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These numbers are created with a formula. Guess the formula for increment and calculate at least 3 next numbers of this sequence.

increment is started with two given first $1$s.

$1,$ $1,$ $2,$ $3,$ $11,$ $44,$ $129,$ $557,$ $2354,$ $7059$

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    $\begingroup$ I just spent ten minutes with my TI-84 and WolframAlpha trying to figure this out. What sorcery is this?! $\endgroup$
    – user9
    Commented Oct 6, 2014 at 18:03
  • $\begingroup$ I'll try it when I can $\endgroup$
    – warspyking
    Commented Oct 6, 2014 at 18:04
  • $\begingroup$ I've spent and hour trying to figure this out D: $\endgroup$
    – warspyking
    Commented Oct 6, 2014 at 19:24
  • $\begingroup$ @Undo, Happy to know... lol. That was my goal to challenge puzzle solvers ;) $\endgroup$
    – Rafe
    Commented Oct 6, 2014 at 19:25
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    $\begingroup$ @Seb, the only hint I can give is: nothing more than elementary school $\endgroup$
    – Rafe
    Commented Oct 6, 2014 at 19:40

3 Answers 3

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I'll write $S_n$ for the $n$-th term in the sequence. The terms you listed satisfy the recurrence $$ S_{n+1}=\begin{cases}4S_n+S_{n-1}-3\text{ if $S_n$ is odd,}\\3S_n-3\text{ if $S_n$ is even.}\end{cases} $$ This predicts that the next three terms are 30587, 129404, 388209.

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  • $\begingroup$ Well done! :-) the equivalent formula for excel is : (A2-1)*4+1+IF(ISODD(A2),A1,A2*-1) for A3 cell $\endgroup$
    – Rafe
    Commented Oct 6, 2014 at 20:48
  • $\begingroup$ Nice one, found the $3S_n -3$ thing one minute before you posted the answer. But I didn't assign it to the even property, I just saw that this rule was applied at every 3 calculations. $\endgroup$
    – Jakube
    Commented Oct 6, 2014 at 20:50
  • $\begingroup$ Congratulations. How long did it take you -_- took me an hour and a half and apparently it "didn't work" $\endgroup$
    – warspyking
    Commented Oct 6, 2014 at 20:51
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After about an hour and a half, this is what I came up with;

1, 1, 2, 3, 11, 44, 129, 557, 2354, 7059, 11961, 19233, 66058

I used a difference table to figure it out:

1, 1, 2, 3, 11, 44, 129, 557, 2354, 7059, ?, ?, ?

0, 1, 1, 8, 33, 85, 428, 1797, ?, ?, ?

1, 0, 7, 25, 52, 375, 1369, ?, ?, ?

-1, 7, 18, 27, 323, 994, ?, ?, ?

8, 11, 9, 296, 671, ?, ?, ?

3, -1, 287, 375, ?, ?, ?

-4, 288, 88, ?, ?, ?

292, -200, ?, ?, ?

-492, ?, ?, ?

Start from the bottom, and work your way up, using the difference table;

1, 1, 2, 3, 11, 44, 129, 557, 2354, 7059, 11961, 19233, 66058

0, 1, 1, 8, 33, 85, 428, 1797, 4902, 7272, 46825

1, 0, 7, 25, 52, 375, 1369, 3105, 21370, 39653

-1, 7, 18, 27, 323, 994, 1736, 17265, 18283

8, 11, 9, 296, 671, 742, 889, 1018

3, -1, 287, 375, 171, 167, 129

-4, 288, 88, -204, -4, -296

292, -200, -292, 200, -292

-492, -492, -492, -492

Note: If I did any of the addition/subtraction incorrectly feel free to edit the answer!

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    $\begingroup$ This is not a solution. You can use it for every other pattern question. Think about first two 1s. How did they produce 2 and 3? they did not made any the same numbers with your solution. - sorry for not tagging you. idk whats wrong with tag part. $\endgroup$
    – Rafe
    Commented Oct 6, 2014 at 20:30
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    $\begingroup$ @Rafe it is still an acceptable sequence of numbers using your original sequence. Just because it isn't the answer you have, don't mean it is wrong. If there is a glass half full of water on the table and you ask me how full/empty it is, and you think it's half empty, but I think it's half full, does that make me wrong $\endgroup$
    – warspyking
    Commented Oct 6, 2014 at 20:32
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    $\begingroup$ So there is always a solution using newton's polynomial to these kind of puzzles. I insist that your solution is not an acceptable answer $\endgroup$
    – Rafe
    Commented Oct 6, 2014 at 20:34
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    $\begingroup$ @Rafe I spent a freaking hour and a half working things out, trying to solve YOUR question. I finally figure out 3 actual numbers that fit the sequence, and you say it's wrong!?!? The numbers work out, I can't believe you can say I'm incorrect... $\endgroup$
    – warspyking
    Commented Oct 6, 2014 at 20:44
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    $\begingroup$ @Rafe Yeah, this is the problem with sequence puzzle, because there can be infinitely many answers. People proposing these kind of questions should be aware of this and not criticizing different answers, perhaps they can say "This is not the answer that I'm looking for, but thank you for your correct answer!" $\endgroup$
    – justhalf
    Commented Oct 7, 2014 at 2:00
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Everyone else's solution is incorrect. The correct solution is to note that the sequence is generated as the values of the following function:

$$f(x)=-\frac{113 x^9}{72576}+\frac{1261 x^8}{20160}-\frac{8917 x^7}{8640}+\frac{4391 x^6}{480}-\frac{821113 x^5}{17280}+\frac{141589 x^4}{960}-\frac{24088019 x^3}{90720}+\frac{1269077 x^2}{5040}-\frac{17357 x}{180}+3$$

Plugging in $x=1$ gives 1, ..., and plugging in 10 yields 7059. The first 20 values, are, in detail:

$$S=\{1,1,2,3,11,44,129,557,2354,7059,13467,6997,-65218,-343550,-1143128,-3097105,-7378745,-1 6030523,-32434384,-61963825\}$$

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  • $\begingroup$ That's a newton polynomial if I'm not mistaken $\endgroup$
    – Rafe
    Commented Oct 7, 2014 at 5:58

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