What's the next number in this sequence?
 

3, 17, 149, 97, 241, 337, 577, 2887, ...

4231.

42

Herr Mayer is correct. Can anyone else guess the pattern?

[QUOTE=grandpascorpion;88084]Herr Mayer is correct. Can anyone else guess the pattern?[/QUOTE]Ah, you wanted us to guess the pattern you chose out of the many patterns that fit the data equally well...


Paul

Well, I thought I supplied enough terms to be unambiguous. For one thing, your answer isn't a prime, like all the other terms in the sequence. (Hint, hint).

But I'll bite ... how did you come up with your answer?

[QUOTE=grandpascorpion;88098]Well, I thought I supplied enough terms to be unambiguous. For one thing, your answer isn't a prime, like all the other terms in the sequence. (Hint, hint).

But I'll bite ... how did you come up with your answer?[/QUOTE]The answer to all questions of this type is 42. It's the answer to the ultimate question of life, the universe and everything.

It is impossible for you to provide enough terms to be unambiguous. No matter how many terms you provide, even a countably infinite number of them, it is always possible to find an uncountably infinite number of reasons to justify any other number as the answer to your question.

Here's another classic. What's the next number in this series: 1, 2, 4, 8, 16?

[spoiler]Answer: 31.
Justification: place 1, 2, 3, 4, 5, 6, ... points on a circle. Join each point to every other point by a straight line. Count the number of areas into which the interior of the circle is divided, paying proper attention to counting degenerate areas where three or more lines meet at an interior point. Alternatively, place the points such that three or more lines never meet at an interior point.
[/spoiler]

Cue discussions of Kolmogorov complexity ...


Paul

Anyways.

The answer I was looking for:

The i-th term in the sequence is the first prime such that one plus that number is a multiple of the i-th prime squared.

3 => 3+1 = 2^2
17 => 17 +1 = 2*3^2

etc.

[QUOTE=grandpascorpion;88109]The answer I was looking for:

The i-th term in the sequence is the first prime such that one plus that number is a multiple of the i-th prime squared.

3 => 3+1 = 2^2
17 => 17 +1 = 2*3^2

etc.[/QUOTE]
So what fabulous prize do I win?

Of course Paul is correct in saying that there are always infinitely many possible anwers to this kind of extend-the-finite-sequence question, but I do believe that this not does preclude a form of Occam's razor from being used, especially when additional context (in this case, it's a prime-number-related forum) is available. In other words, there are similarly infinitely many ways to extend the exampe sequence 1,2,3,4,5,6,..., but most people would agree that the "obvious" or "simplest" next term is 7 rather than 937462927454.

Interestingly, though, "simplest" seems to be a rather arbitrary term here - perhaps one answer is only "simpler" than the others due to it appealing to certain built-in perceptual biases we humans have with respect to these kinds of pattern recognition problems. That is, maybe 7 is only "special" in the foregoing example in the sense of "over the course of many millions of years of evolution, hominids who would have predicted 7 as the next term in the above sequence had a slightly smaller chance of getting eaten than did those who predicted something else."

Anyway, the next several terms in the prime sequence described by Grandpa are as follows:

[i]Sequence is defined as: x_n, n = 0,1,2,..., where x_n is the smallest prime of the form (c_n * p_n^2) - 1, c_n a positive integer.[/i]

The first few terms of the sequence (specifically, for primes p_n < 100) thus defined are

3,17,149,97,241,337,577,2887,4231,10091,7687,16427,3361,3697,39761,56179,6961,44651,35911,70573,10657,49927,41333,63367,56453,...

The associated c_n's are

1,2,6,2,2,2,2,8,8,12,8,12,2,2,18,20,2,12,8,14,2,8,6,8,6,...

Note: For n > 0, c_n must be even, obviously.

Some related questions which may be interesting:

* Are there any n for which c_n has largest factor > p_n, i.e. where the coefficient c_n is less smooth than p_n?

* What is the asymptotic/probabilistic behavior of the c_n as n --> oo ?

Would cyber-satisfaction suffice? :smile:

I understand this argument but as you said, context is key. This forum is about prime numbers and the sequence isn't that obscure if you examine the numbers.

Those are interesting questions. It doesn't look likely for the square case.
For p<14000000, the highest c is only 464.

Maybe for a higher power "x" such a case could be found.

As expected, the high "c" grows much faster with successive powers

x high c prime
3 508 2504681
4 510 309293
5 580 62591
6 522 4113
7 522 7699
8 510 5039
=====================================

I did some rummaging and found the following
C's where the high factor is > p
[code]
x c p
6 26=2*13 7
8 38=2*19 17
10 68=2*2*17 11
11 134=2*67 37 ***
14 14=2*7 5
17 94=2*47 11 ***
17 76=2^2*19 17
17 194=2*97 67 ***
17 254=2*127 103 ***
18 138=2*3*23 13
18 122 = 2*61 31
18 254=2*127 41 ***
19 132=12*11 5
19 66=2*3*11 7
19 86=2*43 37
20 138=2*3*23 5
20 102=2*3*17 13

Other interesting bits:
x c p Comment
5 46=2*23 29 One short
7 134=2*67 103 High P and c-factor
8 464=2^16*29 431 c > p and P is high
11 120=2^3*3*5 7 c is much higher than p
11 38=2*19 19 p equals high factor
13 302=2*151 181 High P and c-factor
15 46=2*23 23 p equals high factor
18 18=2*3*3 3 p = high factor, c=x
[/code]

[QUOTE=ewmayer;88123]Of course Paul is correct in saying that there are always infinitely many possible anwers to this kind of extend-the-finite-sequence question, but I do believe that this not does preclude a form of Occam's razor from being used, especially when additional context (in this case, it's a prime-number-related forum) is available. In other words, there are similarly infinitely many ways to extend the exampe sequence 1,2,3,4,5,6,..., but most people would agree that the "obvious" or "simplest" next term is 7 rather than 937462927454.

Interestingly, though, "simplest" seems to be a rather arbitrary term here - perhaps one answer is only "simpler" than the others due to it appealing to certain built-in perceptual biases we humans have with respect to these kinds of pattern recognition problems.[/QUOTE]
As I said, cue discussions on Kolmogorov complexity.

I would argue that the solution I hid behind spoiler tags to the 1,2,4,8,16 puzzle is just as simple as the "obvious" one which I suspect most people went for.

What is not obvious, to me at least, is why the two solutions differ in their predictions for the next term. That one appears to require much more careful thought.

Paul