Question : What is the next number in the following sequence? (not finite)


My Solution

  • I'm trying a closed form solution

  • I tried computing the differences- they don't converge

  • Tried finding the relationship between their binary equivalents

  • A trivial observation : all numbers are divisible by $6$ except $440$

Edit-1: Seeing the downvotes, I want to clarify something: All I'm trying to do is to find a solution (an expresssion) that represents the sequence

Edit-2 : This question was given by my Math Teacher as a fun exercise in puzzle solving . It is not for a graded homework


closed as too broad by warspyking, Rand al'Thor, GOTO 0, A E, Sp3000 Dec 29 '14 at 17:59

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ This question appears to be off-topic because it would be better suited to Math.SE. $\endgroup$ – A E Dec 29 '14 at 17:11
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    $\begingroup$ @ForIInRange, I think it's something we need a mod to do. I've flagged it. $\endgroup$ – A E Dec 29 '14 at 19:12
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    $\begingroup$ Doing a quick search on Math.SE, it looks like "find the next term in this sequence" questions aren't too well received there $\endgroup$ – Julian Rosen Dec 29 '14 at 19:26
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    $\begingroup$ If it was copied by a teacher on the blackboard and then copied down by you, there is a potential for errors. Could it be oeis.org/A090821 (as suggested by GOTO0)? There would be a 48 missing after 24 and a 420 instead of 440. $\endgroup$ – Florian F Dec 30 '14 at 0:12
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    $\begingroup$ If you are looking for any solution that fits, a simple overfitted polynomial will give you your answer. $\endgroup$ – March Ho Dec 30 '14 at 11:53

The sequence is

4x6, 6x9, 9x10, 10x12, 12x14, 14x15, 15x16, 16x18, 18x20, 20x22, ...

So we have a sequence 4, 6, 9, 10, 12, 14, 15, 16, 18, 20, 22, ... such that

the products of adjacent pairs of terms in this sequence gives the OP's original sequence.

I'd guess the next term in this sequence is 24, so that the answer to the OP's question is

22x24 = 528,

but I'm not certain how to back this up!

One possibility is that there's a mistake in the question and our new sequence should in fact be 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, ..., the sequence of composite numbers, as @GOTO0 suggested. Otherwise, I need to give it more thought.

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    $\begingroup$ I was tempted to think of the product of successive composite numbers, but where are 8 and 21? $\endgroup$ – GOTO 0 Dec 29 '14 at 14:41
  • $\begingroup$ Thank you for your reply. How did you guess 24? $\endgroup$ – Reptilian Dec 29 '14 at 14:43
  • $\begingroup$ @Reptilian Most of the gaps between successive terms in the new sequence seem to be 2. But only most, so I'm not certain of 24. $\endgroup$ – Rand al'Thor Dec 29 '14 at 14:45
  • $\begingroup$ @randal'thor 24 makes sense in this sequence: (2 multiples of 2) (next highest multiple of 3) (3 multiples of 2) (next highest multiple of 3) (4 multiples of 2) (next highest multiple of 3). Of course, it could only be odd multiples of 3, in which case the next multiplier would be 27. $\endgroup$ – Roger Dec 29 '14 at 15:29
  • $\begingroup$ @Roger That really doesn't sound like a neat mathematical rule, as we would expect from this kind of question $\endgroup$ – For I In Range Dec 29 '14 at 19:23

Unless this is the intended answer, in which case it's a bad question anyway, this answer serves to show how broad the question is.

We can always extend a sequence by taking the difference of subsequent elements, which yields a sequence with one fewer number. Once we get a list where all numbers are the same (which will eventually happen if we "bottom out" to get 1 element, but may happen sooner), we can start extending that bottom-most list and propagate the answer up to the top.

24, 54, 90, 120, 168, 210, 240, 288, 360, 440
30, 36, 30, 48, 42, 30, 48, 72, 80
6, -6, 18, -6, -12, 18, 24, 8
-12, 24, -24, -6, 30, 6, -16
36, -48, 18, 36, -24, -22
-84, 66, 18, -60, 2
150, -48, -78, 62
-198, -30, 140
168, 170

In this case we go all the way down and get to the value 2. The next element in such a list is the sum of all the rightmost numbers:

2 + 170 + 140 + 62 + 2 - 22 - 16 + 8 + 80 + 440 = 866

This yields a polynomial fit to your problem which is guaranteed to always produce integers (and, after a while, to always produce distinct increasing integers).

  • $\begingroup$ Sorry warspy, but that's obviously wrong! $\endgroup$ – Rand al'Thor Dec 29 '14 at 14:59
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    $\begingroup$ I meant it's obviously not the intended answer (but I didn't DV!) It shows that the question is bad; sequence questions generally are. Someone should add a comment to this question saying so. (I would have, but h34 is an old enemy of mine IRL, so...) $\endgroup$ – Rand al'Thor Dec 29 '14 at 15:10
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    $\begingroup$ @warspyking while the method of determining a sequence rule based on the differences between terms is usually valid, it does not apply to this case. The final tier of differences that you reached was $-60$, but then you assumed that this was the very final tier possible. In order to be sure that a tier is the final tier of differences, there must be a tier size of larger than $1$. Hope this answered your question! $\endgroup$ – For I In Range Dec 29 '14 at 15:10
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    $\begingroup$ @rand The point of this answer is to show how broad the question is. $\endgroup$ – warspyking Dec 29 '14 at 15:18
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    $\begingroup$ @warspyking - Wittgenstein said the next term in any sequence could be anything you like. I've never considered that to be a very helpful observation. Is that the point you're making? I don't think this question is very interesting, but while you say you're seeking to demonstrate that it's "broad", you don't succeed, if by "broad" you mean that several reasonable answers are possible and one doesn't stand out as more reasonable than the others. Having read the above, I think that must be what you do mean. $\endgroup$ – h34 Dec 29 '14 at 15:36

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