4 ms·
Reminds me of this: 1, 2, 3, 4, 5, 6, 7, 8, 9,... What comes next? 10 if it's the sequence of natural numbers 13 if it's the sequence of N such that N=2^n for
by Moodles 8y ago
Reminds me of this:
1, 2, 3, 4, 5, 6, 7, 8, 9,... What comes next?
10 if it's the sequence of natural numbers
13 if it's the sequence of N such that N=2^n for natural numbers n, where N does not contain a 0
153 if it's sequence of N such that the the sum of the digits of N each raised to the power of the number of digits in N equals N.
- mtreis86 8y agoThose problems always bothered me. I think that for any sequence of numbers there is an infinite number of next-in-sequence solutions regardless of the sequence length or numbers contained. One may be more obvious but you can put any number next and find a pattern that matches. Example - what if those are a sequence of digits in pi.
- cabaalis 8y agoI think "What comes next?" is an incomplete question, without any context. "What comes next in the sequence of (blah)?" is a complete question with full context. But that question wouldn't make anybody feel superior. "Identify as many sequences as you can which fit this set of numbers, and tell me the next number in each sequence" is the non-trick actually being asked.
- jameshart 8y agoI think the only sane approach is to apply a Kolmogorov-shaded Occam's Razor, and select, from the set of possible sequences which fit the pattern, the one which has the simplest generating function.
- CamperBob2 8y agoThat can be tricky in itself, though. 1, 2, 3, 4, .... might be a sequence of positive integers, or it might be a subset of pseudo-random digits emitted by a maximal-length shift register sequence. Someone recently pointed out to me that if you were designing hardware to (e.g.) refresh all memory locations in a system, or otherwise visit a set of 2^n sequential locations in no particular order, then using a maximal-length shift register of the appropriate width is actually simpler than a 'simple' adder or subtractor. That could be seen as a literal interpretation of Kolmogorov complexity.
- pwagland 8y agoWhile I fully agree with the premise that any sequence of numbers can have an essentially infinite choice of 'next number', depending on how you define your sequence, and the simplest way to prove that is simply to given a sequence of _n_ numbers solve the _n+1_ polynomial equation, given that it is incompletely specified then you have an infinite number of solutions. Or just drop in a heavy side step function (https://en.wikipedia.org/wiki/Heaviside_step_function https://en.wikipedia.org/wiki/Heaviside_step_function). However, i'm not sure that I understand your second example. If N=2^n (assuming 2 to the power of n, not 2 xor n) then one would expect the sequence to be 1, 2, 4, 8, 16, etc. If N=2^n using the C xor notation, then we would expect a sequence 3, 0, 1, 6, 7, 4, etc Your third example is pure evil, and I'm glad that I never had you as a maths teacher ;-)
- allenz 8y agoIt's the sequence of n such that 2^n doesn't contain the digit 0. This excludes 10 (1024), 11 (2048), and 12 (4096). See http://oeis.org/A007377 http://oeis.org/A007377
- raincom 8y agoIn philosophy of sciences, it is called 'underdetermination'.
- iopuy 8y agoIn regards to the second example, I understand the sequence but it seems totally arbitrary. It could also be the sequence 11 if it's the sequence of N such that N=2^n for natural numbers n, where N does not contain a 24 But then again that's your point, cheers ;-)
- ChristianGeek 8y agoA if you’re counting in hexadecimal.