17
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Can you work out how this number sequence is built up? Can you work out how it continues?

1, 1, 2, 4, 8, 20, 60, 240, 1500, .......

Hint 1

this is a mathematical puzzle

Hint 2

It does not involve complicated mathematics

From time to time hints will be updated.

Note: this sequence cannot be found in The On-Line Encyclopedia of Integer Sequences

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22
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The next number is

16140

Following this pattern, given the sequence is $a_n$

$a_n = a_{n-1} + a_{n-2} \cdot (a_{n-3} + 1)$

Example: 60 = 20 + 8 * (4 + 1)

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  • $\begingroup$ Great job, well done. -- The solution you have is correct, but expressed ever so slightly differently to the way I had in mind, but it is equivalent. The way I put it together was $a_n = a_{n-1} + a_{n-2} + a_{n-2}.a_{n-3}$. So Well done, congratulations. $\endgroup$ – tom Mar 7 '18 at 10:01
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enter image description here

It seems like I was way off on the pattern I was trying to track. Seems like this is a neat coincidence though.

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    $\begingroup$ What does this mean? I'm confused. $\endgroup$ – NoOneIsHere Mar 6 '18 at 21:52
  • $\begingroup$ looks interesting, but I don't understand this either - can you explain a bit more? $\endgroup$ – tom Mar 7 '18 at 9:58
  • $\begingroup$ It's the distances between the first point and all the other points in the sequence plotted as (x,y) where x = 1, 2, 3. . . and y = the sequence. It seems like distance = y(rounded to the closest even number) $\endgroup$ – Patrick Mar 7 '18 at 14:28
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    $\begingroup$ What you effectively have shown there is that $$Lim_{n \to \infty} \sqrt{(X_n - 1)^2 + (Y_n - 1)^2} \to Y_n$$. Which is another way of saying Y grows much faster than x... or by more than 1 per term. $\endgroup$ – Fifth_H0r5eman Mar 7 '18 at 14:59
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Obligatory nth order polynomial:

$$f(x) = \frac{5}{384}x^8-\frac{4597}{10080}x^7+\frac{19471}{2880}x^6 - \frac{19831}{360}x^5+\frac{308935}{1152}x^4-\frac{1141669}{1440}x^3+\frac{1982957}{1440}x^2-\frac{354937}{280}+465$$

Predicted next five values:

8116, 33619, 112289, 318913, 800021

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  • 1
    $\begingroup$ I dislike people reflexively pointing out that there is always an nth-order polynomial (or spline) that fits n arbitrary points. Anyway the spirit of this question forbade it: "It does not involve complicated mathematics" (yeah you could try to argue that nth-order interpolation doesn't). If the OP wanted to be more lawyerly they could have said "the answer has < n characters and only requires the four elementary operations". $\endgroup$ – smci Mar 7 '18 at 3:34
  • $\begingroup$ "If one plugs it into the given numbers, they all match... the next fives values are [given integers]" I doubt it does, it looks like you're doing rounding, most of your subterms aren't even integers. $\endgroup$ – smci Mar 7 '18 at 3:36
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    $\begingroup$ Ugh, freaking datatypes. I guess I should've done it by hand. Anyways, I've cleaned up the post after realizing how abrasive it was. $\endgroup$ – Jakob Lovern Mar 7 '18 at 4:17
  • $\begingroup$ Your post was perfectly fine, just without the first sentence, also noting that in general that f(n) will give non-integer values. Curious which language you used, Python or what? $\endgroup$ – smci Mar 7 '18 at 4:40
  • $\begingroup$ Interesting solution - sorry not correct, but +1 for the polynomial fitting $\endgroup$ – tom Mar 7 '18 at 9:58

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