7
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Here's a sequence: Sequence

What is the next item? Here are possible answers (but you don't need to look at them to figure it out):

Possible answers

Update

Two answers have pointed out to me that the solution is obvious, and can be explained with a simple pattern. So I will try to add another restriction to help you find the rules I had intended:
Find the rules that explain the transition from any state to the next, without knowing the other states.
I realize this restriction makes the puzzle a bit weird. I apologize.

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  • 2
    $\begingroup$ Is this some form of Conway's Game of Life using a 7-segment display instead of square cells. That would be super awesome. $\endgroup$ – ibrahim mahrir Jan 8 '18 at 2:15
  • $\begingroup$ @ibrahimmahrir yes! That was what I had originally intended (with modified rules). But it turns out that a very simple and obvious pattern emerged, and lead to the correct solution (as discovered in the two current answers). $\endgroup$ – Reinis Mazeiks Jan 8 '18 at 2:21
  • $\begingroup$ should I remove my comment for being a spoiler? $\endgroup$ – ibrahim mahrir Jan 8 '18 at 2:45
  • $\begingroup$ @ibrahimmahrir I've seen Rot13 used in comments to hide spoilers $\endgroup$ – Jay Jan 8 '18 at 2:53
4
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The answer is

E

Given OP's restriction on using global knowledge, my answer is as follows:

The pattern used is a simple one. Look at a line segment's neighbors.
- If they are all the same, then the line segment remains unchanged
- If they are different, then the line segment changes. (To an on state if off and to an off state if on.)

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  • $\begingroup$ Yes! I was just about to post the intended answer, but you found it. :) $\endgroup$ – Reinis Mazeiks Jan 8 '18 at 13:41
4
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There's some ambiguity in the question, because you could make arguments for multiple interpretations... Using the following labels:
taken from wikipedia under CCSA - https://commons.wikimedia.org/wiki/File:7_segment_display_labeled.svg

You could argue the next pattern is E because A & C are always on in each image, B & F alternate on & off in pairs, D & G alternate on/off, E is always off.

But you could also argue I if the C is alternating in groups of four.

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  • $\begingroup$ There is no need to argue about all the segments, just F and B will suffice, because they are the weakest patterns, there is a lack of examples so their pattern could be anything (not enough examples to establish a solid-ish pattern like for the other segments). B for example could be ON ON ON OFF OFF OFF, thus option C is the correct one, ... $\endgroup$ – ibrahim mahrir Jan 8 '18 at 2:05
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    $\begingroup$ What I wanted to say is that all other segments do actually repeat after a defined period, so a pattern is established. B and F don't repeat at all (there is no pattern whatsover). $\endgroup$ – ibrahim mahrir Jan 8 '18 at 2:09
  • $\begingroup$ @ibrahimmahrir you probably meant 2 ON 3 OFF in your first comment? $\endgroup$ – Reinis Mazeiks Jan 8 '18 at 2:32
  • $\begingroup$ @R.M It could be anything, we don't know what's on the left and we dont know what's on the right. There is no borders. It could be 1000 ON and 673 OFF and will still be valid. $\endgroup$ – ibrahim mahrir Jan 8 '18 at 2:44
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    $\begingroup$ @ibrahimmahrir oh, I see what you're saying. As long as a pattern has On On Off Off in it, it is valid. Good point. $\endgroup$ – Reinis Mazeiks Jan 8 '18 at 2:49
3
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I guess the answer is

E
There is a pattern to each segment, and the consistent image for the next pattern is E.

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  • $\begingroup$ You and @Alconja are both correct – the answer is E – but the explanation isn't the one I had intended. That is not your fault, of course. I now realize that this is a flaw in my puzzle; the intended solution wasn't supposed to be based on patterns in individual segments. If you want to, try to find an explanation taking into account my edit. If not, I will just post the intended solution and try to not make this mistake in the future. $\endgroup$ – Reinis Mazeiks Jan 8 '18 at 2:02
0
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Same solution as the others, but including the update (only relying on immediate predecessor state)

I'll use logic notation, using Ajconja's picture. X is the current state of the bar X (dark = on = True), X+ for bar X in the next state.

A = C = True
E = False
G+ = !G
D+ = !D
F+ = (F&G) | (!F&!G)
B+ = (B&G) | (!B&!G)

Or F/B put into words: It turns on in next state if it currently has the same state as G

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