# An unusual and very hard number sequence problem

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.

• You say "Surprising". Is the answer an integer? Jul 29, 2016 at 9:58
• @IAmInPLS No comment :-) Jul 29, 2016 at 10:02

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 divisible by 60, so let's divide them by 60 :
$$\frac{1*a_1}{60} = 1$$
$$\frac{2*a_2}{60} = 3$$
$$\frac{3*a_3}{60} = 7$$
$$\frac{4*a_4}{60} = 15$$
$$\frac{5*a_5}{60} = 31$$
$$\frac{6*a_6}{60} = 63$$
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 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 find 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 :
$$\lim\limits_{n \rightarrow 0} h(n) = \lim\limits_{n \rightarrow 0} \frac{60\dot{}ln(2)\dot{}2^x}{1} = 60\dot{}ln(2)$$

$$a_0 = \lim\limits_{n \rightarrow 0} h(n) = 60\dot{}ln(2) \approx 41,588$$, 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 : • The numbers you have are different from the numbers in the question...? Jul 29, 2016 at 12:09
• @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. Jul 29, 2016 at 12:15
• @hexomino Just edited ;) Jul 29, 2016 at 12:42
• These are not the Mersenne prime numbers: 15 and 63 are not in them and aren't even prime. This is simply the series $2^n-1$. Jul 29, 2016 at 13:11
• 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. :-) Aug 3, 2016 at 6:29

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:
$f(0) = \lim\limits_{x \rightarrow 0} f(x) = \lim\limits_{x \rightarrow 0} \frac{60\dot{}(2^x-1)}{x} = \lim\limits_{x \rightarrow 0} \frac{60\dot{}ln(2)\dot{}2^x}{1} = 60\dot{}ln(2) \approx 41.5888$

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:

n -1 0 1 2 3 4 5 6
an 30 ? 60 90 140 225 372 630
n * an -30 0*? 60 180 420 900 1860 3780
60 * (2n - 1) -30 0 60 180 420 900 1860 3780

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.)

• Last row there should be a 0 instead of a ? Aug 2, 2016 at 9:09
• @Etoplay You're right! Fixed. (BTW, nice answer of your own.) Aug 2, 2016 at 13:30

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.

• 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! Dec 22, 2018 at 3:52

$$30\times2=60$$
$$2\times x+y=90$$
$$60\times2+20=140$$
$$90\times2+45=225$$
$$140\times2+92=372$$
$$225\times2+180=630$$

y,20,45,92,180
$$2\times y+5=20$$
$$20\times2+5=45$$
$$45\times2+2=92$$
$$92\times2-4=180$$

$$y=7.5$$
$$2\times x+7.5=90$$
$$x=41.25$$

• 302 never equals 60. Please explain what's going on with your answer, it seems almost completely unrelated to the problem Aug 31, 2020 at 5:03
• @Cotton Did you miss the *? This answer is giving an algorithm for constructing the sequence. I'm not sure if it's right or wrong, but 30 times 2 equals 60 is correct at least. Aug 31, 2020 at 8:02
• The *’s where not there last I saw it, but they are now Aug 31, 2020 at 13:59

Thinking logically the answer should be 45. Bcoz if we consider the alternative terms then the series follows this pattern:

630-225= 405 225-90= 135 Hence let the missing to term be x. Here Everytime the difference is the multiple of 3. Hence for alternative numbers x should be 45. 90-45 = 45.
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