4
$\begingroup$
                                       ?

                                      233

                                       55

                                       13

                                        3

                     ?, 377, 89, 21, 5, 1  4, 25, 145, 850, ?

                                        7

                                        40

                                       235

                                       1375

                                        ?

What comes next (where the question marks are)?

$\endgroup$
5
$\begingroup$

This is Very Tricky!

Here are the most obvious patterns that come to mind:

Left branch:

Pattern: $ x_n = 4 x_{n-1} + x_{n-2} $

Top branch:

Pattern: $ x_n = 4 x_{n-1} + x_{n-2} $

Right branch:

Pattern: $ x_n = 5 (x_{n-1} + x_{n-2}) $

Bottom branch:

Pattern: $ x_n = 6 x_{n-1} - 5 x_{n-3} $

But then there's the metapattern, that is, the pattern of patterns, that doesn't look quite right:

The top pattern is the same as the one to the left, but the bottom and the right look different.

That doesn't fit the puzzle's aesthetics at all, but what can we do?

As long as we ignore the "obvious pattern" for the right branch, we can also get the exact same values by using the "less obvious" pattern from the bottom branch!

So the initial values in the right branch were deliberately chosen to mislead, and the pattern in there is secretly

the same as the one in the bottom branch!

And then I woke up and took my hand out of the chamber pot:

The patterns are the same. The bottom branch pattern is also $$ x_n = 5 (x_{n-1} + x_{n-2}) $$ and any sequence generated by that pattern automatically will also have the $$ x_n = 6 x_{n-1} - 5 x_{n-3} $$ pattern: we get from the former to the latter by adding an $x_{n-1}$ and subtracting $5(x_{n-2} + x_{n-3})$, which clearly are the same number according to the former equation. So the "obvious" pattern we got for the bottom branch earlier clearly wasn't the obvious one, since the one for the right branch takes only two numbers to initialise. Ouch!

So then let's just fill in the question marks in, while I bow my head in shame:

                                      987

                                      233

                                       55

                                       13

                                        3

>                 1597, 377, 89, 21, 5, 1  4, 25, 145, 850, 4975

                                        7

                                       40

                                      235

                                     1375

                                     8050

I'm leaving the story as it is, since it depicts my thought process, which might bring some entertainment (or at least schadenfreude) to some of you. I'm only hoping there isn't some other even more obvious patter I'm also missing here :-)

|improve this answer|||||
$\endgroup$
  • $\begingroup$ @Βass You are very close to the correct answers but not quite right. $\endgroup$ – Vassilis Parassidis Feb 2 at 16:42
  • $\begingroup$ @VassilisParassidis I don't see how that would work: either my numbers are correct, or they are utter garbage; there's no in-between really. The first case occurs if each branch is an independently solvable, iterative sequence, as I've assumed with reasonably strong evidence (there are three matching steps in every sequence, so there are 12 data points in support of this theory). The other case occurs if the sequences are something else: any semblance between my answers and the correct ones must then surely be completely coincidental. $\endgroup$ – Bass Feb 2 at 17:09
  • $\begingroup$ The sequences work like this: L+Q , 2L+Q From the first relation we obtain the sequence 1,2,5,13,34...….. From the second relation we obtain the sequence 1,3,8,21,55..... Fibonacci did not invent these sequences, he found them. The above sequences have different coefficients. $\endgroup$ – Vassilis Parassidis Feb 2 at 18:24
  • $\begingroup$ @VassilisParassidis Your comment contains many sentences that make sense on their own. However, putting them together, I have absolutely no idea what you are trying to say. (That's mostly because I'm assuming you are not trying to say that my answer is exactly correct, apart from not correctly guessing which of the many interchangeable methods of generating the sequences was used.) $\endgroup$ – Bass Feb 2 at 19:02
  • $\begingroup$ My sequences are correct as you can see in my comment. if I reveal the coefficients of the formulas all the sequences would make perfect sense to you. I am sad I get punished because you cannot answer the question. If you want me to reveal the formulas please ask for that. $\endgroup$ – Vassilis Parassidis Feb 2 at 19:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.