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I think it's $667^3$. $667^3$ can be achieved by taking $667$ $1$'s, followed by $667$ $2$'s, followed by $667$ $3$'s. Let's now prove that it's the maximum we can achieve. Break the sequence into maximal subsequences which are also arthimetic progressions of common difference $1$. Two such progressions don't intersect, so they don't interfere (we can't have $3$-element progressions with elements from two different such subsequences). Assume we have such a subsequence $a_{n_1},\ldots,a_{n_t}$. This contributes $t-2$ to $m$, so we can change the first $\lfloor\frac t3 \rfloor$ numbers into the same number (say $1$), then the next $\lfloor\frac t3\rfloor$ numbers into $2$, and the remeining numbers into $3$. This new sey contributes $\lfloor\frac t3 \rfloor^2(t-2\lfloor\frac t3 \rfloor)$ to $m$, and this is clearly $>t-2$. We do this for all subsequences of the form considered above, until we get, in the end, some $1$'s, some $2$'s, and some $3$'s. Furthermore, if we have $2$ followed by $1$ (or $3$ followed by $2$ or $1$), we can interchange these two and we increase $m$, so after performing all operations of this sort we get $a_i=1,\ \forall i\in\overline{1,a},\ a_i=2,\ \forall i\in\overline{a+1,a+b},\ a_i=3,\ \forall i\in\overline{a+b+1,a+b+c=2001}$. In this case $m=abc$, with $a+b+c=2001$, and after applying $AM-GM$ we find $m=abc\le \left(\frac{2001}3\right)^3=667^3$, Q.E.D.

Does the following work? As grobber has above, the answer is 667^3 with the example of 667 each of ones, twos, and threes. To prove that it is the maximum, take all of the a_i's and split them into 3 groups by residue mod 3. Say there are a, b, c elements in each group, so a+b+c=2001. Note any desired triple (a_i, a_j, a_k) must have one element from each group so there are at most abc elements. Then finish as grobber did.

Not as clean as the modular approach, but... If $a_i\ge a_{i+1}+1$ for some $i$, then switching $a_i$ and $a_{i+1}$ will clearly not decrease $m$. So assume $a_1\le\cdots\le a_{2001}$. Let there be $x_i\ge0$ indices $j$ with $a_j=i$, WLOG with $1\le i\le k$ ($k\ge3$ or else $m=0$). We have $S=\sum_{i=1}^{k}x_i=2001$. Then \[m=\sum_{i=1}^{k-2}x_ix_{i+1}x_{i+2}.\]Assume that $k\ge4$. Then \[m=\left(\sum_{i=1}^{k-4}x_ix_{i+1}x_{i+2}\right)+(x_{k-3}+x_k)x_{k-2}x_{k-1},\]so clearly taking $x_{k-3}\mapsto x_{k-3}+x_k$ and $x_k\mapsto0$ (effectively $k\mapsto k-1$) preserves $S$ and does not decrease $m$. So we can take $k=3$, so that \[x_1x_2x_3\le\left(\frac{x_1+x_2+x_3}{3}\right)^3=667^3.\]

I claim that the answer is $ 667^3 $ which can be obtained by letting $ a_i = 1, a_j = 2, a_k = 3 $ for all $ i \equiv 1 \pmod{3}, j \equiv 2 \pmod{3}, k \equiv 0 \pmod{3} $. Now I shall show that $ 667^3 $ is the highest possible number of triples. Let $ r, s, t $ be the number of $ a_i $ that are congruent to $ 0, 1, 2 \pmod 3 $ respectively. It is clear that since each relevant $ 3 $-element subset forms a complete residue system modulo $ 3 $ so the maximum number of such subsets is $ rst $. Now by AM-GM we have that $ rst \leq \frac{(r + s + t)^3}{27} = 667^3 $ as desired, so we are done.

Cute problem, nice $C1$ First notice that rearranging elements of $A$ leaves $m$ invariant, thus let $a_{2001} \geq a_{2000} \geq ... \geq a_1$. Also observe that translating the whole sequence by an integer value also leaves $m$ invariant, hence let $a_1=1$, and, finally if there exists some $a_i$ such that $a_{i+1}-a_{i} \geq 2 =>$ we can translate the sequence $(a_{i+1},a_{i+2},...,a_{2001})$ such that $a_{i+1}-a_{i}=1$ or $0$ without decreasing $m$. Thus we only need to consider the sequence $(1,2,...,t)$ where $1$ is repeated in $A_m$ ($A$ such that $m$ is maximum) $g_1$ times, $2$ is repeated $g_2$ times, ..., $t$ is repeated $g_t$ times. Thus $g_1+g_2+...+g_t=2001$, and we need to find maximum value of $m=g_1g_2g_3+g_2g_3g_4+...g_{t-2}g_{t-1}g_t$. Since I couldn't find the rather elegant idea of modular classes, we're going to have to do this the long way... First let's say $t \geq 4$. Now consider element $g_t$ of set $(g_1,g_2,...,g_t)$ (what we want to do here is show that we can collapse this $t$-element set to a $t-1$ element one without decreasing $m$)- we know $m=g_1g_2g_3+...+g_{t-5}g_{t-4}g_{t-3} +g_{t-2}g_{t-1}(g_{t-3}+g_{t})$- from this we can clearly see that we can dump all the elements $g_{t}$ of $t$ into $t-3$ without decreasing $m$- hence we only have to consider the case when $t=3$. Thus we have the set $(1,2,3)$, and we want to maximise $m=g_1g_2g_3$ from this... it seems pretty intuitive that putting $g_1=g_2=g_3=667$ should do the trick; this gives the value of $m=667^3$- But wait proving $g_1g_2g_3 \leq 667^3$ is obvious anyways, as $\frac{g_1+g_2+g_3}3 = 667 \geq (g_1g_2g_3)^{\frac{1}{3}}=> 667^3 \geq g_1g_2g_3 =m$, hence we're done. Thus the maximum value of $m$ is $\boxed{667^3}$, and occurs when $a_1=...=a_{667}=1,a_{668}=...=a_{1334}=2, a_{1335}=...=a_{2001}=3$.

I claim the answer is $667^3$ Without loss of generality assume that $a_i$ is nondecreasing. Now, note that replacing $a_0$ by $a_0+3$ will not decrease the number of triples. Thus, by smoothing(or a proof by contradiction) we are left with 3 values. Say that there are x,y,z of these consecutive values, thus there are a total of \[xyz\leq \left(\frac{x+y+z}{3}\right)^3 = 667^3\]Thus, we have at most $667^3$ triples, which is achievable when the first 667 values are 0, the next 667 are 1, and the next 667 are 2.

We claim that the answer is $\boxed{667^3}$. This is achieved by having $a_i=\left\lceil \frac{i}{667} \right\rceil$ for all $1\le i \le 2001$. We prove that there is no higher value of $m$. For every integer $i$, let $b_i$ be the number of indices $1\le j\le 2001$ such that $a_j=i$. Let $M$ be the largest positive integer such that $b_M$ is nonzero. Claim: It suffices to find the maxmium for all sequences such that $b_i$ is nonzero for all $1\le i\le M$. Call such a sequence decent. Proof. Suppose we have a non-decent sequence. Let $Z$ be the least positive integer such that $b_Z=0$. By definition, $Z<M$. Consider the following operation: For all $i$ such that $a_i>Z$, change $a_i$ into $a_i-1$. Notice that this operation cannot decrease $m$, as if the subsequence $a_i,a_j,a_k$ worked before, it still works after the operation. If we repeatedly apply the operation, it must necessarily end as the $a_i$ are finite. However, after it ends, the sequence must have turned decent, but the value of $m$ did not decrease. This means that every non-decent sequence has a decent sequence with at least as high of an $m$-value, proving the claim. $\blacksquare$ Claim: It suffices to find the maximum for all sequences such that only $b_1,b_2,b_3,b_4$ are nonzero. Proof. Suppose we are given a decent sequence that is not in this category, implying $M\ge 5$. Because the sequence is decent $b_{M-4}, b_{M-3}, b_{M-2}, b_{M-1}, b_M$ are all positive. Consider the following operation: For all $i$ such that $a_i=M$, change $a_i$ into $M-3$. By doing this, we change $b_M$ into $0$ and $b_{M-3}$ into $b_M+b_{M-3}$. This means that $m$ has decreased by $b_{M-2} b_{M-1} b_M$ and it has increased by at least $b_{M-4}b_Mb_{M-2}+b_Mb_{M-2}b_{M-1}$, a net positive change. Repeatedly applying this operation yields a sequence with $M=4$ that has an increased $m$-value from the original decent arrangement, proving the claim. $\blacksquare$ Now we simply have $b_1,b_2,b_3,b_4$ such that $b_1+b_2+b_3+b_4=2001$ and we want to maximize $$b_1 b_2 b_3+b_2 b_3 b_4=(b_1+b_4)(b_3)(b_2).$$However, this is easily seen by AM-GM to be $667^3$, and we are done.

let $B$ be a subsequence of $A$ with only terms $0\pmod 3,$ $C$ be one with only terms $1\pmod 3$ and $D$ be one with only terms $2\pmod 3.$ Note that any $(a_i,a_j,a_k)$ must have one term from each subsequence, so there are at most $|B|\cdot |C|\cdot |D|$ such subsequences. By $AM-GM$ this is at most $667^3.$ Our construction is as follows: $a_1=a_2=\dots=a_{667}=1,a_{668}=a_{669}=\dots=a_{1334}=2,a_{1335}=a_{1336}=\dots=a_{2001}=3.$

We claim the answer is $667^3.$ This is attainable by considering $$A=(\underbrace{1,1,\dots,1}_{667},\underbrace{2,2,\dots,2}_{667},\underbrace{3,3\dots,3}_{667}).$$Notice $\{a_i,a_j,a_k\}=\{0,1,2\}\pmod{3}$ so $m=abc,$ where $a,b,c$ are the number of elements congruent to $0,1,2\pmod{3},$ respectively. Notice by AM-GM, $$667=\frac{a+b+c}{3}\ge \sqrt[3]{abc}.$$$\square$

We claim the answer is $667^3$. This is attainable with $A=(1,1, \dots,1,2,2, \dots,2,3,3, \dots, 3)$ where there are $667$ $1$'s, $2$'s, and $3$'s. If the numbers $(a_i,a_j,a_k)$ satisfy the condition then they form a residue class $\pmod{3}$. Now let $a$ be the number of $a_i$ congruent to $0 \pmod{3}$, $b$ be the number congruent to $1 \pmod{3}$, and $c$ be the number congruent to $2 \pmod{3}$. Then we have $a+b+c=2001$ and $$m \le abc \le \left( \frac{a+b+c}{3} \right) ^3 = 667^3,$$and we are done.

We claim the answer is $667^3$. The construction for this is $$A=(\underbrace{1,1,\dots,1}_{667},\underbrace{2,2,\dots,2}_{667},\underbrace{3,3,\dots,3}_{667}).$$Now take $S$ as the sequence of all distinct values in $A$ ordered from least to greatest. If $S$ has two adjacent values $x,y$ that have $y-x>1$, we can shift every element of $A$ that have value of at least $y$ by $-1$, and $m$ will not decrease, since the number of triples counted by $m$ is increased by the number of triples that have one or more elements less than $y$ and one or more elements greater than or equal to $y$. So we can assume that $S$ is an arithmetic progression with common difference $1$. Let $r=|S|$, and take the set $B=\{b_1,\dots,b_r\}$ such that $b_i$ counts the number of times the $i$-th element of $S$ shows up in $A$. Then we have $$m=b_1b_2b_3+b_2b_3b_4+\dots+b_{r-4}b_{r-3}b_{r-2}+b_{r-2}b_{r-1}(b_{r-3}+b_r).$$If we set all the elements of $A$ counted by $b_r$ to the value of the elements of $A$ counted by $b_{r-3}$, i.e. if we add $b_r$ to $b_{r-3}$ and then set $b_r=0$, then we don't change $m$. Thus we can always reduce $B$ to an $(r-1)$-tuple with the same value of $m$, as long as $r\geq 4$. It follows that we only need to find the maximum value of $m$ for $r=3$. By AM-GM, we know that $$m=b_1b_2b_3\leq \frac{(b_1+b_2+b_3)^3}{3^3},$$with equality case $b_1=b_2=b_3$. So $m\leq 667^3$.

The answer is $667^3$ with $A$ having $667$ $1$s, $2$s, and $3$s. Assume WLOG that $A$ is sorted from least to greatest. Think of sequence $B$ where $b_1$ is the number of times $a_1$ occurs in $A$, $b_2$ is the number of times $a_1+1$ occurs in $A$, etc. Let $y$ be the number of elements in $B$. Then, $$m=b_1b_2b_3+b_2b_3b_4+\dots+b_{y-2}b_{y-1}b_y.$$If $y>3$, it is not worse to use the sequence $A'$ that is accompanied with $$B'=(b_2,b_3,b_1+b_4,b_5,\dots,b_y).$$Therefore, it is best if $y=3$, so $B=(b_1,b_2,b_3)$ and we want to maximize $b_1b_2b_3$ when $b_1+b_2+b_3=2001$, which by AM-GM is when $b_1=b_2=b_3=667$, giving our answer.

Let there be $x$ terms $0$ mod 3, $y$ terms $1$ mod 3, and $z$ terms $2$ mod 3. It is easy to see that $m\leq xyz$, and by AM-GM, $xyz \leq 667^3.$ A construction is to let $a_i = \left\lceil \frac{i}{667} \right\rceil$.

Notice that each valid triple $(a_i, a_j, a_k)$ contains exactly one element that is $t\mod 3$ for each $t\in\{0,1,2\}$. Now, let $S_t$ be the sub-multiset of $A$ containing elements congruent to $t\mod 3$. Then, for each $(a,b,c)\in S_0\times S_1\times S_2$, there is at most one valid triple (since the indices must be in increasing order). Hence, \[ m\le 1\cdot |S_0|\cdot |S_1|\cdot |S_2|\le \left(\frac{|S_0| + |S_1| + |S_2|}{3}\right)^3 = \left(\frac{2001}{3}\right)^3 = 667^3.\]Note that equality is achieved when $A$ consists of $667$ $1$'s followed by $667$ $2$'s followed by $667$ $3$'s. Thus the answer is $\boxed{667^3}$.

The answer is $667^3$. This is achieved with 667 1's, 667 2's, and 667 3's in that order. Let $b_n$ denote the number of $i$ such that $a_i=n$. so that the sum of all $b_i$ is 2001. Clearly, the value of the permutation is at most $$\sum_{i=1}^\infty b_ib_{i+1}b_{i+2}.$$ We will now show that $$\sum_{i=1}^\infty b_ib_{i+1}b_{i+2}\leq 667^3.$$ If not all nonzero terms of the $b$ sequences are consecutive, we can "push" them together which increases the value. From now on assume WLOG that all nonzero terms are consecutive. Now, suppose $b_s$ is the first nonzero term. If there are at least 4 nonzero terms in the $b$ sequence, then the transformation $$(b_s,b_{s+3})\rightarrow (0, b_{s+3}+b_s)$$(essentially "merging" $s$ with $s+3$) will either have no effect or increase the value, since the expansion of the sum has $$b_{s+1}b_{s+2}(b_s+b_{s+3}),$$which will stay the same but the increase in $b_{s+3}$ may positively effect future terms, while the decrease in $b_s$ has no adverse effect on previous terms since they are 0 anyways as we are assuming that $s$ is the first nonzero term. Thus, in the optimal case, there are at most 3 nonzero terms. Simply AM-GM gives at most $667^3.$

Wow, this one was an easy problem to guess, and a relatively easy problem to prove in a few minutes, too. I think the answer is $667^3.$ My construction is to let the first 667=2001/3 numbers to be 1, next 667 to be 2, and last 667 to be 3. That yields 667^3 options; Notice $\{a_i,a_j,a_k\}=\{0,1,2\}\pmod{3}$. Let a,b,c be the # of elements 0, 1, and 2 mod 3. By AM-GM, we have $667=\frac{a+b+c}{3}\geq \sqrt[3]{abc}$. There is a (three-way?) bijection between a 0,1, and 2 mod 3 of #s that satisfy this, which means these numbers are represented by a,b, and c, and to maximize m we have our answer. $\blacksquare$

The answer is $667^3$; a construction is to take $667$ 1's, then $667$ 2's, then finally $667$ 3's - it is clear that this yields a total of $667^3$ possible combinations. To show that $667^3$ is maximal, take $\pmod{3}$; let the number of elements in $A$ that are $0 \pmod{3}$, $1 \pmod{3}$, and $2 \pmod{3}$ be $x$, $y$, and $z$ respectively. Note that in each working pair, each element is a member of one of these three sets, and moreover there is at most one working triple for each of the triples with one element in each of the three sets. Therefore, we wish to maximize $xyz$ on $x+y+z = 2001$, which is (by AM-GM) $667^3$.

The idea is to look at mod 3. Any such sequence must contain one of each residue class modulo $3$. Suppose there are $x$ 1's, $y$ 2's, and $z$ 0's under mod $3$. Then the number of triples is bounded above by $xyz$, which is in turn bounded above by $667^3$ by AM-GM. Just fix $x=y=z$ and all the numbers in each class to be equal for construction.

It's optimal to: Have a nondecreasing list, obviously. Have a list of consecutive integers, since otherwise we can move nonconsecutive blocks closer together at no cost. Importantly, consider an optimal sequence that minimizes the number of distinct values in the sequence. Assume FTSOC that there are at least four different numbers in this sequence; let the first four numbers occur $a, b, c$ and $d$ times. But then, we can get a shorter sequence by replacing $d$ with $a+d$ and chopping off all $a$ occurences of the first number, since $bc(a+d) = abc + bcd$, so the number of three consecutive sequences remains the same. Hence there is an optimal sequence which contains $a$ ones, $b$ twos and $c$ threes, with $a+b+c = 2001$. The maximum value of $abc$ is $667^3 = 296740963$ due to AM-GM. whoops i overcomplicated this

I think this is almost a rigid problem. We take $\pmod{3}$. We can see that we can only have a successful pair when we have $0$, $1$, and $2$ $\pmod{3}$ in some order. Specifically we can now see that the maximum value is $\boxed{667^3}$ because of AM-GM. This is possible by consider $667$ ones, twos, and threes in that order. $\blacksquare$

The answer is $667^3$ which can be formed with 667 $1$'s, $2$'s and $3$'s. Now, taking modulo $3$, we see that we wish to maximize $abc$ given $a+b+c = 2001$, which is clearly $667^3$ from AM-GM. $\blacksquare$

We can assume that they are in nondecreasing order (putting the sequence in this order does not decrease $m$) and that the integers that occur are consecutive (if there are no elements equal to $k$ and elements above and below $k$ we can subtract $1$ from each of them and repeat since this does not decrease $m$) and with least term $1$ (just shift everything down) and that there are exactly three different integers (if the maximum integer that occurs is $n \ge 4,$ then let $X_i$ be the number of occurrences of $i,$ then if $X_2X_3 \ge X_{n-2}X_{n-1}$ then we may replace all occurrences of $n$ with $1$ without decreasing $m,$ and if $X_2X_3 \le X_{n-2}X_{n-1}$ then we may replace all occurrences of $1$ with $n$ and shift down without decreasing $m,$ and repeat) and now AM-GM gives the answer is just $667^3$ since we can take $667$ of each of $1,2,3.$

The answer is $m=667^3$. For the construction, we have, \[ \underbrace{1,1,\ldots,1}_{667}; \underbrace{2,\ldots,2}_{667};\underbrace{3,\ldots,3}_{667}. \] Consider the sequence for maximal $m$. Note that arranging the sequence in non-decreasing order does not decrease the value of $m$, since we can map every triplet $(a_i,a_j,a_k)$ from the initial ordering to the new ordering. \[ \underbrace{a_1,a_1,\ldots,a_1}_{b_1};\underbrace{a_2\ldots,a_2}_{b_2};\ldots;\underbrace{a_n,\ldots,a_n}_{b_n}. \] We relax the condition of $b_i = \in \mathbb N$ for a while to analyze the inequality. If $a_1$ is not a part of any triple, then we can increase $a_1$ to $a_2 - 1$ to contradict the maximality. SImilarly, $a_n$ must be part of some triple. If $a_{i+1} - a_i \ge 2$ for some minimal $i$, then we can decrease the values of elements of $b_{i+1},b_{i+2},\ldots,b_n$ to contradict the maximality by increasing $m$ by setting $a_{i+1} = a_i + 1$. So now the sequence is like, \[ \underbrace{a_1,a_1,\ldots,a_1}_{b_1};\underbrace{a_1+1,\ldots,a_1}_{b_2};\ldots;\underbrace{a_1 + (n-1), \ldots, a_1 + (n-1)}_{b_n}. \] So $m = b_1b_2b_3 + b_2b_3b_4 + \cdots + b_{n-2}b_{n-1}b_n$. If $b_i \neq b_{i+1}$, note that we can change $m$ from $b_1b_2b_3 + \cdots + b_ib_{i+1}(b_{i-1} + b_{i+2}) + \cdots + b_{n-2}b_{n-1}b_n$ by replacing $b_i,b_{i+1} \rightarrow \dfrac{b_i + b_{i+1}}{2}$ to increase $m$ contradicting maximality. Therefore we deduce $b_i = b_j$ for all $i,j$, that is $b_i = \dfrac{2001}{n}$. Thus $m = (n-2)\left(\dfrac{2001}{n}\right)^3 = 2001 ^3 \cdot \dfrac{n-2}{n^3}$. Now note that $\dfrac{n-2}{n^3}$ is decreasing $\implies n=3$ for maximum $m$. Therefore $m = \left(\dfrac{2001}{3}\right)^3 = 667^3$ and thus we are done.

Let a sub-sequence which satisfies all the required conditions of the problem be called a xoinky tuple. Then, we need to find the maximum number of xoinky tuples in a 2001-element sequence. We engage in some smoothing. We order the elements in increasing order as $a_1\leq a_2 \leq \dots \leq a_{2001}$. First, note that $a_1=1$ or else, we can shift all the terms down by $a_1-1$ and it cause no effect to the number of xoinky tuples. Now, say that we have terms in this sequence with value $n-1$ and $n+1$ but no terms with value $n$ for some $n\geq 2$. Then, we can shift all the terms greater than $n-1$ by 1. Note that this causes no change to xoinky tuples of which any term with value $n+1$ is the smallest, as all terms larger than $x$ are also shifted and thus the difference between terms remains unchanged. Clearly no xoinky tuples exist where $n+1$ is the middle term or largest term (no terms of $n$ exist which is a contradiction). Thus, these operations do not decrease the number of xoinky tuples of this sequence. Now, let $x_i$ be the number of terms of value $i$. Say we have terms larger than $3$. Then, we have $x_4 >0$ and $x_5,x_5\geq 0$. Now, we simply shift terms of value 4,5 and 6 down by 3. Now, note that, the number of xoinky subsets before this operation is \[x_1x_2x_3+x_2x_3x_4+x_3x_4x_5+x_4x_5x_6+f\]and after the operation is \[(x_1+x_4)(x_2+x_5)(x_3+x_6)+f\]for some fixed value $f$. (The operation is $x_1,x_2,x_3,x_4,x_5,x_6,\dots \mapsto x_1+x_4,x_2+x_5,x_3+x_6,0,0,0,\dots$ where no change occurs to the terms after the dots) But, when $x_5,x_6>0$ \begin{align*}(x_1+x_4)(x_2+x_5)(x_3+x_6) &= x_1 x_2 x_3 + x_2 x_4 x_3 + x_1 x_5 x_3 + x_4 x_5 x_3 \\ & + x_1 x_2 x_6 + x_2 x_4 x_6 + x_1 x_5 x_6 + x_4 x_5 x_6 \\ &> x_1x_2x_3+x_2x_3x_4+x_3x_4x_5+x_4x_5x_6 \end{align*}When $x_6=0$ but $x_5>0$, \begin{align*} (x_1+x_4)(x_2+x_5)(x_3+x_6) &= x_1 x_2 x_3 + x_2 x_4 x_3 + x_1 x_5 x_3 + x_4 x_5 x_3 \\ &> x_1 x_2 x_3 + x_2 x_3 x_4 + + x_3 x_4 x_5 \\ & = x_1x_2x_3+x_2x_3x_4+x_3x_4x_5+x_4x_5x_6 \end{align*}When, $x_5,x_5=0$ we have \begin{align*} (x_1+x_4)(x_2+x_5)(x_3+x_6) &= x_1 x_2 x_3 + x_2 x_3 x_4 \\ & = x_1x_2x_3+x_2x_3x_4 &= x_1x_2x_3+x_2x_3x_4+x_3x_4x_5+x_4x_5x_6 \end{align*}So, we see that upon performing the described operation, the number of xoinky subsets does not decrease. Now, we perform the initial operation, of reducing the gap of 3 between the terms of value 3 and value 7 (if such exist) and are left where we started. Thus, we find that in the optimal configuration, we are left with terms with value 1, 2 and 3. Then, we simply resort to AM-GM. Note that, \[m=x_1x_2x_3 \leq \left(\frac{x_1+x_2+x_3}{3}\right)^3 = \left(\frac{2001}{3}\right)^3 = 667^3\]which is indeed achievable by considering 667 terms of value 1,2 and 3 each.

We claim the answer is $\boxed{667^3}$ which can be obtained by $667$ $1$'s, $667$ $2$s, and $667$ $3$s in that order. Now note that if $a_i > a_{i+1}$ then switching $a_i$, and $a_{i+1}$ does not decrease $m$. There fore we can assume $a_i \le a_{i+1}$ for all $i$. Now define $b_k$ as the number of elements in our $2001$-tuple that have value $k$, so $\sum_{k \ge 1} b_k =2001$. Note that $$m= \sum_{k \ge1} b_k b_{k+1} b_{k+2}$$Suppose $r \le a_i \le r+n-1$ for all $i$, then note that by am-gm $$m= \sum_{k \ge1} b_k b_{k+1} b_{k+2} \le \frac{(n-2)}{n^3} 2001^3$$However the function $f(n)=\frac{n-2}{n^3}$ is maximized at $n=3$ over the positive domain by calculus, so $m \le \frac{1}{3^3}2001^3=667^3$ and we are done.

I claim the answer is $667^3$, achievable by $667$ ones, twos, and then threes. Clearly it is optimal to sort increasing, and then make everything consecutive. Thus this is just an algebra problem, optimize $\sum b_ib_{i + 1}b_{i + 2}$, where $b_i$ is the number of times $i$ appears in the sequence. We can obviously see $b_i$ does not have any zeroes between positive values, otherwise we could just shift down and get a higher sum, so we can see that $b_i$ is a block of positive values, and then zeroes we don't care about. Now we show that some optimal set of $b_i$ has block size $3$. Of course, if $b_{i + 3}$ is positive and the $i+ 3$ is the highest index with positive $b$, set $b_{i + 3}$ as $0$ and replace $b_{i}$ with $b_i + b_{i + 3}$. The total loss is exactly $b_{i + 1}b_{i + 2}b_{i + 3}$, gain is at least that. Thus we just need to optimize $xyz$ where $x + y + z = 2001$, but this is trivial by AM-GM.

Notice that if there are any two elements in the sequence $a_i$ and $a_{i+1}$ so that $a_i > a_{i+1}$, we can swap the two elements, which does not reduce $m$. Repeat this process until the sequence is nondecreasing. Now, if any two elements $a_i$ and $a_{i+1}$ exist so that $a_{i+1} - a_{i} = k > 1$, we can subtract $k-1$ from all elements after $a_i$ by $k-1$, which does not reduce $m$. After this, we get that all consecutive numbers in the sequence either differ by $1$ or are equal to eachother. Now split the sequence $a_i$ into groups of numbers $g_1$, $g_2$, $\dots$, $g_c$ so that all of the numbers in each group are equal. For example, a sequence $1, 1, \dots, 2, 2$ would be split into $g_1 = (1, 1, \dots)$ and $g_2 = (2, 2)$. Let $n_i$ be the number of elements in $g_i$. Now $m$ can be written as $n_1n_2n_3 + n_2n_3n_4 + \dots n_{c-2}n_{c-1}n_c$. Notice that $\sum_{i=1}^c n_i = 2001$. We can merge $g_1$ and $g_2$ to get $m = (n_1+n_2)(n_3)(n_4) + (n_3n_4n_5) + \dots$, which does not decrease $m$ as $n_1n_3n_4 \geq n_1n_2n_3$ due to the groups being in increasing order. We continuously do this until we are left with $3$ groups $g_1$, $g_2$, and $g_3$. Now, a simple AM-GM argument makes choosing $n_1 = n_2 = n_3 = 667$, so the answer is $667^3$.

The answer is $\boxed{667^3}$, which is obtained by having $667$ ones, then $667$ twos, and finally $667$ threes. It is evident that in a sequence maximizing $m$, the sequence is arranged in nondecreasing order. Thus, we can assume WLOG that $a_1 \le a_2 \le \dots \le a_{2001}$. Suppose that the number $k$ be used $x_k$ times; hence, \[\sum_{k \ge 1} x_k = 2001.\] We wish to maximize \[m = x_1x_2x_3+x_2x_3x_4+x_3x_4x_5+\dots = (x_1+x_4)x_2x_3+x_3x_4x_5+\dots.\] This implies that $x_k+x_{k+3}$ is fixed for all $k$. In other words, the maximal $m$ requires at most $3$ distinct numbers in its corresponding sequence. We then wish to maximize \[x_1x_2x_3 \le \left(\frac{x_1+x_2+x_3}{3} \right)^3 = 667^3,\] as desired.

(We replace $2001$ with $3n$) We claim that the answer is $n^3$. The equality holds when: $a_1 = a_2 = \cdot = a_n = 1, a_{n+1} = a_{n+2} = \cdots = a_{2n} = 2, a_{2n+1} = \cdots = a_{3n}=2$. Now, we will prove that it's the maximum. For any sequence $(a_1, \cdots, a_{3n})$, if $a_i > a_j$ for $i < j$ then removing $a_j$ from $j$th position and place it before $a_i$. Notice that, this greedy step increases (or doesnt change) the value of $m$. Thus, we have $(a_1, \cdots, a_{3n})$ to be sorted in increasing order. Now, notice that: if we have $a_{k+1} - a_{k} \ge 2$, then decreasing all the numbers $a_{k+1}, a_{k+2}, \cdots, a_{3n}$ by $c= a_{k+1}-a{k}-1$ doesnt decrease the value of $m$. Thus, we can have $a_{k+1} - a_{k} \le 1$ for all $1 \le k \le 3n$. Let $a_1=a$ and $a_{3n}=a+k$. Amongst the sequence $(a_1, \cdots, a_{3n})$, let $a+i$ occur $l_i$ times. Thus, we are for maximising $$S = l_0 l_1 l_2 + \cdots + l_{k-2} l_{k-2} l_k$$Notice that smoothing any two variables $l_k, l_{k+1}$ closer increase $S$. Thus, $S$ is maximum when all of $l_i$ are equal to $\frac{3n}{k+1}$ with $S \le \frac{(k-1) 27 n^3}{(k+1)^3}$. Note that $k \ge 2$ and therefore, for $k=2$, we have: $$S_{max} = n^3$$and we are done.

We show by induction that for the general $n,$ it is always optimal (not necessarily strictly optimal) to constrain the sequence among $3$ consecutive numbers. The base cases, $n=1, 2, 3$ are trivial. Now suppose that the proposition holds for all $n \leq k$ for some $k \geq 3.$ Then consider $n+1$ numbers. For the sake of a contradiction, assume that our proposition does not hold for $n+1$ numbers. Clearly to optimize $m$ we should put the first $n$ numbers among three consecutive numbers, WLOG let them be $1, 2, 3.$ Then the $n+1$th number is either $0$ or $4,$ WLOG it is $4.$ Then suppose that there are $a$ $1$s, $b$ $2$s, and $c$ $3$s. Then $$m = abc+bc \cdot 1 = bc(a+1).$$Hence this case is equivalent to if the $n+1$th number was just $1$ in the first place. Thus the induction is complete, so suppose that we have $a$ $1$s, $b$ $2$s, and $c$ $3$s. Then $m=abc \leq \left(\frac{a+b+c}{3}\right)^3 = 667^3$ by AM-GM. For attainability, we can let $a=b=c=667.$