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Subject:
Number series
Category: Miscellaneous Asked by: dozer69-ga List Price: $10.00 |
Posted:
05 Apr 2006 21:17 PDT
Expires: 05 May 2006 21:17 PDT Question ID: 715997 |
What is next in the series? 1, 2, 3, 4, 7, 10, 13, 16, 15, ? | |
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Subject:
Re: Number series
Answered By: sublime1-ga on 05 Apr 2006 22:36 PDT |
dozer69... While there's no guarantee that there isn't more than one possible solution, the following consistencies suggest that the next number is 14: 1 is added to the initial number 3 times: 1 +1 = 2 +1 = 3 +1 = 4 3 is added to that number 4 times: 4 +3 = 7 +3 = 10 +3 = 13 +3 = 16 -1 is added to that number to give you 15. The established pattern suggests that that would be repeated 5 times, so that the full series would be: 1, 2, 3, 4, 7, 10, 13, 16, 15, 14, 13, 12, 11 After that, it's anybody's guess, though the most logical pattern would be to add -3 six times: 1, 2, 3, 4, 7, 10, 13, 16, 15, 14, 13, 12, 11, 8, 5, 2, -1, -4, -7. The most obvious pattern, then, is: Add 1 three times. Add 3 four times. Add -1 five times. Add -3 six times. Beyond that, there's no certainty. If anything is unclear or needs further elaboration, let me know... sublime1-ga |
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Subject:
Re: Number series
From: ansel001-ga on 06 Apr 2006 02:04 PDT |
I have some trouble buying your answer Sublime. It doesn't seem very, well, sublime. It seems forced. Here is a link to a database of integer sequences. http://www.research.att.com/~njas/sequences/ If you enter a few consecutive numbers from a sequence it will give you a list of possible sequences that contain those numbers. It is a huge database, but doesn't contain your particular sequence. I'm afraid I don't have any insight to an answer to your question at the present time. |
Subject:
Re: Number series
From: indistinguishable-ga on 06 Apr 2006 02:44 PDT |
ansel001-ga, typically these kind of sequences can be solved by thinking about them; databases generally are not needed. The next number in the sequence is almost certainly 14. After that, it gets a bit more complicated as the sequence doesn't give a whole lot of information. Working with what's there, sublime1-ga certainly provided a plausible explanation of what comes after the 14. 3 times +1 4 times +3 5 times -1 6 times -3 --> this is the most probable number following the -1, but it could really be any other number. Other possibility is: 3 times +1 4 times +3 3 times -1 (note, this still gives 14 as the next number in the sequence that was posted) 4 times -3 --> I think it would be less likely to be another number than -3 in this case. |
Subject:
Re: Number series
From: ansel001-ga on 06 Apr 2006 03:24 PDT |
Indistinguishable, I am well aware that you can solve sequences by thinking about them and that is what I attempted first. If possible I like to find an elegent solution to a problem. The answers you and Sublime provided certainly are possible and may well be correct, but from an elegence standpoint I don't like them. I consider them to be unsatisfying, but not necessarily incorrect answers. Come to think of it, I have run across sequence puzzles with answers like those you gave, and I always thought they were stupid sequences. However, that doesn't mean your answers are incorrect. I included the link to "The Online Encyclopedia of Integer Sequences" because I find it to be interesting. |
Subject:
Re: Number series
From: ansel001-ga on 06 Apr 2006 03:37 PDT |
I forgot to mention one other thing. Usually the sequences are infinite sequences that can be defined in ways such as: The Nth term in the series is N^3 or The next term in the series is the sum of the previous two terms However, the possible solutions that have been proffered are finite sequences. The rules suggested only give a few terms and then stop with no hint as to what, if anything, comes next. |
Subject:
Re: Number series
From: indistinguishable-ga on 06 Apr 2006 12:09 PDT |
Well I prefer having any solution over not having an elegant solution. Given that there are no other suggested solutions currently, this 'forced', 'unelegant', and 'unsatisfying' (using your terms) solution will have to do. |
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