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I heard this number sequence problem about 25 years ago from a maths lecturer.

What number should go in the place of the question mark in this mathematical sequence:

$..., 30, ?, 60, 90, 140, 225, 372, 630, ...$

I generally dislike number sequence puzzles, but the (intended) answer to this puzzle is so surprising that I feel it should be more widely known. I don't know the origin of this puzzle, so it would be great if anyone has any info on it.

If nobody gets anywhere with it, I will start dropping hints in the comments below to get the ball rolling.

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    $\begingroup$ Don't add hints in the comments: add them in spoilers in the uestion itself. $\endgroup$ – Mithrandir Jul 29 '16 at 9:54
  • $\begingroup$ You say "Surprising". Is the answer an integer? $\endgroup$ – IAmInPLS Jul 29 '16 at 9:58
  • $\begingroup$ @Mithrandir If that's preferred, no problem. $\endgroup$ – Jaap Scherphuis Jul 29 '16 at 10:01
  • $\begingroup$ @IAmInPLS No comment :-) $\endgroup$ – Jaap Scherphuis Jul 29 '16 at 10:02
  • $\begingroup$ Are you sure, that the first number should be $30$ and not $-30$? $\endgroup$ – The Dark Truth Jul 29 '16 at 13:18
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Okay, I found something.

We must consider :

the numbers as $a_n$, with $n = -1,..., 6$ thus $a_0$ is the one we're supposed to find.

We have :

$a_{-1} = 30$
$a_0 = ?$
$a_1 = 60$
$a_2 = 90$
$a_3 = 140$
$a_4 = 225$
$a_5 = 372$
$a_6 = 630$

Then :

If we look at the sequence $n*a_n$, we get :
$-1*a_{-1} = -30$
$?$
$1*a_1 = 60$
$2*a_2 = 180$
$3*a_3 = 420$
$4*a_4 = 900$
$5*a_5 = 1860$
$6*a_6 = 3780$
For $n>0$, these numbers are all dividible by 60, so let divide them by 30 :
$\frac{1*a_1}{60} = 1$
$\frac{2*a_2}{60} = 3$
$\frac{3*a_3}{60} = 7$
$\frac{4*a_4}{60} = 15$
$\frac{5*a_5}{60} = 31$
$\frac{6*a_6}{60} = 63$
which are the famous Mersenne numbers, and can be noted : $b_n=2^n - 1$

Thus :

For $n>0$, we have : $n*a_n = 60*b_n$

Finally :

It works also for $a_{-1}$ also, since $-1*a_{-1} = -30 = 60*(2^{-1}-1) = 60*(1/2-1)$.
I think that @Etoplay found the final answer, nonetheless I will add it here in order to have a full answer written.
The formula to found the values $a_n$ is thus : $a_n = h(n) = \frac{60\dot{}b_n}{n} = \frac{60\dot{}(2^n-1)}{n}$, and we are looking for $h(0)$.
We can now use L'Hôpital's rule with $f(n) = 60\dot{}(2^n-1)$ and $g(n) = n$. We have :
$\lim\limits_{n \rightarrow 0} h(n) = \lim\limits_{n \rightarrow 0} \frac{f(n)}{g(n)} = \lim\limits_{n \rightarrow 0} \frac{f'(n)}{g'(n)}$

With $f'(n) = 2^x\dot{}ln(2)$, using the fact that $2^x = (e^{ln(2)})^x = e^{x\dot{}ln(2)}$ and $g'(n) = 1$, we find that :
$\lim\limits_{n \rightarrow 0} h(n) = \lim\limits_{n \rightarrow 0} \frac{60\dot{}ln(2)\dot{}2^x}{1} = 60\dot{}ln(2)$

And the answer is :

$a_0 = \lim\limits_{n \rightarrow 0} h(n) = 60\dot{}ln(2) \approx 41,588$, which is the number to be found instead of the question mark in the question.
We can see it on this plot of $h$, courtesy of @Chris in the comments :

enter image description here

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    $\begingroup$ The numbers you have are different from the numbers in the question...? $\endgroup$ – dcfyj Jul 29 '16 at 12:09
  • $\begingroup$ @IAmInPLS I think you should clarify that you are talking about the sequence $a_-1$, $a_0$, $a_1$, etc. Unfortunately I don't know how to add index -1. $\endgroup$ – rhsquared Jul 29 '16 at 12:15
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    $\begingroup$ @hexomino Just edited ;) $\endgroup$ – IAmInPLS Jul 29 '16 at 12:42
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    $\begingroup$ These are not the Mersenne prime numbers: 15 and 63 are not in them and aren't even prime. This is simply the series $2^n-1$. $\endgroup$ – The Dark Truth Jul 29 '16 at 13:11
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    $\begingroup$ Well done! I have accepted your answer. Although Etoplay came up with the final part, I know you got there first and edited it out again to give other people a go. :-) $\endgroup$ – Jaap Scherphuis Aug 3 '16 at 6:29
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Based on the other answers the formula for the values is $f(x)=$

$\frac{60\dot{}(2^x-1)}{x}$

(with the first value is for $x=-1$).

The question is what is $f(0)$.

If we assume that $f$ is continuous then we get with the help of L'Hospital's Rule:
$f(0) = \lim\limits_{x \rightarrow 0} f(x) = \lim\limits_{x \rightarrow 0} \frac{60\dot{}(2^x-1)}{x} = \lim\limits_{x \rightarrow 0} \frac{60\dot{}ln(2)\dot{}2^x}{1} = 60\dot{}ln(2) \approx 41.5888$

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I'd say:

Any number can go in the place of the question mark!

Let's index the numbers $a_{n}$, $-1 \le n \le 6$, with $a_0$ the missing one.

Now let's look at the following table:

$\begin{array}{|c|cccccccc|} \hline n & -1 & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline a_n & 30 & ? & 60 & 90 & 140 & 225 & 372 & 630 \\ \hline n \cdot a_n & -30 & 0 \cdot ? & 60 & 180 & 420 & 900 & 1860 & 3780 \\ \hline 60 \cdot (2^n - 1) & -30 & 0 & 60 & 180 & 420 & 900 & 1860 & 3780 \\ \hline \end{array}$

So the sequence can be defined as $n \cdot a_n = 60 \cdot (2^n - 1)$, and this holds whatever the question mark is (because $0 \cdot ? = 0$).

(This of course builds on IAmInPLS's answer, but as he didn't quite get there I allowed myself to put my own version.)

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    $\begingroup$ Last row there should be a 0 instead of a ? $\endgroup$ – Etoplay Aug 2 '16 at 9:09
  • $\begingroup$ @Etoplay You're right! Fixed. (BTW, nice answer of your own.) $\endgroup$ – Angkor Aug 2 '16 at 13:30
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I picked up the pattern that for all the known terms in the sequence, calculating the log of a term with the base being the term that proceeds it will be approximately equal to 1.09, losing it's degree of accuracy of 1.09 the bigger the terms are. Thus, I found my answer to be approximately 42.

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  • $\begingroup$ Hi, and welcome to Puzzling! Why don't you take the tour and earn another badge? For number sequence patterns we tend to prefer answers with a definitive conclusion, but it's nice to see you've had a go! (Hint: the actual answer is within that range, and there's a reason for the correlation.) Hope you can go on to post more answers! $\endgroup$ – boboquack Dec 22 '18 at 3:52

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