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Another variant of my previous question, this one might be easier (or harder).

What is the next number in the sequence $1, 2,6, 16, ...$?

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.

First hint:

enter image description here

Second hint:

enter image description here

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  • $\begingroup$ As the "hints" are required to solve the puzzle (lest it be far too broad) they probably should not be spoilered (and possibly not even referred to as hints...) $\endgroup$
    – Will
    Commented Sep 20, 2016 at 21:53
  • $\begingroup$ @Will, someone edited and added spoilers. I decided to keep them. The point is - ONE hint is enough, any of the two. I'll clarify $\endgroup$
    – Yuriy S
    Commented Sep 20, 2016 at 21:53
  • $\begingroup$ Technically, they might have been intended as clues rather than hints, especially as 4 numbers alone are never enough for a sequence. $\endgroup$
    – humn
    Commented Sep 20, 2016 at 22:09
  • $\begingroup$ @humn, the word 'hints' was used because it goes well with the 'honeycomb' $\endgroup$
    – Yuriy S
    Commented Sep 20, 2016 at 22:09
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    $\begingroup$ You are a puzzler and a poet. $\endgroup$
    – humn
    Commented Sep 20, 2016 at 22:10

2 Answers 2

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On the basis of the second hint the answer is evidently

42

unless I have miscalculated. The idea is that you

proceed clockwise and inward to fill the triangle, starting with a 1 at top left and always filling each cell with the sum of the adjacent numbers you have so far. The black cell is the last one to be filled.

I think (but haven't checked) that the idea of the first hint is that

you start with a 1 and form triangles by a clockwise outward spiral, and each time you form a new triangle larger than its predecessor you record the number in its final cell.

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  • $\begingroup$ You are right. As for the first hint, it's probably harder to figure out $\endgroup$
    – Yuriy S
    Commented Sep 20, 2016 at 22:02
  • $\begingroup$ Having got the second, the first seems pretty straightforward, and I've amended my answer to say what I think is going on there. I haven't checked the actual arithmetic to verify that it works, but it seems like it should. $\endgroup$
    – Gareth McCaughan ♦
    Commented Sep 20, 2016 at 22:06
  • $\begingroup$ Well, there are other ways to describe the same thing (e.g., the number of steps you go before turning increases by 1 each time, and there's a shaded cell at the end of each run of same-direction steps) and what's simplest is a matter of taste. Or do you mean my description is actually wrong? $\endgroup$
    – Gareth McCaughan ♦
    Commented Sep 20, 2016 at 22:30
  • $\begingroup$ I think we may be at cross purposes. The after-the-comma stuff is just saying where the black cells go, which is not the same in the first and second hints. $\endgroup$
    – Gareth McCaughan ♦
    Commented Sep 20, 2016 at 23:32
  • $\begingroup$ Gareth, I want to thank you for your insight, since you are right, the sequence is formed by the last numbers of a finished triangle, and this rule also works for hexagons from the previous question and for any ofther cases I considered. Now I understand this a little better $\endgroup$
    – Yuriy S
    Commented Sep 21, 2016 at 11:34
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Taking the baton from Gareth McCaughan’s solution to the bottom honeycomb hint, pictured as...

   
... where trails begin at the top left corners and proceed in triangular clockwise inward spirals...

...here is how the top honeycomb hint not only gets to the same...

...answer = 42...

...but also...

...must proceed beyond it before all the honeycomb hint cells have honey.
           
This time a triangular clockwise outward spiral begins at the non-black-cell with 1.

As described by Gareth McCaughan...

...each black cell is presumed to be empty before the trail gets there, and each cell to be filled in gets the sum of the numbers already filled in for its neighbors.

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    $\begingroup$ I accepted Gareth's answer, since he was the first and gave the solution in principle, even if without illustrations. I hope you don't mind. You filled all the blanks exactly right $\endgroup$
    – Yuriy S
    Commented Sep 21, 2016 at 11:37
  • $\begingroup$ Good, just wanted to join the fun once the dust had settled $\endgroup$
    – humn
    Commented Sep 21, 2016 at 11:40

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