Obligatory Six Afraid of Seven Joke

First off, note that $x_n\mid x_{n+1}$ . Now, suppose that $x_5 < 12$ . This means $x_3+x_4 > 12$ . We know that $x_3 \le x_4$ since $x_3\mid x_4$ so $x_4 > 6$ . But since $x_4\mid x_5$ , the conditions tell us we must have $x_4=x_5$ . This means that $x_2+x_3=1$ , but since they are positive integers, there is a contradiction.

Now we take it by cases. If $x_5=12$ , then we see that $x_3+x_4=12$ . Since $x_3\le x_4$ we must either have $x_4=6$ or $x_4=12$ . If $x_4=12$ , then we arrive at the same contradiction as in the first case. If $x_4=6$ then $x_3=6$ , and then we arrive at the conclusion $x_1+x_2=1$ which is a contradiction.

If $x_5=16$ , then $x_3+x_4=9$ . Since $x_3\le x_4$ then $x_4\ge 4.5$ . Since the factors of $16$ are $1,2,4,8,16$ , we can either have $x_4=8$ or $x_4=16$ . If $x_4=16$ then we arrive at the same contradiction as the first case. If $x_4=8$ then $x_3=1$ . Then, $x_2=1$ and $x_1=7$ . This case actually works!

Thus, we have that $x_7=144\cdot (16+8)=\boxed{3456}$ .

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In hindsight, not only does the sequence starting $7,1,1,\ldots$ work, the sequence $2,1,2,\ldots$ also works: $2,1,2,6,18,144,3456\ldots$

A nice coincidence that it arrived at the same value for $x_7$ !

• I'll divide this solution in parts, hoping to help you folks to understanding it. First one: expanding .
• We have $144 = x_{5}(x_{4} + x_{3})$ , $x_{5} = x_{4}(x_{3} + x_{2})$ and $x_{4} = x_{3}(x_{2} + x_{1})$ .
• Expanding it, we have:
  $144 = (x_{3})(x_{2} + x_{1})(x_{3} + x_{2})[x_{3}(x_{2} + x_{3}) + x_{3}] =$
  $(x_{3})(x_{2} + x_{1})(x_{3} + x_{2})(x_{3})(x_{2} + x_{1} + 1)$ .
  
  
• Now, factoring, we can also replace 144 as $2*2*2*2*3*3$ .

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• So far, we have $2*2*2*2*3*3 = (x_{3})(x_{2} + x_{1})(x_{3} + x_{2})(x_{3})(x_{2} + x_{1} + 1)$ . Now we begin the second part: checking cases .
• We can easily see that $x_{1}$ , $x_{2}$ and $x_{3}$ are all lower than 12 , otherwise $x_{6}$ would be much bigger than 144.
• We can also see that 144 is divisible by both $(x_{2} + x_{1})$ and $(x_{2} + x_{1} + 1)$ . In other words, it's divisible by two consecutive numbers. Since we state that all three incognites are lower than 12, we have 3 cases: (2, 3) , (3, 4) and (8, 9) .

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• Third part: checking case one (2, 3) .
• Let's suppose $(x_{2} + x_{1}) = 2$ . Also: since all x s are integers, we state $(x_{1} = x_{2}) = 1$ .
• Replacing then, we have: $2*2*2*2*3*3 = (x_{3})(2)(x_{3} + 1)(x_{3})(3)$ . $2*2*2*3 = (x_{3})(x_{3} + 1)(x_{3})$ .
• The only combination we could have afterwards is $x_{3} = 2$ and $(x_{3} + 1) = 6$ , which is ABSOLUTELY false. So, case one doesn't work.

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• Fourth part: case two (3, 4) .
• As we did in the last step, let's suppose $(x_{2} + x_{1}) = 3$ .
• Replacing, we find: $2*2*2*2*3*3 = (x_{3})(3)(x_{3} + x_{2})(x_{3})(4)$ . $2*2*3 = (x_{3})(x_{3} + x_{2})(x_{3})$ .
• So, as the only combination we could possibly have, we accomplish $x_{3} = 2$ and $x_{2} = 1$ . Since $(x_{2} + x_{1}) = 3$ , $x_{1} = 2$ .
• Replacing in the other equations, we have $x_{4} = 6$ and $x_{5} = 18$ .
• FINALLY , we conclude that $x_{7} = x_{6}(x_{5} + x_{4}) = 144(18 + 6) = 3456$ .

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• Guys, I know this one get a little messy and complex and big, so feel free to ask any questions you want and i'll do my best to answer you all.