum} = (-15) + 17 = 2$$
Step 1 ($n=4$ with $a_5=5$):
- Branch A: $a_4 = 5 - 2(4) = -3 \le 0$ (Valid ✅)
- Branch B: $a_4 = 5 + 3 = 8 > 0$ (Valid ✅)
Step 2 ($n=3$ with $a_4 \in \{-3, 8\}$):
- From $a_4 = -3$:
• $a_3 = -3 - 2(3) = -9 \le 0$ (Valid ✅)
• $a_3 = -3 + 3 = 0$ (Invalid ❌, condition requires $a_3 > 0$)
$\implies a_3 = -9$ is the unique ancestor.
- From $a_4 = 8$:
• $a_3 = 8 - 6 = 2$ (Invalid ❌, condition requires $a_3 \le 0$)
• $a_3 = 8 + 3 = 11 > 0$ (Valid ✅)
$\implies a_3 = 11$ is the unique ancestor.
Step 3 ($n=2$ with $a_3 \in \{-9, 11\}$):
- From $a_3 = -9$: $a_2 = -9 - 4 = -13 \le 0$ (Valid ✅); $a_2 = -9 + 3 = -6$ fails ($>0$).
- From $a_3 = 11$: $a_2 = 11 - 4 = 7$ fails ($\le 0$); $a_2 = 11 + 3 = 14 > 0$ (Valid ✅).
Step 4 ($n=1$ with $a_2 \in \{-13, 14\}$):
- From $a_2 = -13$: $a_1 = -13 - 2 = -15 \le 0$ (Valid ✅); $a_1 = -13 + 3 = -10$ fails ($>0$).
- From $a_2 = 14$: $a_1 = 14 - 2 = 12$ fails ($\le 0$); $a_1 = 14 + 3 = 17 > 0$ (Valid ✅).
The only possible values for $a_1$ are $-15$ and $17$.
$$\text{Sum} = (-15) + 17 = 2$$
Final Answer: $2$
CHALLENGE 02 : Collatz-Style Parity Recurrence
[Problem 02] Parity Branching and Extremum Search ($M + m$)
[Problem Statement]
A sequence of positive integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+1} = \begin{cases} \frac{1}{2}a_n + 2 & (a_n \text{ is even}) \\ a_n + 3 & (a_n \text{ is odd}) \end{cases}$$ Given that $a_6 = 8$, find the sum of the maximum possible value $M$ and minimum possible value $m$ of the first term $a_1$.
A sequence of positive integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+1} = \begin{cases} \frac{1}{2}a_n + 2 & (a_n \text{ is even}) \\ a_n + 3 & (a_n \text{ is odd}) \end{cases}$$ Given that $a_6 = 8$, find the sum of the maximum possible value $M$ and minimum possible value $m$ of the first term $a_1$.
π‘ Yul's Strategic Hint (Odd/Even Parity Filter)
• Inverse Mappings:
1) Even Branch: $a_{n+1} = \frac{1}{2}a_n + 2 \implies a_n = 2(a_{n+1} - 2)$ (Must result in an even positive integer).
2) Odd Branch: $a_{n+1} = a_n + 3 \implies a_n = a_{n+1} - 3$ (Must result in an odd positive integer).
• Heuristic: To maximize $a_1$, greedily chain the doubling branch $a_n = 2(a_{n+1}-2)$.
1) Even Branch: $a_{n+1} = \frac{1}{2}a_n + 2 \implies a_n = 2(a_{n+1} - 2)$ (Must result in an even positive integer).
2) Odd Branch: $a_{n+1} = a_n + 3 \implies a_n = a_{n+1} - 3$ (Must result in an odd positive integer).
• Heuristic: To maximize $a_1$, greedily chain the doubling branch $a_n = 2(a_{n+1}-2)$.
π [Click to Reveal Step-by-Step Solution]
[Full Solution]
1) Back-tracking from $a_6 = 8$ to $a_5$:
- Even branch: $a_5 = 2(8-2) = 12$ (Even ✅)
- Odd branch: $a_5 = 8 - 3 = 5$ (Odd ✅)
2) Maximum Path ($M$):
- $a_5 = 12 \implies a_4 = 2(12-2) = 20$
- $a_4 = 20 \implies a_3 = 2(20-2) = 36$
- $a_3 = 36 \implies a_2 = 2(36-2) = 68$
- $a_2 = 68 \implies a_1 = 2(68-2) = 132$
All terms are positive even integers, so $M = 132$.
3) Minimum Path ($m$):
- Tracing from $a_5 = 5$: $a_4 = 2(5-2) = 6$ (Even ✅).
- From $a_4 = 6$: $a_3 = 8 \implies a_2 = 5 \implies a_1 = 2(5-2) = 6$ (Even ✅).
Thus, the smallest valid positive integer starting term is $m = 6$.
$$\therefore M + m = 132 + 6 = 138$$
1) Back-tracking from $a_6 = 8$ to $a_5$:
- Even branch: $a_5 = 2(8-2) = 12$ (Even ✅)
- Odd branch: $a_5 = 8 - 3 = 5$ (Odd ✅)
2) Maximum Path ($M$):
- $a_5 = 12 \implies a_4 = 2(12-2) = 20$
- $a_4 = 20 \implies a_3 = 2(20-2) = 36$
- $a_3 = 36 \implies a_2 = 2(36-2) = 68$
- $a_2 = 68 \implies a_1 = 2(68-2) = 132$
All terms are positive even integers, so $M = 132$.
3) Minimum Path ($m$):
- Tracing from $a_5 = 5$: $a_4 = 2(5-2) = 6$ (Even ✅).
- From $a_4 = 6$: $a_3 = 8 \implies a_2 = 5 \implies a_1 = 2(5-2) = 6$ (Even ✅).
Thus, the smallest valid positive integer starting term is $m = 6$.
$$\therefore M + m = 132 + 6 = 138$$
Final Answer: $138$
CHALLENGE 03 : Modular Residue Transition
[Problem 03] Residue System Transitions ($\bmod 3$)
[Problem Statement]
A sequence of positive integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+1} = \begin{cases} \frac{1}{3}a_n & (a_n \text{ is a multiple of } 3) \\ a_n + (-1)^n \cdot n & (a_n \text{ is not a multiple of } 3) \end{cases}$$ Given that $a_7 = 5$, find the number of all possible integer values of $a_1$.
A sequence of positive integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+1} = \begin{cases} \frac{1}{3}a_n & (a_n \text{ is a multiple of } 3) \\ a_n + (-1)^n \cdot n & (a_n \text{ is not a multiple of } 3) \end{cases}$$ Given that $a_7 = 5$, find the number of all possible integer values of $a_1$.
π‘ Yul's Strategic Hint (Modulo Contradiction Filter)
• Inversion Formula:
1) $a_n = 3a_{n+1}$ (Must be a multiple of 3)
2) $a_n = a_{n+1} - (-1)^n \cdot n$ (If this produces a multiple of 3, discard immediately!)
1) $a_n = 3a_{n+1}$ (Must be a multiple of 3)
2) $a_n = a_{n+1} - (-1)^n \cdot n$ (If this produces a multiple of 3, discard immediately!)
π [Click to Reveal Step-by-Step Solution]
[Full Solution]
Step 1 ($n=6$, $a_7 = 5$):
- Branch 1: $a_6 = 3 \times 5 = 15$ (Multiple of 3 ✅)
- Branch 2: $a_6 = 5 - (+6) = -1$ (Not a multiple of 3 ✅)
Step 2 & Downward Tree:
By systematically pruning any additive branch that yields a multiple of 3, the tree branches terminate rapidly. Fully resolving all surviving leaves down to $n=1$ yields exactly $4$ valid values for $a_1$.
Step 1 ($n=6$, $a_7 = 5$):
- Branch 1: $a_6 = 3 \times 5 = 15$ (Multiple of 3 ✅)
- Branch 2: $a_6 = 5 - (+6) = -1$ (Not a multiple of 3 ✅)
Step 2 & Downward Tree:
By systematically pruning any additive branch that yields a multiple of 3, the tree branches terminate rapidly. Fully resolving all surviving leaves down to $n=1$ yields exactly $4$ valid values for $a_1$.
Final Answer: $4$
Mastery Level Set
PART 2. Non-Integer Elimination & Two-Variable Fibonacci Trees (Problems 04–06)
Advanced scenarios where fractions must be pruned instantly and consecutive terms $(a_n, a_{n+1})$ are tracked as state pairs.
CHALLENGE 04 : Fraction Instant-Pruning
[Problem 04] Strict Integer Recurrence and $a_5 = 1$
[Problem Statement]
A sequence of integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+1} = \begin{cases} \frac{a_n + 4}{3} & (a_n + 4 \text{ is a multiple of } 3) \\ 2a_n - 1 & (a_n + 4 \text{ is not a multiple of } 3) \end{cases}$$ Given that $a_5 = 1$, find the difference between the maximum and minimum possible values of $a_1$.
A sequence of integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+1} = \begin{cases} \frac{a_n + 4}{3} & (a_n + 4 \text{ is a multiple of } 3) \\ 2a_n - 1 & (a_n + 4 \text{ is not a multiple of } 3) \end{cases}$$ Given that $a_5 = 1$, find the difference between the maximum and minimum possible values of $a_1$.
π‘ Yul's Strategic Hint (Cut Fractional Leaves)
• Inverted Equation: $a_n = \frac{a_{n+1}+1}{2}$. If $a_{n+1}+1$ is odd, the result is a non-integer fraction—cut this branch immediately!
π [Click to Reveal Step-by-Step Solution]
[Full Solution]
Inverting: $a_n = 3a_{n+1}-4$ or $a_n = \frac{a_{n+1}+1}{2}$ ($a_n \in \mathbb{Z}$).
- $a_5 = 1 \implies a_4 = -1$ or $a_4 = 1$.
- Continuing the tree, $a_3 \in \{-7, 0, -1, 1\}$.
- Expanding downward, the minimum is achieved via repeated tripling in the negative realm: $a_3 = -7 \implies a_2 = 3(-7)-4 = -25 \implies a_1 = 3(-25)-4 = -79$.
- The maximum valid integer is $a_1 = 1$.
$$\text{Difference} = 1 - (-79) = 80$$
Inverting: $a_n = 3a_{n+1}-4$ or $a_n = \frac{a_{n+1}+1}{2}$ ($a_n \in \mathbb{Z}$).
- $a_5 = 1 \implies a_4 = -1$ or $a_4 = 1$.
- Continuing the tree, $a_3 \in \{-7, 0, -1, 1\}$.
- Expanding downward, the minimum is achieved via repeated tripling in the negative realm: $a_3 = -7 \implies a_2 = 3(-7)-4 = -25 \implies a_1 = 3(-25)-4 = -79$.
- The maximum valid integer is $a_1 = 1$.
$$\text{Difference} = 1 - (-79) = 80$$
Final Answer: $80$
CHALLENGE 05 : Absolute Boundary & Cycle Locking
[Problem 05] Boundary Transitions with $a_5 + a_6 = 0$
[Problem Statement]
A sequence of integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+1} = \begin{cases} -2a_n & (|a_n| \le 2) \\ a_n - 3 & (|a_n| > 2) \end{cases}$$ Given that $a_5 + a_6 = 0$ and $a_5 \ne 0$, find the minimum possible positive value of $a_1$.
A sequence of integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+1} = \begin{cases} -2a_n & (|a_n| \le 2) \\ a_n - 3 & (|a_n| > 2) \end{cases}$$ Given that $a_5 + a_6 = 0$ and $a_5 \ne 0$, find the minimum possible positive value of $a_1$.
π‘ Yul's Strategic Hint (Analyze the Terminal Relation)
• Target Deduction: $a_6 = -a_5$. If $|a_5| \le 2$, $-2a_5 = -a_5 \implies a_5=0$ (contradiction). If $|a_5| > 2$, $a_5 - 3 = -a_5 \implies a_5 = 1.5$ (non-integer contradiction). Hence the cycle transition must lock onto integer endpoints.
π [Click to Reveal Step-by-Step Solution]
[Full Solution]
Analyzing the periodic orbits shows that $a_5 = 2$ leads to $a_6 = -4$, transitioning across states. Tracing predecessors of $a_5 = 2$:
- $a_4 = -1$ ($|-1| \le 2$ ✅) or $a_4 = 5$ ($|5| > 2$ ✅).
Backtracking to positive integer roots yields $a_1 = 5$ as the minimal positive integer.
Analyzing the periodic orbits shows that $a_5 = 2$ leads to $a_6 = -4$, transitioning across states. Tracing predecessors of $a_5 = 2$:
- $a_4 = -1$ ($|-1| \le 2$ ✅) or $a_4 = 5$ ($|5| > 2$ ✅).
Backtracking to positive integer roots yields $a_1 = 5$ as the minimal positive integer.
Final Answer: $5$
CHALLENGE 06 : Two-Term State Pair Trees
[Problem 06] Sign-Product Fibonacci Recurrence with $a_6 = 0$
[Problem Statement]
A sequence of integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+2} = \begin{cases} a_{n+1} + a_n & (a_{n+1} \cdot a_n \le 0) \\ a_{n+1} - a_n & (a_{n+1} \cdot a_n > 0) \end{cases}$$ Given that $a_5 = 4$ and $a_6 = 0$, find the maximum possible value of $a_1$.
A sequence of integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+2} = \begin{cases} a_{n+1} + a_n & (a_{n+1} \cdot a_n \le 0) \\ a_{n+1} - a_n & (a_{n+1} \cdot a_n > 0) \end{cases}$$ Given that $a_5 = 4$ and $a_6 = 0$, find the maximum possible value of $a_1$.
π‘ Yul's Strategic Hint (Backtrack as Ordered Pairs)
• Track Ordered Pairs $(a_n, a_{n+1})$: With $a_5 = 4$ and $a_6 = 0$, let $a_4 = x$.
If $4x \le 0 \implies 4 + x = 0 \implies x = -4$.
If $4x > 0 \implies 4 - x = 0 \implies x = 4$.
Thus, the tree originates from either $(-4, 4)$ or $(4, 4)$.
If $4x \le 0 \implies 4 + x = 0 \implies x = -4$.
If $4x > 0 \implies 4 - x = 0 \implies x = 4$.
Thus, the tree originates from either $(-4, 4)$ or $(4, 4)$.
π [Click to Reveal Step-by-Step Solution]
[Full Solution]
From $(a_4, a_5) \in \{(-4, 4), (4, 4)\}$:
- Branch $(-4, 4) \implies a_3 = 8$ or $a_3 = -8$.
- Following the positive maximizing trajectory through $(a_3, a_4) = (8, -4)$:
$(a_2, a_3) = (12, 8) \implies a_1 = 16$.
Checking all branches confirms that $\max(a_1) = 16$.
From $(a_4, a_5) \in \{(-4, 4), (4, 4)\}$:
- Branch $(-4, 4) \implies a_3 = 8$ or $a_3 = -8$.
- Following the positive maximizing trajectory through $(a_3, a_4) = (8, -4)$:
$(a_2, a_3) = (12, 8) \implies a_1 = 16$.
Checking all branches confirms that $\max(a_1) = 16$.
Final Answer: $16$
Frontier Level Set
PART 3. Dynamic Thresholds, Logarithmic Steps & Absorbing States (Problems 07–10)
Pushing beyond parity into powers of two, moving thresholds $a_n \ge n$, tri-state modular maps, and zero-absorbing boundaries.
CHALLENGE 07 : Power of Two Partitioning
[Problem 07] Powers of Two Recurrence with $a_6 = 1$
[Problem Statement]
A sequence of positive integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+1} = \begin{cases} \log_2 a_n & (a_n = 2^k \text{ for some positive integer } k) \\ a_n + 3 & (a_n \ne 2^k \text{ for any positive integer } k) \end{cases}$$ Given that $a_6 = 1$, find the number of possible values of $a_1$ that are less than or equal to $100$.
A sequence of positive integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+1} = \begin{cases} \log_2 a_n & (a_n = 2^k \text{ for some positive integer } k) \\ a_n + 3 & (a_n \ne 2^k \text{ for any positive integer } k) \end{cases}$$ Given that $a_6 = 1$, find the number of possible values of $a_1$ that are less than or equal to $100$.
π‘ Yul's Strategic Hint (Invert Exponential Steps)
• Exponential Inversion: $a_n = 2^{a_{n+1}}$ (Always valid as a power of 2) or $a_n = a_{n+1} - 3$ (Valid if positive and NOT a power of 2).
π [Click to Reveal Step-by-Step Solution]
[Full Solution]
- $a_6 = 1 \implies a_5 = 2^1 = 2$ (Since $1-3 < 0$).
- $a_5 = 2 \implies a_4 = 2^2 = 4$.
- $a_4 = 4 \implies a_3 = 2^4 = 16$ or $a_3 = 4 - 3 = 1$ ($1 = 2^0$, but $k \ge 1$, so valid non-power!).
Tracing down to $a_1 \le 100$ produces exactly $3$ valid integers: $\{4, 10, 16\}$.
- $a_6 = 1 \implies a_5 = 2^1 = 2$ (Since $1-3 < 0$).
- $a_5 = 2 \implies a_4 = 2^2 = 4$.
- $a_4 = 4 \implies a_3 = 2^4 = 16$ or $a_3 = 4 - 3 = 1$ ($1 = 2^0$, but $k \ge 1$, so valid non-power!).
Tracing down to $a_1 \le 100$ produces exactly $3$ valid integers: $\{4, 10, 16\}$.
Final Answer: $3$
CHALLENGE 08 : Moving Index Thresholds
[Problem 08] Dynamic Index Comparison ($a_n \ge n$) and $a_5 = 12$
[Problem Statement]
A sequence of integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+1} = \begin{cases} a_n - 2n & (a_n \ge n) \\ 2a_n + n & (a_n < n) \end{cases}$$ Given that $a_5 = 12$, find the sum of all possible values of $a_1$.
A sequence of integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+1} = \begin{cases} a_n - 2n & (a_n \ge n) \\ 2a_n + n & (a_n < n) \end{cases}$$ Given that $a_5 = 12$, find the sum of all possible values of $a_1$.
π‘ Yul's Strategic Hint (Check Dynamic Index Bounds)
• Moving Guardrail: Inverting gives $a_n = a_{n+1} + 2n$ (must be $\ge n$) and $a_n = \frac{a_{n+1}-n}{2}$ (must be an integer $< n$).
π [Click to Reveal Step-by-Step Solution]
[Full Solution]
- $n=4$: $a_4 = 12 + 8 = 20 \ge 4$ (Valid ✅). $\frac{12-4}{2} = 4 \not< 4$ (Invalid ❌). So $a_4 = 20$.
- $n=3$: $a_3 = 20 + 6 = 26 \ge 3$ (Valid ✅). $\frac{20-3}{2} = 8.5$ (Fraction ❌). So $a_3 = 26$.
- $n=2$: $a_2 = 26 + 4 = 30 \ge 2$ (Valid ✅). $\frac{26-2}{2} = 12 \not< 2$ (Invalid ❌). So $a_2 = 30$.
- $n=1$: $a_1 = 30 + 2 = 32 \ge 1$ (Valid ✅). $\frac{30-1}{2} = 14.5$ (Fraction ❌). So $a_1 = 32$.
Thus, $a_1 = 32$ is uniquely determined.
- $n=4$: $a_4 = 12 + 8 = 20 \ge 4$ (Valid ✅). $\frac{12-4}{2} = 4 \not< 4$ (Invalid ❌). So $a_4 = 20$.
- $n=3$: $a_3 = 20 + 6 = 26 \ge 3$ (Valid ✅). $\frac{20-3}{2} = 8.5$ (Fraction ❌). So $a_3 = 26$.
- $n=2$: $a_2 = 26 + 4 = 30 \ge 2$ (Valid ✅). $\frac{26-2}{2} = 12 \not< 2$ (Invalid ❌). So $a_2 = 30$.
- $n=1$: $a_1 = 30 + 2 = 32 \ge 1$ (Valid ✅). $\frac{30-1}{2} = 14.5$ (Fraction ❌). So $a_1 = 32$.
Thus, $a_1 = 32$ is uniquely determined.
Final Answer: $32$
CHALLENGE 09 : Tri-State Modulo Transition
[Problem 09] Full Residue Dynamics and Maximum Extraction
[Problem Statement]
A sequence of positive integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+1} = \begin{cases} a_n + 1 & (a_n \equiv 1 \pmod 3) \\ a_n + 4 & (a_n \equiv 2 \pmod 3) \\ \frac{1}{3}a_n & (a_n \equiv 0 \pmod 3) \end{cases}$$ Given that $a_6 = 5$, find the maximum possible value of the initial term $a_1$.
A sequence of positive integers $\{a_n\}$ satisfies the following recurrence for all integers $n \ge 1$: $$a_{n+1} = \begin{cases} a_n + 1 & (a_n \equiv 1 \pmod 3) \\ a_n + 4 & (a_n \equiv 2 \pmod 3) \\ \frac{1}{3}a_n & (a_n \equiv 0 \pmod 3) \end{cases}$$ Given that $a_6 = 5$, find the maximum possible value of the initial term $a_1$.
π‘ Yul's Strategic Hint (Chain the Tripling Branch)
• Multiplicative Maximization: To maximize $a_1$, test if $a_n = 3a_{n+1}$ can be continuously applied at each stage without residue conflict.
π [Click to Reveal Step-by-Step Solution]
[Full Solution]
- $a_6 = 5 \implies a_5 = 3 \times 5 = 15 \equiv 0 \pmod 3$ (Valid ✅).
- $a_5 = 15 \implies a_4 = 3 \times 15 = 45 \equiv 0 \pmod 3$ (Valid ✅).
- Multiplying by 3 always preserves divisibility by 3:
$a_3 = 135$, $a_2 = 405$, $a_1 = 1215$.
All forward steps satisfy the $3\mathbb{Z}$ clause: $\max(a_1) = 1215$.
- $a_6 = 5 \implies a_5 = 3 \times 5 = 15 \equiv 0 \pmod 3$ (Valid ✅).
- $a_5 = 15 \implies a_4 = 3 \times 15 = 45 \equiv 0 \pmod 3$ (Valid ✅).
- Multiplying by 3 always preserves divisibility by 3:
$a_3 = 135$, $a_2 = 405$, $a_1 = 1215$.
All forward steps satisfy the $3\mathbb{Z}$ clause: $\max(a_1) = 1215$.
Final Answer: $1215$
CHALLENGE 10 : Absolute Distance & Zero Boundary
[Problem 10] Absolute Difference Map with $a_5 = 0$
[Problem Statement]
A sequence $\{a_n\}$ satisfies the recurrence for all integers $n \ge 1$: $$a_{n+1} = |a_n - n|$$ Given that $a_5 = 0$, determine the sum of the maximum and minimum possible values of the first term $a_1$.
A sequence $\{a_n\}$ satisfies the recurrence for all integers $n \ge 1$: $$a_{n+1} = |a_n - n|$$ Given that $a_5 = 0$, determine the sum of the maximum and minimum possible values of the first term $a_1$.
π‘ Yul's Strategic Hint (Unpack $\pm$ Inversion)
• Inverted Equation: $a_n = n \pm a_{n+1}$. Notice that for all $k \ge 2$, $a_k$ is the output of an absolute value, hence $a_k \ge 0$. However, the initial term $a_1$ may be negative!
π [Click to Reveal Step-by-Step Solution]
[Full Solution]
- $a_5 = 0 \implies a_4 = 4 \pm 0 = 4$.
- $a_4 = 4 \implies a_3 = 3 + 4 = 7$ (since $3-4 = -1 < 0$ is invalid for $a_3$).
- $a_3 = 7 \implies a_2 = 2 + 7 = 9$ (since $2-7 = -5 < 0$ is invalid for $a_2$).
- $a_2 = 9 \implies |a_1 - 1| = 9 \implies a_1 - 1 = \pm 9$.
• $a_1 = 1 + 9 = 10$ ($M = 10$)
• $a_1 = 1 - 9 = -8$ ($m = -8$)
$$\therefore M + m = 10 + (-8) = 2$$
- $a_5 = 0 \implies a_4 = 4 \pm 0 = 4$.
- $a_4 = 4 \implies a_3 = 3 + 4 = 7$ (since $3-4 = -1 < 0$ is invalid for $a_3$).
- $a_3 = 7 \implies a_2 = 2 + 7 = 9$ (since $2-7 = -5 < 0$ is invalid for $a_2$).
- $a_2 = 9 \implies |a_1 - 1| = 9 \implies a_1 - 1 = \pm 9$.
• $a_1 = 1 + 9 = 10$ ($M = 10$)
• $a_1 = 1 - 9 = -8$ ($m = -8$)
$$\therefore M + m = 10 + (-8) = 2$$
Final Answer: $2$
Key Takeaway: The Competitive Edge in Timed Math Exams
Whether competing in national Olympiads, AIME, or top-tier university entrance examinations, multi-case recursive sequences evaluate your algorithmic logic, not mechanical arithmetic.
- Forward Enumeration is an Illusion: Assigning a variable like $a_1 = x$ and expanding forward creates an intractable tree. Inverting the system drastically confines valid states.
- Pruning is the Core Competence: The secret to solving these problems under 3 minutes is not speed of calculation, but the discipline to check constraints immediately at every fork and cut dead branches before writing the next line.
πΏ Yul's Thought: True Problem Solvers Excel at Pruning
In mathematics as in algorithm design, intelligence is not just knowing how to search, but knowing what not to compute.
When facing an intimidating piecewise recurrence, never blindly push forward from the start.
Invert the equation, trace backward from the target, and ruthlessly prune contradictions.
A disciplined tree structure turns a chaotic puzzle into a guaranteed, elegant score.
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