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First sequence.

$1-2-3-5-5-8-7-11-9-4-?-?-13$

second.

$1-99-4851-156849-?-71523144$

I can't find the number in that question marks. I mean I can't find the pattern of these sequences.

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1  
oeis.org/… –  Jyrki Lahtonen Jul 20 '11 at 9:20

2 Answers 2

The second is the progression of the binomial coefficients for 99. So the nth term is ${99 \choose n}$. Nonobvious.

Or maybe it's a progression that just happens to share these values for the first couple indices. Really, it could be a polynomial. As we have 5 known points, we could fit a quartic and come up with an answer. Or we could choose any number of quintics. These questions are so arbitrary, continuing to pervade the idea that every question has exactly one correct answer.

The problem is that while many questions do have a correct answer, that is not at all characteristic of every question. I wouldn't even go so far as to say that those questions that are patently true or false are any better or worse than others. They're just different. But we should not expect every question to have a single correct answer. Nor should we always expect questions to have easy solutions, or pretty solutions, or elegant solutions.

To misquote someone: God doesn't need elegant math. He integrates empirically.

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1  
Which is why these questions are best posed with multiple choice answers. –  Daniel Freedman Jul 20 '11 at 9:39
    
+1. you are perfectly right. –  4545454545SI Jul 20 '11 at 9:51
    
@Daniel Freedman: Multiple choice doesn't really help against smartasses who know Lagrange interpolation — for any answer, you can still find a polynomial interpolating those points. –  Ilmari Karonen Jul 21 '11 at 17:17

I think there is miss typing in your sequence.

Check this out.

$\ \ 1-\ \ 2-\ \ 3-\ \ 4-\ \ 5-\ \ 6-\ \ 7-\ \ 8-\ \ 9-10-11-12-13$

$\ \ 0-\ \ 0-\ \ 0-\ \ 1-\ \ 0-\ \ 2-\ \ 0-\ \ 3-\ \ 0-\ \ 4-\ \ 0-\ \ 5-\ \ 0$

$\ \ 1-\ \ 2-\ \ 3-\ \ 5-\ \ 5-\ \ 8-\ \ 7-11-\ \ 9-14-11-17-13$

So the answer should be $11$ and $17$.

$1\times (1+99)=1\times 100$

$2\times (99+4851)=99\times 100$

$3\times (4851+156849)=4851\times 100$

$4\times (156849+?)=156849\times 100$

$5\times (?+71523144)=?\times 100$

$n\times (a_{n}+a_{n+1})=a_{n}\times 100$

so the answer is $3764376$.

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1  
oeis.org/A063914 –  Martin Sleziak Jul 20 '11 at 9:38
    
Jyrki Lahtonen in his comment above mentioned the site in the connection with the other sequence, so I tried to find there this one, too. BTW I liked your answer. –  Martin Sleziak Jul 20 '11 at 9:44

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