6
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$1,20,171,?,9625,57756,288775$

Please find the missing number in the series.

I have already tried with squares, cubes and progressions but couldn't find the problem.

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17
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Starting with $1$, you proceed as follows:

Add $1$, and multiply the result by $10$.

This gives you $20$.

Subtract $1$, and multiply the result by $9$.

This gives you $171$.

Add $1$, and multiply the result by $8$.

This gives you $1376$.

Subtract $1$, and multiply by $7$.

This gives you $9625$. Etc.


How do you get to this result?

Look at the last three numbers given. If you take the ratio of the last two values ($288775/57756$), you get a result very close to $5$. If you take the ratio $57756/9625$ You get a result very close to $6$. From there on it's just some fiddling...

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  • 4
    $\begingroup$ Awesome!!....missed taking the ratios! $\endgroup$ – G.One Jan 3 '15 at 9:44
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Partial Answer:

Let

$1=a_1$, $20=a_2$, etc $\ldots$

then

it seems like for some number $n$ that belongs to the group of numbers $\{1,2,3,\ldots\}$ (i.e. a natural number),

we have that

$a_{n+1}-a_n$ is prime!

$$a_2-a_1=20-1=19\tag{$n=1$}$$ $$a_3-a_2=171-20=151\tag{$n=2$}$$ Skipping $n=3$ because of that unknown number, $$a_4-a_3=57756-9625=48131\tag{$n=4$}$$ $$a_5-a_4=288775-57756=231019\tag{$n=5$}$$

Let

the $m^\text{th}$ prime number be denoted as $p_m$,

then we have

a pattern! $$19,151,(a_3-a_2), 48131, 231019$$ $$\Downarrow$$ $$p_8, p_{36}, (a_3-a_2), p_{4957}, p_{20524}$$ And now the pattern becomes $$8, 36, ?, 4957, 20524$$

But I can't find a pattern. One thing I noticed is that

$$4957=\left(44-3^3\right)^3+44$$ and the close similarity, $$36 = \left(29-3^3\right)^3+29-1$$ (and, something pretty cool: $37^2=\left(38-3^3\right)^3+38$).

But, although similar, they do not match quite enough for me to say they belong to a kind of pattern. Overall, however, it is remarkably interesting :P

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  • 1
    $\begingroup$ hmm...nice try! $\endgroup$ – G.One Dec 8 '18 at 7:40

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