3
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3 4 9 8 9 8 13 14 13 ?

So what is the next number?

  1. Hint

    the actual difference between numbers is just 1 and the sequence actually starts with 1.

  2. Hint

    Spelling is the key!

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  • $\begingroup$ Your sequence is too short, and/or consists of not enough numbers, for anyone to guess it without hints. (In fact, given no context at all, ? might be any number, as pointed out by Fimpellizieri’s answer.) $\endgroup$ – Lynn Feb 13 '16 at 22:39
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    $\begingroup$ it is not that complicated @Lynn . just need different point of view than just numbers in a sequence... i will give a hint soon... this is regular sequence with a little tweak in it. $\endgroup$ – Oray Feb 13 '16 at 22:42
  • $\begingroup$ you say spelling is key. just for clarity the numbers present are: three four nine eight nine eight thirteen fourteen thirteen ? $\endgroup$ – Bort Feb 15 '16 at 14:10
  • $\begingroup$ @Bort check hint 1 and hint 2 together before spelling the sequence... $\endgroup$ – Oray Feb 15 '16 at 14:18
  • $\begingroup$ I have an idea that almost works but it requires the 14 being 12 instead. This isn't the case by any chance, is it? $\endgroup$ – hexomino Feb 15 '16 at 18:08
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Going for the obvious here.... (because someone has to..)

14.

The whole thing is very arbitrary. Goes from 3 -> 4 then jumps up 6 and sequences 1 down 9 -> 8 -> 9 -> 8 and then jumps up 6 again (from 8 to 14: just wrote this for the sake of some sort of pattern) and sequences up 13->14->13->14.

Very disappointed if this is the answer :-{

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  • $\begingroup$ gladly no :) this is not the answer $\endgroup$ – Oray Feb 14 '16 at 10:50
  • $\begingroup$ @Oray ...phew :-] $\endgroup$ – Ben Feb 14 '16 at 10:53
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I think the answer is

12

The formula for generating the $n$th term in the sequence is

$n$ + 2*(number of letters in English spelling of $n$) - 4

For example the first term in the sequence is

1 + 2*3 - 4 = 3

The 9th term is

9 + 2*4 - 4 = 13

And so the 10th term is

10 + 2*3 - 4 = 12

In relation to the hints

The underlying sequence is the positive integers beginning at 1 and generation of the sequence depends on the spelling of the numbers.

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  • $\begingroup$ this is very close, actually the answer is right but the methodology is not... luckily i believe u got the right answer to be honest :) u are not supposed to extract 4... it is so weird u got it because of the number of vowels and consonants in the spellings :p $\endgroup$ – Oray Feb 15 '16 at 23:01
  • $\begingroup$ Interesting. I was pursuing a line of thought which involved numbers of vowels and consonants but couldn't get it to fit without some cumbersome rules. Can you tell me this, what is the smallest value of n such that the nth term in your sequence is different to that in mine. This may help me with the reasoning. $\endgroup$ – hexomino Feb 15 '16 at 23:50
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    $\begingroup$ ok number+#ofvowels*#ofconsonant try it... $\endgroup$ – Oray Feb 17 '16 at 8:04
  • $\begingroup$ Yeah, that makes sense. Working it out, our two sequences match up if either there are two consonants or two vowels in the word. So the first time there is a difference will be 11 (yours is 20, mine 19). Although, we should have the same result for n = 12, 22, 26, 32, 36, 42, 46, 52, 56, 62, 66, 70, 80 and 90. $\endgroup$ – hexomino Feb 17 '16 at 10:13
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Consider the following polynomial:

$$P(x) = 898-k-\frac{507793}{210}\cdot x-\frac{5}{192}\cdot kx^6+\frac{3013}{17280}\cdot kx^5-\frac{95}{128}\cdot kx^4+\frac{4523}{2268}\cdot kx^3-\frac{6515}{2016}\cdot kx^2+\frac{7129}{2520}\cdot kx+\frac{1092977}{420}\cdot x^2+\frac{1}{362880}\cdot kx^9-\frac{1}{8064}\cdot kx^8+\frac{29}{12096}\cdot kx^7-\frac{5}{4032}\cdot x^9-\frac{7552313}{5040}\cdot x^3+\frac{41377}{80}\cdot x^4-\frac{107677}{960}\cdot x^5+\frac{77}{5}\cdot x^6-\frac{4363}{3360}\cdot x^7+\frac{103}{1680}\cdot x^8$$

It is a straightforward calculation to check that:

  • $P(1) = 3$
  • $P(2) = 4$
  • $P(3) = 9$
  • $P(4) = 8$
  • $P(5) = 9$
  • $P(6) = 8$
  • $P(7) = 13$
  • $P(8) = 14$
  • $P(9) = 13$
  • $P(10) = k$

So it might be that there are multiple solutions?

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  • $\begingroup$ it is actually much easier than you think... but a little different. i will give a hint tomorrow if cannot be solved. $\endgroup$ – Oray Feb 13 '16 at 22:41
  • $\begingroup$ And I thought that working with such a long polynomial is illegal...:p $\endgroup$ – manshu Feb 14 '16 at 7:43
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It's not 100% convincing, but worth a try:

3 4 9 8 9 8 13 14 13 ?
Numbers 3, 6, 9: 9, 8, 13 (no idea. Typo? :)
Numbers 2, 5, 8: 4, 9, 14 (+5)
Numbers 1, 4, 7: 3, 8, 13 (+5)
Number 10: 13 + 5 = 18

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I thought the question would be solved after giving Hint 1 and Hint 2 but I could not get the right one.

Methodology

Since the sequence starts with 1 and increases by 1 (Hint 1) and spelling was important (Hint 2), You needed to check the spelling of 1, which is one. There are two vowels and one consonant and the generalization is found from here;

Equation

Number + (# of Vowels)x(# of Consonants)

The Answer

For 10, ten has two consonants and one vowel so 10+2x1=12.

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