13
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Find the next number in the sequence $1,7,74,\dots$

First hint:

enter image description here

Second hint:

enter image description here

To clarify: the 'hints' are really clues. They are hidden, because any of the two alone could be enough to solve the puzzle. So, anyone who is interested, can look up only one hint, or both.

There is a simple rule for both cases, all the blanks can be filled and the sequence can be continued ad infinitum.

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  • $\begingroup$ Refreshingly large clues/data ratio! (just 3 actual data and 2000 words worth of clues) $\endgroup$ – humn Sep 20 '16 at 19:44
  • $\begingroup$ @humn, thanks then. This is my first post here $\endgroup$ – Yuriy S Sep 20 '16 at 19:47
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    $\begingroup$ @humn, I made it up, and I haven't seen this sequence (or the arrays as a whole) anywhere $\endgroup$ – Yuriy S Sep 20 '16 at 20:10
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    $\begingroup$ Looks like a new take on Minesweeper. :) $\endgroup$ – Ian MacDonald Sep 20 '16 at 20:17
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    $\begingroup$ If you want to post another puzzle, go ahead and do it in a new question! $\endgroup$ – Deusovi Sep 20 '16 at 21:22
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Your answer is, after some minor corrections:

1017.

Because:

enter image description here

The rule is that

you fill out the honeycomb by simply adding the sum of the surrounding cells into the next cell clockwise.

The second hint can be filled out analogously:

enter image description here

It would be interesting to see in another solution what is the mathematics behind this rule. I suspect nice recursions lurking behind the scenes.

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  • $\begingroup$ Indeed, as 129+12+14=155. Thanks. $\endgroup$ – Matsmath Sep 20 '16 at 20:02
  • $\begingroup$ Now it is correct. Would you like to fill in the hexagons below as well? $\endgroup$ – Yuriy S Sep 20 '16 at 20:12
  • $\begingroup$ I see now that it follows (perhaps) exactly the same rule. I will do that. $\endgroup$ – Matsmath Sep 20 '16 at 20:14
  • $\begingroup$ Matsmath, I added another case, I hope you don't mind $\endgroup$ – Yuriy S Sep 20 '16 at 20:37
  • $\begingroup$ If the answer to that puzzle is different from this, then I would suggest you to start another post on that. Perhaps there are others who would like to pursue that challenge. $\endgroup$ – Matsmath Sep 20 '16 at 20:41

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