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You are given the sequence:

$$\frac{1}{4}, 1, \frac{9}{4}, 2 + \frac{2}{\sqrt{3}}, \frac{5}{2} + \frac{5}{2 \sqrt{2}}, 3 + 3 \sqrt{\frac{1}{10} (5 + \sqrt{5})}$$

Transcription: one fourth, then one, then nine fourths, then two plus the fraction two over the square root of three, then five halves plus the fraction five over the product of two and the square root of two, then three plus the product of three and the square root of the product of one tenth and the sum of five and the square root of five

What is the next number? Explain

Hint: The next number is rational.

Explanation hint: this is a series of coefficients of a constant. To be clear, I mean that there exists a constant which, if multiplied by each these numbers, gives the actual values of interest. Chris gives a helpful example: "a sequence which is the area of circles with radius n would be pi, 4pi, 9pi, etc. The coefficients of the constant would just be 1, 4, 9, etc" But pi is not the correct constant for this particular sequence.

New hint: the correct constant is pi^2

This is my own puzzle

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    $\begingroup$ I added a text transcription of the sequence because it's in MathJax, which can be difficult for screen readers to parse. $\endgroup$ – bobble Nov 16 '20 at 2:40
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    $\begingroup$ Well, I know what the next number is, but the way I have of calculating them goes singular for the first two terms in the sequence so clearly I need a different perspective somehow... $\endgroup$ – Gareth McCaughan Nov 16 '20 at 3:03
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    $\begingroup$ a quick wolfram-alpha-ing revealed ... nothing at all. $\endgroup$ – merrybot Nov 16 '20 at 4:51
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    $\begingroup$ Conor, as you may see in comments to Paul Panzer's answer there's some confusion about what you mean by "a series of coefficients of a constant". That may be deliberate -- you may be being deliberately vague so as not to give things away -- but if you intended your meaning to be clearer than it seems to have been, you might consider clarifying it... $\endgroup$ – Gareth McCaughan Nov 19 '20 at 14:22
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    $\begingroup$ @ConorHenry You might want to also post your own answer as well with your intended solution to put us all out of our misery. ;-) $\endgroup$ – Chris Nov 30 '20 at 9:09
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Partial answer

Note: I have had this sat in the edit-box for about the last 18 hours, because I didn't feel that it was enough to post (since it conspicuously fails to explain the first two terms, and I'm pretty sure something more "concrete" is intended). But I see someone else has posted a partial with even less in it, so I'm posting this too. Of course I have no way of proving that it's been sat here for the last 18 hours, nor indeed that the other poster didn't also do what he did much earlier and sit on it for a while before posting.


This sequence is $a_0,a_1,\dots$ where for $n>1$ $a_n$ equals

$\frac{n+1}2\left(1+\frac1{2\sin{\pi/n}}\right)$

and if this continues then the next number will be

exactly 7.

But

it's easy to see that this formula blows up when $n=0$ or $n=1$. To get the given value when $n=1$ we would need $1+\frac1{2\sin\pi}=1$ -- i.e., we would need to pretend that $1/0=0$ or perhaps that $\sin\pi=\infty$. To get the given value when $n=0$ we would need $\frac12\left(1+\frac1{2\sin0}\right)=\frac14$ -- i.e., we would need to pretend that $\sin0=-1$.

I guess

there's some geometrical interpretation that makes those slightly surprising things happen naturally. Suppose we consider a regular $n$-gon of side 1 and divide it into $2n$ right-angled triangles in the obvious way. All the angles at the centre are $\pi/n$, and their "far" sides are $1/2$, so the distances from centre to vertex are $\frac1{2\sin\pi/n}$. So that term is the circumradius of a regular $n$-gon of side length 1. Of course there's no such thing as a regular 0-gon or 1-gon, but maybe we can think instead of $n$ points equally spaced around a circle, with successive ones being distance 1 apart, or something like that. So, e.g., if we interpret $frac1{2\sin\pi/n}$ as "radius of smallest circle containing $n$ points regularly spaced around a circle, with adjacent points 1 unit apart" then this now gives the right answer for $n=1$, but for $n=0$ we would need that radius to be $-1$ which still seems a bit arbitrary. (And of course it would be nicer to have a natural geometrical interpretation for the whole expression rather than just one part of it.)

Conor asks for the next next term. I will be astonished if it isn't

$4\left(1+\frac1{2\sin\pi/7}\right)$ which so far as I know doesn't have any very nice expression in radicals (unless you allow yourself things like $(-1)^{2/7}$) but is approximately 8.6095.

But if this is correct, it doesn't give me any more help finding an expression that deals nicely with n=0 and n=1 -- it's just applying the formula I already had.

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  • $\begingroup$ I'm very impressed. I'm open to giving some additional hints, but I don't know if/when that would be appropriate. $\endgroup$ – Conor Henry Nov 16 '20 at 21:06
  • $\begingroup$ @Gareth You commented that you you have an idea that can explain the sequence except for n=0 and n=1, that was 18 hours ago. Evidently you had the solution already at that time. $\endgroup$ – Florian F Nov 16 '20 at 21:33
  • $\begingroup$ I've edited the question $\endgroup$ – Conor Henry Nov 16 '20 at 21:40
  • $\begingroup$ Your guess is not correct. Hope that helps. $\endgroup$ – Conor Henry Nov 16 '20 at 22:14
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    $\begingroup$ Actually, I bet that one's wrong too. I suspect there's some sort of packing-points-into-a-circle thing going on and the reason why the pattern breaks down at n=7 is that that's when you can put a point in the middle of your n-1 points and have it be no closer to them than they are to one another. $\endgroup$ – Gareth McCaughan Nov 17 '20 at 11:18
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Most of the work has already been done by @Florian F and @Gareth McCaughan. Based on their insights here is a geometric reading that I believe covers all cases:

First, lets get one bit out of the way that feels a bit less natural than the rest: The prefactor $\frac {n+1}2$, lets just get rid of it and use the transformed values $\frac 1 2, 1 ,\frac 3 2, 1 + \frac 1 {\sqrt 3}$ etc.

Let's assume we are dealing with discs of diameter one. Given n such discs what is the radius of the smallest circle we can pack them into? For $n=0,1,2$ it is trivially $0,\frac 1 2, 1$ which is one half less than the corresponding values in the series. These values may be interpreted as the minimum distance from the center of the circle to the center of any disc not intersecting the circle.

For $n=3,...,6$ the best circle packing available is arranging the discs in a circle https://en.wikipedia.org/wiki/Circle_packing_in_a_circle . Comparing the diameters of minimal enclosing circles given at this link we recover the puzzle sequence by dividing by two (diamter to radius), adding $\frac 1 2$ (to get the minimum distance for outside disc centers) and, finally, multiplying by $\frac {n+1} 2$.

Based on this I claim that the subsequent terms are

$n=7 \mapsto 8$, $n=8 \mapsto \frac 9 2 \left ( 1 + \frac 1 {2\sin \frac \pi 7} \right )$, $n=9 \mapsto 5 \left ( 1 + \sqrt {\frac {2 + \sqrt 2} 2} \right )$ etc.

Re OP's latest hint:

My guess would be that the constant in question is $\pi$. I assume the trigonometry (including $\pi$ itself in term $8$) apparently needed to calculate the terms can in principle be sidestepped using something like roots of unity.

If you like this answer and haven't already done so make sure to also check out Florian's and especially Gareth's answers who had essentially worked it all out including the packing idea.

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  • $\begingroup$ Nice. Am I right in my reading that the (n+1)/2 factor is just an arbitrary multiplier in your solution - ie it isn't coming from the geometric description? $\endgroup$ – Chris Nov 17 '20 at 15:40
  • $\begingroup$ @Chris Yep, that's correct. $\endgroup$ – Paul Panzer Nov 17 '20 at 16:11
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    $\begingroup$ @Chris maybe yeah. But might as well be some technicality to do with OP's using them as coefficients to compute some constant. $\endgroup$ – Paul Panzer Nov 17 '20 at 16:28
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    $\begingroup$ D'oh! I thought about packing points into a circle (which makes n=1 work) but somehow never thought of making them discs instead to fix up n=0. Which is ironic since my PhD thesis was on circle packings :-). $\endgroup$ – Gareth McCaughan Nov 17 '20 at 16:49
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    $\begingroup$ I assume coeffiicient of a constant means that you multiple each term by a constant that gives the actual value you care about. ie a sequence which is the area of circles with radius n would be pi, 4pi, 9pi, etc. The coefficients of the constant would just be 1, 4, 9, etc. I might be wrong on that but its certainly what I assumed when I read "coefficients of a constant". $\endgroup$ – Chris Nov 19 '20 at 10:45
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The 7th term is

7

Because ...

Let's rewrite the terms:
$\frac{1}{2} (1 - \frac{1}{2})$
$\frac{2}{2} (1 + 0)$
$\frac{3}{2} (1 + \frac{1}{2})$
$\frac{4}{2} (1 + \frac{1}{\sqrt{3}})$
$\frac{5}{2} (1 + \frac{1}{\sqrt{2}})$
$\frac{6}{2} (1 + \sqrt{ \frac{5+\sqrt{5}}{10} })$

... and from there ...

The last square root reminds me of the proportions you find in a decagon.

Checking my trigonometry table I find that
$\frac{1}{2} \frac{1}{\sin(\pi/5)} = \frac{1}{2} \sqrt{\frac{10+2\sqrt{5}}{5}} = \sqrt{\frac{5+\sqrt{5}}{10}}$
and I verify that
$\frac{1}{2} \frac{1}{\sin(\pi/3)} = \frac{1}{2} \frac{2\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$

The general formula seems to be
$a_n = \frac{n+1}{2} (1 + \frac{1}{2} \frac{1}{\sin(\pi/n) })$

This formula indeed works for all $n>1$ but not $n=0$ and $n=1$ where there is a division by $0$.

So as others have discovered, the values for $n>1$ can be explained. The initial $\frac{1}{4}$ and $1$, however, remain unexplained.
I suspect the values are some distance or proportion found in a regular n-gon. Except that, well, a 0-gon or a 1-gon doesn't mean much. So it can't explain the first 2 terms.
But if the formula is correct where it applies, we can compute the next element of the sequence. It is $a_6$ which evaluates to $7$.

PS: It seems this is not the intended answer. The OP added a hint showing that the general formula given here doesn't work for the term after that.

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    $\begingroup$ Your formula obviously (and in a superficial and easily-fixed way) doesn't match the given one -- I guess you made the obvious simplification of dividing by (n+1) or something before looking further. In any case, it seems to me that a large part of the point of the puzzle must be to explain those first two terms, perhaps by finding another way of looking at the sequence that handles those special cases naturally. $\endgroup$ – Gareth McCaughan Nov 16 '20 at 21:01
  • $\begingroup$ Yep, I was a bit sloppy copying it from my text notepad to mathjax. I fixed that. In the meantime I saw you found the same answer as I intended. $\endgroup$ – Florian F Nov 16 '20 at 21:47
  • $\begingroup$ I realise that the parenthetical bit in my comment could be read with exactly the opposite of my intended sentiment. I meant "but not in any way that means you haven't actually found the correct formula" rather than "in a way only an idiot could have missed". (No such thing as the latter, as I know from the many many many dumb things I do myself.) $\endgroup$ – Gareth McCaughan Nov 16 '20 at 21:57
  • $\begingroup$ No problem. To me you were just pointing out a small mistake, like a typo. $\endgroup$ – Florian F Nov 16 '20 at 22:10
  • $\begingroup$ That was how I meant it. Good! $\endgroup$ – Gareth McCaughan Nov 17 '20 at 11:19
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As determined by other answers, the next term is

7

I mentioned this is a sequence of coefficients of the constant pi^2.

$$\pi^2\left(\frac{1}{4}\right), \pi^2(1),\pi^2\left(\frac{9}{4}\right), \pi^2\left(2 + \frac{2}{\sqrt{3}}\right), \pi^2\left(\frac{5}{2} + \frac{5}{2 \sqrt{2}}\right), \pi^2\left(3 + 3 \sqrt{\frac{5 + \sqrt{5}}{10}}\right), \pi^2(7) $$

explanation:

The sequence gives the minimized combined volume of a group of congruent tori with tube diameter = 1, arranged such that one torus must interlock with every other torus in the set. Perhaps this could be better stated. The 0th term is 1/4, because $\pi^2/4$ is the volume of a Horn Torus with an inner diameter of 0 and tube diameter of 1. The second term is two times the volume of a torus whose tube diameter is 1, and whose inner diameter is also 1, thus, the combined volume of two interlocking tori with inner diameter minimized. As Gareth McCaughan noticed, the remaining terms relate to optimal circle packing, because of the way the minimized primary diameter of the tori must grow as additional tori are linked to the initial torus. The third term gives the combined volume of 3 tori whose inner diameter is equal to twice their tube diameter of 1, and so on. Calculating additional terms beyond 7 is far beyond my abilities and I assume would be very difficult to prove.

This was my first math puzzle so I apologize for some of my beginner mistakes that may have made this challenge frustrating for participants. I am accepting Gareth McCaughan's answer as he was in my opinion the closest.

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