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What term will come next: 5, 6, 8, 11, 45, 74, 101?

I saw this question here on Quora. I am thinking everything, but cannot seem to find a pattern. I am thinking, 5(3rd prime), two numbers based on some pattern, 11(5th prime), another two numbers on that pattern, 101(26th prime).

Still looks very far fetched.

Any help is appreciated.

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closed as too broad by AJL, dennisdeems, McMagister, Deusovi, CodeNewbie Sep 1 '15 at 13:54

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.

  • $\begingroup$ I don't know $\endgroup$ – Raystafarian Aug 5 '15 at 16:45
  • $\begingroup$ See my post below for a proper mathematical answer. If it is not satisfying, see math.stackexchange.com/q/111440/21820 for some striking examples of false patterns that show that there is really no such thing as finding the next number in a sequence. $\endgroup$ – user21820 Sep 1 '15 at 14:30
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I think this series has less to do with primes. What you can try is check whether the differences of some terms are in A.P. or not.
For example:

6-5=1
8-6=2
11-8=3

since 1,2,3 are in AP. you can proceed with this fact.

[edit]: Now, 45-11=34 so it breaks the AP pattern.
But we have:
74-45=29
101-74=27
Since 34,29,27 form a series( diference 5,2 and then 1 as 5/2=2 and 2/2=1. by integer division like in C/C++)
Thus the next term is: 101+ {27-(2/2)}= 127

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  • $\begingroup$ I'm lost on your conclusion. you proved that there are 2 groups of 3 numbers that form 2 series [hidden in the form of 7 numbers] (which as far as I can tell are unrelated), but I don't understand how you got the 7th number. $\endgroup$ – Kingrames Sep 1 '15 at 11:22
  • $\begingroup$ @Kingrames, I am taking 101 as "the 7th number". the difference between consecutive terms 11,45,74,101 are 34,29,27 respectively. $\endgroup$ – ABcDexter Sep 1 '15 at 18:56
  • $\begingroup$ I meant the 7th number in the series 1, 2, 3, 34, 29, 27, X. $\endgroup$ – Kingrames Sep 1 '15 at 18:57
  • $\begingroup$ Ok, then read this update please. $\endgroup$ – ABcDexter Sep 1 '15 at 19:11
  • $\begingroup$ @Kingrames, I am taking 101 as "the 7th number". It is already given in the series, so fitting it in a logical pattern is what I did. What I did might seem a bit awkward, but it's what first came to my mind. $\endgroup$ – ABcDexter Sep 1 '15 at 19:18
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There is not enough information to determine the answer. The simplest explanation is that the sequence is:

$5,6,8,11,45,74,101,101,101,101,\cdots$ (remaining constant all the way)

The reason is that any other explanation will almost certainly require a longer description of the sequence than this one. So by Occam's razor this one would be preferable.

My answer can be easily made mathematically precise via information theory, specifically one can define Kolmogorov complexity. To be fair we cannot choose a language specially designed for a particular answer otherwise its complexity can be made to be arbitrarily small. If we hence choose any mainstream Turing complete language we will almost certainly find that the sequence I gave above has the smallest program generating it.

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    $\begingroup$ Occam's razor simply says that we should make no more assumptions than are necessary. It doesn't say we should cast aside assumptions because they don't suit us. A condition implicit in the question is that the terms in the series are generated by some principle that can be deduced by analysis of the first six terms. Occam would certainly not recommend disregarding this condition. $\endgroup$ – dennisdeems Sep 1 '15 at 13:50
  • $\begingroup$ I don't know why my comment got deleted. My answer is mathematically correct but people who do not know any mathematics downvote for no reason and even remove my comment. "deduced by analysis" is totally imprecise and cannot be made so without something equivalent to Kolmogov complexity that would support my answer rather than dennisdeems' popular comment. $\endgroup$ – user21820 Sep 3 '15 at 14:20
  • $\begingroup$ @user21820 It's quite bold to call people here without mathematical knowledge. You should some of the top math puzzles here. In any case, the issue your post is not its mathematical incorrectness (which it is not, it is correct by all means), but this isn't a valid use of Occam's razor. According to your argument, any mathematical work on any sequence (Fibonacci, geometric, arithmetic and plenty others ) is pointless. $\endgroup$ – Rohcana Sep 11 '15 at 22:34
  • $\begingroup$ @Anachor: I am bold when I know I am certainly correct, and I do not care about popular opinion after I know exactly where it is wrong. I'll assume you honestly do not understand why Occam's razor does apply here and give a brief explanation. The question asks "what comes next in the sequence" with zero information about what kind of sequence it is. Not one bit. So the easiest explanation is exactly the one I gave. In contrast, we study those sequences you name precisely because they arise in nature or in many applications, in which case we never need to ask what is next. [continued] $\endgroup$ – user21820 Sep 12 '15 at 8:03
  • $\begingroup$ @Anachor: Instead we have already defined those sequences so what comes next is already determined! Nothing to do with Occam's razor. If instead you want to ask why we define those sequences, it is simply because they actually come up in the real world or in applications. In the case of the real world, we define sequences among other things that we hope can model some aspect of the world. We still choose the simplest according to Occam's razor. For instance if you see something going in a straight line for a period of time and know nothing else, you rightly guess it goes on that way. $\endgroup$ – user21820 Sep 12 '15 at 8:06

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