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This is one of Brilliant's daily challenges. What I see is four arithmetic progressions. I did my calculations according to the formulas:

$$S_{n1} = \frac n2(2a + (n − 1) × d_1),$$ $d_1=1$. $$ S_{n2} = \frac n2(2a + (n − 1) × d_2),$$ $d_2=2$.

$X = S_{n1} + S_{n2} - 3$(so that I don't count the central dot four times) = $78 \cdot 2 + 144 \cdot 2 - 3 = 441$. The problem is Brilliant doesn't have such an answer. Did I make a mistake somewhere?

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The first several centered hexagonal numbers are

$$1,7,19,37,61…$$

What is the 12th centered hexagonal number?

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2 Answers 2

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First method:
Let’s do it by their method. Look at one of their highlighted triangles. Each triangle contains $\displaystyle \frac{n(n+1)}{2}$ dots, where there are $n$ hexagonal rings outside the central dot. Since there are 6 such triangles, the total number of dots is (including the central dot): $\displaystyle 6\frac{n(n+1)}{2}+1=\fbox {3n(n+1)+1}$ where $n=0,1,2,…$ .

Second method:
Notice the pattern by counting the no. of dots in each row and summing it up for each n:$$n=0:\ 1$$$$n=1: \ 2+3+2=7$$$$n=2: \ 3+4+5+4+3=19$$ etc. So for $n=k$ we have Number of dots = $$\color{blue}{(k+1)}+(\color{blue}{(k+1)}+1)+…+ (\color{blue}{(k+1)}+(k-1))+\color{red}{(\color{blue}{(k+1)}+k)}+(\color{blue}{(k+1)}+(k-1))+… +\color{blue}{(k+1)}$$$$=2[(k+1)+((k+1)+1)+…+ ((k+1)+(k-1))]+ ((k+1)+k)$$ The part in square brackets is an AP with starting term k+1, common difference = 1 and number of terms = k. So the expression is $$2\left[\frac{k}{2}(2(k+1)+(k-1))\right]+(2k+1)$$$$=k(3k+1)+(2k+1)=3k^2+3k+1=\fbox{3k(k+1)+1}; k=0,1,2,3,…$$

To re-index values to have n starting from 1, replace n with (n-1). Thus the $nth$ “centered hexagonal number” is $$\color{green}{3n(n—1)+1}.$$ where n=1,2,3,…

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  • $\begingroup$ Actually if you can observe a little bit harder then you may count dots along vertical lines too. $\endgroup$ Jul 3 at 5:05
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    $\begingroup$ The 12th hexagonal number then, is 397. $\endgroup$ Jul 3 at 6:36
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    $\begingroup$ Where was my mistake? $\endgroup$ Jul 4 at 2:08
  • $\begingroup$ What do the APs represent in your attempt? What is “a” here? What sequence are you considering when writing d=2 for the second AP? It seems from the subtraction of 3 that you have divided the whole diagram into four parts. Please clarify. $\endgroup$ Jul 4 at 3:48
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    $\begingroup$ I’ve seen your edit. The mistake is that your “n” has different values for both the APs. Like, for the vertical triangles, it would be 1+2+3+4, so that n=4. However, in the horizontal part, if you’re not counting the central dot, it is 1+3+5, so n=3. Also note that the central part is being counted only twice and not four times, i.e. in the vertical upper and lower triangles. $\endgroup$ Jul 5 at 10:17
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For now, we will not add the single dot in the center.


Let $f(n)$ be the formula for the number of dots of the $n$th hexagon. Keep in mind that this is not the centered hexagonal number yet.

Notice that there are always six dots for each hexagon. Also, notice that for the $n$th hexagonal number, in a particular side excluding the vertices, there are $n - 1$ dots. Since hexagons have six sides, we have $6n - 6$ dots.

Hence, we have \begin{align*} f(n) &= 6n - 6 + 6 \\ f(n) &= 6n \end{align*}

Now, let $P(n)$ be the centered hexagonal number without the single dot in the center. We can see that $$P(n) = f(0) + f(1) + f(2) + \cdots + f(n).$$ Simplifying the right-hand side in terms of $n$, \begin{align*} P(n) &= f(0) + f(1) + f(2) + \cdots + f(n) \\ P(n) &= 6(0) + 6(1) + 6(2) + \cdots + 6n \\ P(n) &= 6(0 + 1 + 2 + \cdots + n) \\ P(n) &= 6\left(\frac{n(n + 1)}{2}\right) \\ P(n) &= 3n(n + 1) \end{align*}

Adding the dot in the center, we now have $P(n) = 3n(n + 1) + 1$. In this formula, the dot is also a hexagon, although its "size" makes it essentially a point. Since we are looking for the 12th hexagonal number, by replacing $n$ by $11$ (since the $+1$ is the $12$th hexagonal number), we get $397$.

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  • $\begingroup$ The first hexagonal number is 1, but your formula yields 7. IMHO, you need to re-index it. $\endgroup$ Jul 5 at 10:42
  • $\begingroup$ @insipidintegrator With that issue, I believe there's no need for reindexing as long as the definitions are correct. In my formula, $n = 0$ is the start. I defined it in that way because it is consistent to the diagram. A 'size-0' hexagon is essentially a point, so $P(0) = 1$. The only problem is that I haven't stated $n=0$ as the start. $\endgroup$
    – soupless
    Jul 5 at 10:55
  • $\begingroup$ Found it. Will edit now. $\endgroup$
    – soupless
    Jul 5 at 10:58

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