Given $2017$ positive real numbers $a_1,a_2,\dots ,a_{2017}$. For each $n>2017$, set $$a_n=\max\{ a_{i_1}a_{i_2}a_{i_3}|i_1+i_2+i_3=n, 1\leq i_1\leq i_2\leq i_3\leq n-1\}.$$Prove that there exists a positive integer $m\leq 2017$ and a positive integer $N>4m$ such that $a_na_{n-4m}=a_{n-2m}^2$ for every $n>N$.
Problem
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Tags: algebra
jeneva
26.03.2017 18:41
It's somehow remind me of IMO2010#6 .
guangzhou-2015
27.03.2017 08:00
any solution
bobthesmartypants
05.04.2017 22:25
Define the quality of the number $a_i$ where $1\le i\le 2017$ as $a_i^{1/i}$.
Let $a_M$ where $1\le M\le 2017$ be the term with highest quality. If there are multiple terms with the same highest quality, let $M$ be the smallest integer so that $a_M$ that has the highest quality.
Step 1: $a_n$ is the product of an odd number of $a_i$ where $1\le i\le 2017$.
Proof: Simple by construction: if $a_n = a_{i_1}\cdots a_{i_k}$ and $i_k > 2017$, then by defintion $a_{i_k} = a_{i_k'}a_{i_{k+1}'}a_{i_{k+2}'}$, so eventually all indices will be reduced to the range $[1, 2017]$.
Call this product of $a_n$ into terms with indices in the range $[1,2017]$ its expansion.
Step 2: $a_m$ appears in the expansion of $a_n$ less than $M$ times, for $m\ne M$ and all large enough $n$.
Proof: Suppose $a_m$ appears at least $M$ times; then $a_n =a_m^M \cdot A$ for some integer $A$ equalling the rest of the expansion of $a_n$. Note that since $a_M$ has the highest quality, $a_m^{1/m} \le a_M^{1/M} \implies a_m^M \le a_M^m$, then $a_n = a_M^m \cdot A$ gives an expansion of $a_n$ with greater or equal value. Thus, we may assume that all expansions contain $a_m$ less than $M$ times for all $m\ne M$.
Step 3: for all large enough $B$, $a_{B+2M} = a_{B}\cdot a_M^2$.
Proof: Consider the set of $B$ such that $B\equiv c\pmod{2M}$, for some constant $0\le c < 2M$. By Step 2, we may find an element $B'$ in this set such that the product of the terms $a_m$ in its expansion where $m\ne M$ is maximal; let $a_{B'}=A\cdot a_M^k$ for integer $k$, and $A$ the rest of the expansion of $a_{B'}$. Now suppose $a_{B'+2M} = A'\cdot a_M^{k'} > A\cdot a_M^{k+2}$ for integer $k'$, and $A'$ is the rest of the expansion of $a_{B'+2M}$. Since $A$ is maximal, $A' < A\implies k' > k+2 \ge 2$. Then $A'\cdot a_M^{k'-2} > a\cdot a_M^k = a_{B'}$. However, since the indices of the expansion of $A'$ add up to $B'+2M - k'M$, the indices of the expansion of $A'\cdot a_M^{k'-2}$ add up to $B'$ so we have found a larger expansion of $a_{B'}$, contradiction. Thus, $a_{B'+2M} = A\cdot a_M^{k+2} = a_{B'}\cdot a_M^2$.
Inductively, we get that $a_{B'+2kM} = a_{B'}\cdot a_M^{2k}$, so $a_{B+2M} = a_B\cdot a_M^2$ for all $B \ge B'$.
In particular, by taking $N-4M$ to be the maximum of all the $B'$ over all cases of $B\equiv c\pmod{2M}$, we have $a_{n+2M} = a_n\cdot a_M^2$ for all $n \ge N-4M$. But then clearly $a_na_{n-4M} = a_{n-4M}^2 \cdot a_M^4 = a_{n-2M}^2$ for all $n\ge N$, and we are done. $\Box$
CJA
05.08.2017 03:58
any one have orther solutions?
bobthesmartypants
14.10.2017 01:54
My solution is a little hard to understand and I think it has a few small holes in it, so I'll try to post a more detailed solution.
Lemma 1: $a_n$ can be written as the product of an odd number ($>1$) of elements from the set $\{a_1, a_2, \ldots, a_{2017}\}$.
Proof: just use the given definition for any terms $a_n$ with $n > 2017$. Define the representation of $a_n$ in terms of numbers from $\{a_1, a_2, \ldots, a_{2017}\}$ as the expansion of $a_n$.
Lemma 2: Given any odd-sized ($>1$) multiset $\{i_1, i_2, \ldots , i_k\}$ with elements from $\{1, 2, \ldots, 2017\}$ satisfying $i_1+i_2+\cdots +i_k = n$, $$a_n\ge a_{i_1}a_{i_2}\cdots a_{i_k}.$$
Proof: Note that the statement for $k=3$ is true by definition. We can arrive at the expansion $a_{i_1}a_{i_2}\cdots a_{i_k}$ by using the lemma for $k=3$ repeatedly until the desired expansion is achieved, so the lemma is true for all odd $k$.
Define the quality of the number $a_i$ where $1\le i\le 2017$ as $a_i^{1/i}$.
Let $a_M$ where $1\le M\le 2017$ satisfy that $M$ is the smallest integer such that $a_M$ has the highest possible quality.
Lemma 3: For all $1\le m\le 2017$, $a_m^M \le a_M^m$.
Proof: Equivalently, $a_m^{1/m} \le a_M^{1/M}$ which is true by maximality of $a_M$'s quality.
Lemma 4: The number of times $a_m$ appears in the expansion of $a_n$ for $m\ne M$ is less than $2M$.
Proof: Suppose $a_n = a_m^{2M}\cdot A$, where $A$ is the rest of the expansion of $a_n$. By Lemma 3, $$a_m^{2M} = (a_m^M)^2 \le (a_M^m)^2 = a_M^{2m}$$so $a_n = a_m^{2M}\cdot A \le a_M^{2m}\cdot A $ and by Lemma 2, $a_M^{2m}\cdot A \le a_n$, so in fact equality holds and we can exchange $a_m^{2M}$ with $a_M^{2m}$ without changing the value of $a_n$. After all such exchanges occur, all exponents of $a_m$ where $m\ne M$ will be less than $2M$ as desired.
In particular, Lemma 4 tells us the exponents of the expansion of $a_n$ are bounded except for the exponent of $a_M$.
Lemma 5: For all large enough $B$, $a_B+2M = a_B \cdot a_M^2$.
Proof: Let $c$ be a constant modulo $2M$. Consider the set $\mathcal{B}$ of $B$ such that $B\equiv c\pmod{2M}$. Denote $a_B = a_M^k\cdot A$ where $A$ is the part of the expansion of $a_B$ that does not contain any $a_M$. By Lemma 4, there exists a smallest number $B_0$ such that the expansion $a_{B_0} = a_M^{k_0} \cdot A_0$ maximizes $A_0$. I claim that $$a_B = a_{B_0} \cdot a_M^{(B-B_0)/M}$$for all $B\ge B_0$ with $B\in\mathcal{B}$. We use induction.
Base case: trivially, $a_{B_0}= a_{B_0}$.
Induction step: suppose $a_{B_0 + 2iM} = a_{B_0}\cdot a_M^{2i}$, and we want to show $a_{B_0 + 2(i+1)M} = a_{B_0}\cdot a_M^{2(i+1)}$. Suppose for the sake of contradiction that the expansion of $a_{B_0 + 2(i+1)M}$ is $$a_{B_0 + 2(i+1)M} = a_M^k\cdot A > a_{B_0+2iM} \cdot a_M^2 = a_{B_0} \cdot a_M^{2(i+1)}= a_M^{k_0 + 2(i+1)} \cdot A_0.$$Since $A_0$ is maximal, $A_0\ge A$ so $k\ge k_0 + 2(i+1) \ge 2$. Thus, $$a_M^{k-2} \cdot A > a_M^{k_0 + 2i} \cdot A_0 = a_M^{2i}\cdot a_{B_0} = a_{B_0+2iM}.$$But by Lemma 2, $a_M^{k-2} \cdot A \le a_{B_0+2iM}$ contradiction, so $$a_{B_0 + 2(i+1)M} = a_{B_0+2iM} \cdot a_M^2 =a_{B_0} \cdot a_M^{2(i+1)} $$as desired.
In particular, for all $B\ge B_0$ such that $B\in \mathcal{B}$, $$a_{B+4M}a_B = a_{B_0}\cdot a_M^{(B+4M-B_0)/M} \cdot a_{B_0} \cdot a_M^{(B-B_0)/M} = a_{B_0}^2 \cdot a_M^{2(B-B_0+2M)/M} = a_{B+2M}^2$$which is exactly the problem statement we wish to prove. Since this works for any $c$ that we choose, simply take $m=M$ and $N$ equals the maximum $B_0$ over all $c$'s for the original problem statment, done.
Lam.DL.01
14.10.2017 14:07
bobthesmartypants wrote: My solution is a little hard to understand and I think it has a few small holes in it, so I'll try to post a more detailed solution.
Lemma 1: $a_n$ can be written as the product of an odd number ($>1$) of elements from the set $\{a_1, a_2, \ldots, a_{2017}\}$.
Proof: just use the given definition for any terms $a_n$ with $n > 2017$. Define the representation of $a_n$ in terms of numbers from $\{a_1, a_2, \ldots, a_{2017}\}$ as the expansion of $a_n$.
Lemma 2: Given any odd-sized ($>1$) multiset $\{i_1, i_2, \ldots , i_k\}$ with elements from $\{1, 2, \ldots, 2017\}$ satisfying $i_1+i_2+\cdots +i_k = n$, $$a_n\ge a_{i_1}a_{i_2}\cdots a_{i_k}.$$
Proof: Note that the statement for $k=3$ is true by definition. We can arrive at the expansion $a_{i_1}a_{i_2}\cdots a_{i_k}$ by using the lemma for $k=3$ repeatedly until the desired expansion is achieved, so the lemma is true for all odd $k$.
Define the quality of the number $a_i$ where $1\le i\le 2017$ as $a_i^{1/i}$.
Let $a_M$ where $1\le M\le 2017$ satisfy that $M$ is the smallest integer such that $a_M$ has the highest possible quality.
Lemma 3: For all $1\le m\le 2017$, $a_m^M \le a_M^m$.
Proof: Equivalently, $a_m^{1/m} \le a_M^{1/M}$ which is true by maximality of $a_M$'s quality.
Lemma 4: The number of times $a_m$ appears in the expansion of $a_n$ for $m\ne M$ is less than $2M$.
Proof: Suppose $a_n = a_m^{2M}\cdot A$, where $A$ is the rest of the expansion of $a_n$. By Lemma 3, $$a_m^{2M} = (a_m^M)^2 \le (a_M^m)^2 = a_M^{2m}$$so $a_n = a_m^{2M}\cdot A \le a_M^{2m}\cdot A $ and by Lemma 2, $a_M^{2m}\cdot A \le a_n$, so in fact equality holds and we can exchange $a_m^{2M}$ with $a_M^{2m}$ without changing the value of $a_n$. After all such exchanges occur, all exponents of $a_m$ where $m\ne M$ will be less than $2M$ as desired.
In particular, Lemma 4 tells us the exponents of the expansion of $a_n$ are bounded except for the exponent of $a_M$.
Lemma 5: For all large enough $B$, $a_B+2M = a_B \cdot a_M^2$.
Proof: Let $c$ be a constant modulo $2M$. Consider the set $\mathcal{B}$ of $B$ such that $B\equiv c\pmod{2M}$. Denote $a_B = a_M^k\cdot A$ where $A$ is the part of the expansion of $a_B$ that does not contain any $a_M$. By Lemma 4, there exists a smallest number $B_0$ such that the expansion $a_{B_0} = a_M^{k_0} \cdot A_0$ maximizes $A_0$. I claim that $$a_B = a_{B_0} \cdot a_M^{(B-B_0)/M}$$for all $B\ge B_0$ with $B\in\mathcal{B}$. We use induction.
Base case: trivially, $a_{B_0}= a_{B_0}$.
Induction step: suppose $a_{B_0 + 2iM} = a_{B_0}\cdot a_M^{2i}$, and we want to show $a_{B_0 + 2(i+1)M} = a_{B_0}\cdot a_M^{2(i+1)}$. Suppose for the sake of contradiction that the expansion of $a_{B_0 + 2(i+1)M}$ is $$a_{B_0 + 2(i+1)M} = a_M^k\cdot A > a_{B_0+2iM} \cdot a_M^2 = a_{B_0} \cdot a_M^{2(i+1)}= a_M^{k_0 + 2(i+1)} \cdot A_0.$$Since $A_0$ is maximal, $A_0\ge A$ so $k\ge k_0 + 2(i+1) \ge 2$. Thus, $$a_M^{k-2} \cdot A > a_M^{k_0 + 2i} \cdot A_0 = a_M^{2i}\cdot a_{B_0} = a_{B_0+2iM}.$$But by Lemma 2, $a_M^{k-2} \cdot A \le a_{B_0+2iM}$ contradiction, so $$a_{B_0 + 2(i+1)M} = a_{B_0+2iM} \cdot a_M^2 =a_{B_0} \cdot a_M^{2(i+1)} $$as desired.
In particular, for all $B\ge B_0$ such that $B\in \mathcal{B}$, $$a_{B+4M}a_B = a_{B_0}\cdot a_M^{(B+4M-B_0)/M} \cdot a_{B_0} \cdot a_M^{(B-B_0)/M} = a_{B_0}^2 \cdot a_M^{2(B-B_0+2M)/M} = a_{B+2M}^2$$which is exactly the problem statement we wish to prove. Since this works for any $c$ that we choose, simply take $m=M$ and $N$ equals the maximum $B_0$ over all $c$'s for the original problem statment, done.
Nice solution , bobthesmartypants
Lam.DL.01
14.10.2017 14:08
When I got the actual tests, I can just prove 2 theorem and fail the test
Attachments:
alexiaslexia
16.05.2021 09:22
This might be a not necessarily wordy Solution of a relatively simple sequence problem; but this presented Solution does not use the $\textit{strictly decreasing argument}$, but instead opted to completely characterize the $a_n$s. (Also this is a struggle to write up smh, it's possible for a mistake to lurk somewhere due to the abstract nature of the argument.) In the first part of this Solution, we will not explicitly consider the elements of $\{a_i\}_{i \geq 2018}$. Specifically, we will be focusing on the make-believe or hypothetical $\{b_i\}$ (which is undoubtedly related to $\{a_i\}$) where $b_i = a_i$ for $1 \leq i \leq 2017$ and $b_n$ may consist of any product \[ b_n = b_{i_1}b_{i_2}b_{i_3} \]as long as the sum $i_1+i_2+i_3$ equals to $n$. From this, define a $\textit{term}$ to be a product of $a_i$s (not necessarily different), a term's $\textit{cardinality}$ to be the number of $a_i$ inside it (a.k.a. $\sum_{i=1}^{2017} e_i$); and a term's $\textit{value}$ to be the number \[ \sum_{i=1}^{2017} i \cdot e_i \]where the term is represented as $\prod_{i=1}^{2017} a_i^{e_i}$. Next, define a term to be $\textit{reachable}$ iff we can construct a sequence $\{b_i\}$ so that $b_K$'s term is exactly that for some $K$. Note that we do not speak of reachable $\textit{numbers}$ but reachable $\textit{terms}$; a number may be representable in two ways (or equalling to two terms) with one of the two being reachable and the other being not reachable. For simplicity, denote the set of terms which is representable as $b_K$ to be $B_K$ ($K$ fixed). We first throw the direct conclusions out the door. $\color{green} \rule{6.9cm}{2pt}$ $\color{green} \clubsuit$ $ \boxed{\textbf{Invariant-ing and Extremal-ling.}}$ $\color{green} \clubsuit$ $\color{green} \rule{6.9cm}{2pt}$ Every element of $B_K$ must have odd cardinality and value $K$. Moreover, the value of $a_K$ is the maximum of all possible $\textit{reachable}$ terms in $B_K$. $\color{green} \rule{25cm}{0.2pt}$ $\color{green} \spadesuit$ $\boxed{\textbf{Proof.}}$ $\color{green} \spadesuit$ The first statement is easily proven by induction: the first $2017$ terms do indeed have cardinality $1$ and value according to their indices; and assuming $b_{i_1}, b_{i_2},b_{i_3}$ all satisfy the condition, it is easily inferred that $b_n$ does, too. The second is also reasonably simple to prove by induction: assuming $a_{i_1},a_{i_2}$ and $a_{i_3}$ are the maximal of of $B_{i_k}, 1 \leq k \leq 3$, then \[ a_n = \max{(a_{i_1}a_{i_2}a_{i_3})}, \ \forall i_1+i_2+i_3 = n \]is also the maximum of all products of three terms from the sets $B_{i_1}, B_{i_2},B_{i_3}$ with index sum $= n$.
It is easily provable that $a_K$'s size is more than or equal any reachable term with value $K$. Indeed, consider the sequence $\{b_i\}$ with $b_K$ equalling to the reachable term. Since it's provable by induction that $a_i \geq b_i$ for every $i$, then we get $a_K \geq b_K$.
Moreover, the elusive $a_K$ also belongs to the set $B_K$. As a result, $a_K$ is exactly equal to the maximal element of $B_K$; i.e. the greatest reachable term there is.
We are done for the first part. $\blacksquare$ Next, we will state the necessary and sufficient condition for a term to be reachable. $\color{red} \rule{3.7cm}{2pt}$ $\color{red} \clubsuit$ $\color{red} \boxed{\textbf{The Criterion.}}$ $\color{red} \clubsuit$ $\color{red} \rule{3.7cm}{2pt}$ Let $T$ be the set of all terms. Define the function \[ R_3: T \rightarrow \mathbb{N} \]where \[ R_3(t) = R_3(a_{i_1} \cdot \ldots \cdot a_{i_m}) = M_1 + M_2 + M_3 \]where $t \in T$'s term representation equals the expression in parentheses (in the middle) with not necessarily different indices; and $M_1, M_2, M_3$ denoting the max, second max and third max of the multiset $\{i_1,i_2,\ldots,i_m\} (1 \leq i_j \leq 2017)$.
I wanted to analyze the $\text{rad}$ equivalent of a term: where $\text{rad}$ is the usual NT function. However, while this function does register the terms $a_1,\ldots,a_{2017}$ as $\textit{primes}$, but its purpose is to identify the $\textit{overall reach}$ of the term; and that's obtained by summing the three largest indices.
Then, $R_3$ got its name.
We Claim that $t$ is reachable iff its cardinality is odd and $R_3(t) \geq 2018$. $\color{red} \rule{25cm}{0.2pt}$ $\color{red} \spadesuit$ $\color{red} \boxed{\textbf{Proof.}}$ $\color{red} \spadesuit$ Let $t$ satisfies $R_3(t) \geq 2018$. From this, form the sequence $\{b_i\}$ leading to \[ b_{\sum_{j = 1}^{m} i_j} = t \]First, form $b_{M_1+M_2+M_3} = a_{M_1}a_{M_2}a_{M_3}$. This can be done since $M_1+M_2+M_3 \geq 2018$. Further, if $b_k$ is (already) formed with \[ M_1 + M_2 + M_3 + \ \text{some remaining terms} \ = k \]we construct $b_{k + \ \text{two terms not selected yet}}$ by selecting $i_1 = k$, and $i_2,i_3$ equal to the two terms not selected. Since $m$ is odd, the desired $t$ will be a part of $\{b_i\}$. $\color{red} \rule{25cm}{0.2pt}$ On the flipside, let $t$ be reachable. We will prove the second statement by strong induction on $K$ (i.e. the value of $t$). Note that the case for $K = 2018$ is easily provable. Assume this statement is true for $K = k$. Then, letting $b_{k+1} = t$, consider the $i_1,i_2,i_3$ forming it. If at least one of $i_1,i_2,i_3$ is at least $2018$; or \[ \max{(i_1,i_2,i_3)} \geq 2018 \]then we can apply our induction hypothesis on its maximum, to get that \[ R_3(b_{k+1}) = R_3(b_{i_1}b_{i_2}b_{i_3}) \geq R_3(b_{\max{(i_1,i_2,i_3)}}) \geq 2018 \]Otherwise, let all of $i_1, i_2, i_3$ to be among the primary bases of $\{b_i\}$. Then, the value of $R_3(t)$ is exactly $k+1$; and we clearly can infer that $k+1 \geq 2018$. Conclusively, we get $R_3(t) \geq 2018$ for whatever representation $t$ has in $\{b_i\}$. $\blacksquare$ $\blacksquare$
If $t_1,t_2 \in T$ have the same value, and same image under $R_3$, then $t_1$ is reachable iff $t_2$ is.
In the last Section we prove the reachable-ness and uniformity of the large $a_i$s. Now we define some quality check counters to eliminate the blatantly incorrect terms (a.k.a. the reachables of $\{b_i\}$ that won't make the cut for $\{a_i\}$.) Call a term to be a $\textit{impossibly maximal}$ term if there must exist another reachable term with the same value that is greater than it. Call all other terms $\textit{possibly maximal}$, then define the set $P_K$ to be the set of possibly maximal terms. $\color{magenta} \rule{7.9cm}{2pt}$ $\color{magenta} \clubsuit$ $ \boxed{\textbf{Computations of Maximal Reachables.}}$ $\color{magenta} \clubsuit$ $\color{magenta} \rule{7.9cm}{2pt}$ We Claim that the set of possibly maximal $\textit{numbers}$ are finite. So, while there may be two primary base with equal quality (defined right after this), their end products (manifested as their $\textit{numbers}$) have finite variations. $\color{magenta} \rule{25cm}{0.3pt}$ $\color{magenta} \spadesuit$ $\boxed{\textbf{Proof.}}$ $\color{magenta} \spadesuit$ Define the $\textit{quality}$ of a primary base $a_i$ ($1 \leq i \leq 2017$) to be $a_i^{1/i}$. Call the index $C$ to be $\textit{good}$ if $a_C$ has the largest quality among all primary bases; and if there are more than one good index, select one (randomly) to be the good index. Now, we prove that all members of $P_K$ can be represented as \[ a_C^k \cdot \prod_{a_i \: \text{base}, i \ne C} a_i^{e_i}, \ \text{with} \ e_i \leq 2C+2 \]implying $P_K$'s cardinality to be finite. Suppose otherwise, and there exists an exponent $a_i^{e_i}$ of $t \in P_K$ where $i \ne C$ and $e_i \geq 2C+3$. Then, consider the term \[ t' = t \cdot \dfrac{a_C^{2i}}{a_i^{2C}} \]By simple ineq, we can prove that $t' \geq t$ (the ineq is strict if $i$ is not of maximum quality), and $t'$ is reachable with the same value. So, either we can prove that $t \not\in P_K$, or we can rename the representation of $t$ into something where all $a_i, i \ne C$ have a small exponent. $\color{blue} \rule{2.8cm}{2pt}$ $\color{blue} \diamondsuit$ $\color{blue} \boxed{\textbf{Finishing.}}$ $\color{blue} \diamondsuit$ $\color{blue} \rule{2.8cm}{2pt}$ Pick $m = C$. Now, we prove that for $K \geq N$, then \[ \dfrac{a_{K+2C}}{a_{K}} = a_C^2 \]Indeed, select \[ N = (2C+2) \cdot (1+2+\ldots+2017-C) + 1 \]For our final definition, for every $t \in P_K$, define $t$'s $\textit{quality loss}$ to be $\frac{a_C^{K/C}}{t}$. We will prove that $P_K$ and $P_{K+2C}$'s maximum terms have equal quality loss. This is relatively simple: $a_{K+2C}$ must have a $a_{C}^2$ in its representation, or there must exist an $a_i^{e_i}$ with $e_i \geq 2C+3$. So, \[ \dfrac{a_{K+2C}}{a_C^2} \in P_K \]Secondly. we can prove that $a_K \cdot a_C^2 \in B_{K+2C}$ (so $\max{(P_{K+2C})} = \max{(B_{K+2C})} \geq a_K \cdot a_C^2$). From here, we can get \[ \dfrac{a_K}{a_{K+2C}} \geq \dfrac{1}{a_C^2}, \dfrac{a_{K+2C}}{a_K} \geq a_C^2 \]This proves the statement. We are done. $\blacksquare$ $\blacksquare$ $\blacksquare$
Since the explanation is kinda long and scary, I'll mostly try to explain the relevance of categorizing $B_K$ with $R_3$ here.
In its core, the main goal of all those notations constricting the possible $t \in B_K$ is to handle all possible cases of $\{a_i\}$'s construction --- as a sequence which has 2017 degrees of freedom tend to be a bit wild.
$\color{green} \rule{25cm}{2pt}$
$\color{red} \text{Sec 1: What invariants/patterns are keeping the sequence in check?}$ Since $a_{i_1},a_{i_2}$ and $a_{i_3}$ is slightly too complicated to start with, I started with changing $2017$ into $2$ and $3$ into $2$. You can try this for yourself, but by standard bashing we get $m = 1$ if $a_1 = a \geq \sqrt{a_2} = \sqrt{b}$, and $m = 2$ otherwise.
Next, I tried to expand a bit by letting the primary base to be $a_1,a_2,a_3$. Quickly, this got out of hand, since
\[ a_6 = \max{(a^3c,a^2b^2,abc,b^3,c^2)} \]and this can be split in five separate cases (and this doesn't guarantee how the later terms go!)
Changing $3$ to $4$ brought a better sense of clarity, since
\[ a_7 = \max{(a^3d,a^2bc,abd,ac^2,b^2c,cd)} \]and all terms whose value equals $7$ are included in $a_7$'s consideration, except those with $a$ exponent $4$ or greater. Here I mistakenly concluded that $t$ is reachable iff its $a$ exponent is at most $K-4$. (This is true only when the primary base consists of four terms.)
Since the $2$ case is significantly less complicated than the $3$ case, I switched with it right after. Fixing the base to have three terms, $a_7$ and $a_8$'s value consists of these options:
\begin{align*}
a_7 &= \max{(a^4c,a^3b^2,ac^2,bc^2)} \\
a_8 &= \max{(a^3bc,a^2b^3,a^6b,a^3bc,bc^2)}
\end{align*}and this just becomes more complicated. While almost all terms are included there, I wondered why some (e.g. $a^2c^2$) is not fit for $a_8$. Retracing back to $a_1,a_2,a_3$ brings the answer: all the reachable terms must have either $3$ or $5$ cardinality.
$\color{green} \rule{25cm}{2pt}$
$\color{red} \text{Sec 2: Periods and Constructions.}$ Keeping in mind that $m = C$ for $a_C$ which provides the most $\textit{value}$, I was working from $\color{green} \clubsuit$ $ \boxed{\textbf{Invariant-ing and Extremal-ling}}$ $\color{green} \clubsuit$ (without the second assertion) and $\color{blue} \diamondsuit$ $\color{blue} \boxed{\textbf{Finishing}}$ $\color{blue} \diamondsuit$ to establish some sort of pattern among all $a_i$s.
Here I looked back at the problem condition: $N > 4m$ seems too superficial to be informative, and the condition $a_na_{n-4m} = a_{n-2m}^2$ is also very general. After changing it to be
\[ \dfrac{a_n}{a_{n-2m}} = \dfrac{a_{n-2m}}{a_{n-4m}} \]it became a bit clearer: if it can be proven that $\frac{a_n}{a_{n-2m}} = a_C^2$ for large $n$, we are done.
One could just finish by establishing that the quality loss of $a_i$ decreases afterward by picking
\[ a_{K+2C} = a_K \cdot a_C \cdot a_C \]but I didn't think of this earlier --- and I explicitly went on proving the possible quality loss to be finite (not decreasing).
Notes.If one goes down this path, it can be proven that all quality loss are in the form of
\[ \prod_{i = 1}^{2017} \left(\dfrac{q_C}{q_i}\right)^{e_i} \]with $q_i$ being the quality of $a_i$. Since this value is bounded by
\[ \text{max q. loss of} \ a_{2018},a_{2019},\ldots,a_{2018+2C-1} \]we can establish that the possibilities of quality loss is finite, since $\frac{q_C}{q_i} \geq 1$.
Since an infinite sequence with finitely many possibilities are constant after a certain point, then all $2C$ subsequences are basically constant after some point, and we are done.
Strangely enough, the Criterion came later than the idea of $\textit{recycling}$ $a_i^{2C}$; the addition of $2$ is done to preserve the parity of $t$'s cardinality. However, if I were to explicitly change $t$ into $t'$ to prove $\{a_i\}$'s uniformity, I'd need an explicit criterion on which term is actually formable.
Changing the base from $\{a_1,a_2,a_3\}$ into $\{a_1,\ldots,a_{69}\}$ does the trick: I can construct the initial push of $a_{M_1+M_2+M_3}$ whenever their sum is big enough (a.k.a. $70$ or more) but otherwise, no push can be strong enough to bring, for example, $a_1^{1969}a_2^{7}a_3^{6}a_4^{5}$ into a $b_{2021}$.
Finalizing this argument creates the function $R_3$ and a necessary-sufficient way to determine whether $t$ is included on $a_K$'s consideration (and the $b_K$ is just a substitute to remove the constrictive $\max{(a_{i_1}a_{i_2}a_{i_3})}$ out of the equation for the intro.)
Note: I came up with the second assertion in $\color{green} \clubsuit$ $ \boxed{\textbf{Invariant-ing and Extremal-ling}}$ after establishing the rest to confirm my hunch of $a_K$ being the maximal of all the reachables. This detail is probably the most intuitive of all the Solution and Motivation; yet, proving it takes significant brain power.
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Number1048576
07.05.2023 22:52
Let $b_i = ln(a_i)$ for all $i \in \mathbb{N}$ (since the $a_i$ are positive we can do this). Then $b_n = max(b_{i_1} + b_{i_2} + b_{i_3})$.
Let $b_k$ be the member of the $b_i$ with $i \leq 2017$ and with $k$ as large as possible such that the quantity $\frac{b_k}{k}$ is maximised, and write $b_k = ck$, where $c$ is some constant. Then note that for all $i > 2017$, $b_i \leq ci$. Also, let $x_n = cn - b_n$.
Note that if $x_j$ is the smallest value which ever appears in the sequence where $j \equiv b \pmod {2k}$ for some fixed $b$, then since $x_j \leq x_{j+2km} \leq x_j + 2x_k = x_j$, we must have $x_{j+2km} = x_j$ for all $m \in \mathbb{N}$. Therefore, since every number in the $x_i$ can be expressed as the sum of numbers in the $x_i$ with $i \leq 2017$, and since there are only a finite number of different numbers which can be formed as the sum of those numbers which are larger than the smallest number in the $x_i$ with $i \leq 2k$, such a minimum value must be reached, so eventually the $x_i$ are periodic $\pmod {2k}$, which means that the $b_i$ eventually satisfy $b_{n} - b_{n-2m} = b_{n-2m} - b_{n-4m}$, so the $a_i$ eventually satisfy $\frac{a_n}{a_{n-2m}} = \frac{a_{n-2m}}{a_{n-4m}}$, as required.