7
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$1, 1, 0, -2, 4, 18, 328, 107638 ,?$

This problem can be complex, use your imagination.

Hint:

The wording is deliberate

Hint 2:

The first and second elements of the sequence are not equal

Hint 3:

The missing number is in the billions range

Hint 4:

Instead of squaring think more like two brackets multiplied together

Hint 5:

If i was to write out the sequence as $a(n+1)=...$ then the right hand side would include a $Re()$ function

Hint 6:

The first number in the sequence is $1$ , the second number is $1+i$

Hint 7:

The fifth number is $4-2i$

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  • 4
    $\begingroup$ Just as a note you should not really be putting a hint into the first edit of the question. Hints are great for when nobody seems to have a clue where to start and can be added in later if the question seems to be proving overly hard for people. Putting it in before anybody has even seen the question is being unfair to the community and particular those who would have worked it out without the hint. $\endgroup$ – Chris Mar 7 '18 at 15:16
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    $\begingroup$ Thanks for the feedback. This is my first question so I didn't realise that was the etiquette, I shall bear that in mind for the future. $\endgroup$ – Jon.G Mar 7 '18 at 15:29
4
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Finally... I think I got this :"

As mentioned on the puzzle, the first and the second element are not equal.
Indeed, the exact values for each elements are:

$1 + 0i$
$1 + 1i$
$0 + 2i$
$-2 + 0i$
$4 - 2i$
$18 - 4i$
$328 - 54i$
$107638 - 17384i$

The numbers on the sequence is:

The real part of the (complex) number, which is denoted by $Re()$ function.
For example, $Re(1 + 0i) = 1$, $Re(0 + 2i) = 0$, and $Re(107638 - 17384i) = 107638$.

So, what is the pattern? The pattern is:

$A_n = A_{n-1} \times (Re(A_{n-1}) + i)$

For example,
$A_7 = 328 - 54i$
$A_8 = A_7 \times (Re(A_7) + i)$
$A_8 = (328 - 54i) \times (328 + i)$
$A_8 = 107638 - 17384i$

Finally, what is the answer? The answer is:

$A_9 = A_8 \times (Re(A_8) + i)$
$A_9 = (107638 - 17384i) \times (107638 + i)$
$A_9 = 11585956428 - 1871071354i$

So, it is $Re(A_9) = \fbox{11585956428}$.

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  • $\begingroup$ Ahhh I'm so happy someone got it, do you think this puzzle was too difficult? $\endgroup$ – Jon.G Mar 28 '18 at 15:11
  • $\begingroup$ And oh God, this is one of the great (pure) math sequence puzzle >< ... I hope this solution will gain more upvotes for this puzzle! $\endgroup$ – athin Mar 28 '18 at 15:11
  • $\begingroup$ @Jon.G Yes, this is a complex one but I really enjoyed it, thanks! :D $\endgroup$ – athin Mar 28 '18 at 15:12
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    $\begingroup$ @athin Pun completely intentional? :) $\endgroup$ – Rubio Mar 30 '18 at 16:25
  • $\begingroup$ I'm surprised the first 6 digits of my guess are the same! :o $\endgroup$ – malioboro Apr 25 '18 at 16:14
3
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Partial Answer

Noting that

$18=4^2 + 2$, $328=18^2 + 4$(also, $2^2=4$. coincidence?), and $107638=324^2 + 54$.

can’t seem to find a pattern, though

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  • $\begingroup$ I've added a hint that might help $\endgroup$ – Jon.G Mar 13 '18 at 10:20
  • $\begingroup$ I noted that as well (also that 18*3=54) and spent too much time trying to figure out the pattern that doesn't appear to exist $\endgroup$ – MetaZen Mar 13 '18 at 22:38
3
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My solution didn't answer all hints, but I'll try it first, is it

11585940356

inspired by @TrojanByAccident's answer we can get:

$1$
$1 = 1^2 + 0$
$0 = 1^2 + -1$
$-2 = 0^2 + -2$
$4 = -2^2 + 0$
$18 = 4^2 + 2$
$328 = 18^2 + 4$
$107638 = 328^2 + 54$

so it shows a pattern $a[n] = a[n-1]^2 + somenumber$

and also from TrojanByAccident's answer we can break the somenumber to:

$1$
$1=1^2+0$
$0 = 1^2 + 1*-1$
$-2 = 0^2 + 1*-2$
$4 = -2^2 + 0$
$18 = 4^2 + -2*-1$
$328 = 18^2 + 4*1$
$107638 = 328^2 + 18*3$

so now it shows a pattern $a[n] = a[n-1]^2 + a[n-2]*anothersomenumber$

The problem is, we can't find a pattern in the 0 part of anothersomenumber. So I just guess this part, the pattern of anothersomenumber:

? -1 -2 ? -1 1 3

I guess it is $b[n]=b[n-1]-b[n-3]$

so it gives you: 0 -1 -2 -2 -1 1 3 4

so the next number is:

$107638^2 + 328*4 = 11585940356$

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  • $\begingroup$ Not the right answer but a nice attempt $\endgroup$ – Jon.G Mar 22 '18 at 11:18

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