1
$\begingroup$

I remember reading that 'the next number in a sequence of numbers can be anything. It is all about finding the a relation between previous numbers such that the required number becomes next in sequence'.

For e.g: take the sequence 3,5,7...

  1. The next number can be 9, if we look at the sequence as A.P with common difference of 2.
  2. The next number can be 11, if we look at the sequence as sequence of prime numbers...

Basically it can be anything, if we can think of a relationship between the numbers...

I can't remember the name of the paradox. Can somebody help me?

Thanks, tecMav.

  • 0
    Maybe you want to read about "Lawlike and lawless sequences", "Random Sequences" and "Choice Sequences" to find your paradox name. However, those terms refer to solid facts.2012-11-01

2 Answers 2

2

Among all "programs" producing an infinite sequence with the given beginning there is a shortest one. Occam's razor tells you to use this program in order to produce the next term in the sequence.

  • 0
    And a minor point: Only computable sequences can be produced by programs, so Kolmogorov complexity doesn't apply to non-computable sequences. And only definable sequences can be subject to Occam's razor, so Occam's razor doesn't apply to non-definable sequences.2014-06-06
0

What Mark Bennet,xavierm02 and in a different way André Nicolas are saying is that whatever the next term in the sequence is, there is a relation that gives you that first terms of the sequence (this relation can be (a Lagrange) polynomial $f(n)$).

For example suppose that you have the sequence $3,5,7,\ldots .$ The next term can be $100$ and the relation giving your sequence can be $f(n)$ with $f=\frac{91}{6}x^3-91x^2+\frac{1013}{6}x-90$.