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Can you find the next number in the following sequence:

3, 6, 9, 8, 13, 20, 17, 14, 21, ?

If you're operating differently you probably prefer to solve the following sequence:

2, 0, 0, 12, 35, 384, 135, 32, 315, ?


Hint 1

The missing numbers are not the same.

Hint 2

Both sequences are finite

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  • 2
    $\begingroup$ Thanks @Rubio. I've changed the puzzle and added a sequence I wanted to use as a hint later. Do you think this one's better or should I try it in the sandbox first? $\endgroup$ – swit Apr 13 '18 at 12:23
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    $\begingroup$ This seems better now, yes; thanks for being open to feedback :) $\endgroup$ – Rubio Apr 13 '18 at 18:54
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    $\begingroup$ Might be useful to someone: desmos.com/calculator/m17pq62ero $\endgroup$ – Just a browsing guest Apr 13 '18 at 21:44
  • $\begingroup$ Would it always be the same if it starts with the same number? Or could there be multiple sequences that start with 3, using this pattern? $\endgroup$ – Lily Potter Apr 15 '18 at 5:56
  • $\begingroup$ This sequence is neither arithmetic or geometric?? $\endgroup$ – CR241 Jul 27 '18 at 23:29
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Observation 1)

Split the sequence into $(3, 6, 9, 8)$ - three increasing one decreasing, $(13, 20, 17, 14)$ - two increasing two decreasing and $(21, ?, \cdots,\cdots)$ - one increasing three decreasing. So the question mark must be less than $21$.

Observation 2)

The consecutive differences are $3,3,3,1,5,7,3,3,7$. They are no more than $10$ and $1$ occurs once, $3$ occurs five times, $5$ occurs once and $7$ occurs twice. Except $7$, the number of occurrences are odd, so the difference between $21$ and the question mark is either $7$ or $17$. If it is the latter, that introduces an extra $1$ which breaks the odd rule.

Solution

Therefore, the missing number is $21-7=14$.

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