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The relationship is $\operatorname{gcd}(a_1,a_2,\dots,a_t)\cdot\operatorname{lcm}(a_1,a_2,\dots,a_t)=a_1\cdot a_2\cdot\dots\cdot a_t$. I think this holds for every sequence of length less than $3$, although I'm not sure how to start with this one.

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    $\begingroup$ Let $p$ be a prime. With what exponent does it appear on the left and on the right hand sides? $\endgroup$ – Daniel Fischer Nov 15 '14 at 21:18
  • $\begingroup$ $\prod_{k=1}^n a_k = \text{gcd}(a_1, \cdots, a_n)^{n-1}\text{lcm}(a_1, \cdots, a_n)$ $\endgroup$ – Petite Etincelle Nov 15 '14 at 21:19
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    $\begingroup$ @PetiteEtincelle That's not correct. $\endgroup$ – Daniel Fischer Nov 15 '14 at 21:20
  • $\begingroup$ @DanielFischer I don't see why, can you enlighten me? $\endgroup$ – Petite Etincelle Nov 15 '14 at 21:22
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    $\begingroup$ @Petite: it's not correct because you can easily write down counterexamples. For example, $2 \cdot 3 \cdot 6 \neq \gcd(2, 3, 6)^2 \text{lcm}(2, 3, 6)$; the LHS is $36$ but the RHS is $12$. $\endgroup$ – Qiaochu Yuan Nov 15 '14 at 21:24
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Let $$a_i = \prod_{k=1}^np_k^{a_{k,i}}$$ where $a_{k,i} \geq 0$. $$\gcd(a_1,a_2,\ldots,a_t) = \prod_{k=1}^np_k^{\min_ia_{k,i}}$$ $$\text{lcm}(a_1,a_2,\ldots,a_t) = \prod_{k=1}^np_k^{\max_ia_{k,i}}$$ Hence, we need $$\sum_{i=1}^t a_{k,i} = \max_ia_{k,i} + \min_ia_{k,i}$$ If $t=2$, this always holds true. For $t>2$, note that this would mean $a_{k,i}=0$ for $t-2$ numbers. This also means that $\min_i a_{k,i} = 0$ and hence $a_{k,i}=0$ for $t-1$ numbers. Hence, for any $t\geq3$, we need the numbers to be pairwise relatively prime.

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  • $\begingroup$ Wouldn't that be equivalent with the gcd for any three numbers being $1$? $\endgroup$ – user190865 Nov 15 '14 at 21:37
  • $\begingroup$ @user190865 I updated my argument. $\endgroup$ – Adhvaitha Nov 15 '14 at 21:46
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It is

$$gcd(2,4,8)=2$$ and

$$lcm(2,4,8)=8.$$

However,

$$lcm(2,4,8)gcd(2,4,8)=14\ne 64 =2\cdot 4\cdot 8.$$

So, for a sequence of length $3$ is false. You can construct a sequence having as many terms as you want adding $2'$s to the previous one.

Now, for sequences of two terms it is right. Can you prove it?

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  • $\begingroup$ There are arbitrarily long sequences for which the equality holds. $\endgroup$ – Daniel Fischer Nov 15 '14 at 21:19
  • $\begingroup$ I have said that it is possible to construct a sequence of arbitrary length ($\ge 3$) that doesn't satisfy the equality. $\endgroup$ – mfl Nov 15 '14 at 21:22
  • $\begingroup$ Yes. And I have added that it is possible to construct arbitrarily long sequences for which the equality holds. Together that implies that apart from the sufficient "length $\leqslant 2$", the length is not a criterion to determine whether it holds or not. $\endgroup$ – Daniel Fischer Nov 15 '14 at 21:24
  • $\begingroup$ @DanielFischer You are right. I have misunderstood your comment. $\endgroup$ – mfl Nov 15 '14 at 21:27
  • $\begingroup$ I have to admit that I haven't formulated it in a way that makes misunderstanding it particularly difficult. $\endgroup$ – Daniel Fischer Nov 15 '14 at 21:31

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