Editing Decagonal number

{{Short description|Figurate number representing a decagon}}
In [[mathematics]], a '''decagonal number''' is a [[figurate number]] that extends the concept of [[triangular number|triangular]] and [[square number]]s to the [[decagon]] (a ten-sided polygon). However, unlike the triangular and square numbers, the patterns involved in the construction of decagonal numbers are not rotationally symmetrical. Specifically, the ''n''-th decagonal numbers counts the dots in a pattern of ''n'' nested decagons, all sharing a common corner, where the ''i''th decagon in the pattern has sides made of ''i'' dots spaced one unit apart from each other. The ''n''-th decagonal number is given by the following formula
: <math>d_n = 4n^2 - 3n</math>.
The first few decagonal numbers are:
: [[0 (number)|0]], [[1 (number)|1]], [[10 (number)|10]], [[27 (number)|27]], [[52 (number)|52]], [[85 (number)|85]], [[126 (number)|126]], [[175 (number)|175]], [[232 (number)|232]], [[297 (number)|297]], 370, 451, 540, 637, 742, 855, 976, [[1105 (number)|1105]], 1242, 1387, 1540, 1701, 1870, 2047, 2232, 2425, 2626, 2835, 3052, 3277, 3510, 3751, [[4000 (number)|4000]], 4257, 4522, 4795, 5076, 5365, 5662, 5967, 6280, 6601, 6930, 7267, 7612, 7965, 8326 {{OEIS|id=A001107}}.

The ''n''th decagonal number can also be calculated by adding the square of ''n'' to thrice the (''n''−1)th [[pronic number]] or, to put it algebraically, as 
: <math>D_n = n^2 + 3\left(n^2 - n\right)</math>.

== Properties ==

* Decagonal numbers consistently alternate [[Even and odd numbers#Parity in mathematics|parity]].
* <math>D_n</math> is the sum of the first <math>n</math> natural numbers congruent to 1 mod 8.
* <math>D_n</math> is number of divisors of <math>48^{n-1}</math>.
* The only decagonal numbers that are square numbers are 0 and 1.
* The decagonal numbers follow the following recurrence relations:
:<math>D_n=D_{n-1}+8n-7 , D_0=0</math>
:<math>D_n=2D_{n-1}-D_{n-2}+8, D_0=0,D_1=1</math>
:<math>D_n=3D_{n-1}-3D_{n-2}+D_{n-3}, D_0=0, D_1=1, D_2=10</math>

==Sum of reciprocals==
The [[sums of reciprocals|sum of the reciprocals]] of the decagonal numbers admits a simple closed form:
<math display=block>
\sum_{n=1}^{\infty}\frac{1}{4n^{2}-3n}+\sum_{n=1}^{\infty}\frac{1}{n\left(4n-3\right)}=\ln\left(2\right)+\frac{\pi}{6}.
</math>

===Proof===
This derivation rests upon the method of adding a "constructive zero":
<math display=block>
\begin{align}
\sum_{n=1}^{\infty}\frac{1}{n\left(4n-3\right)} & {} =\frac{4}{3}\sum_{n=1}^{\infty}\left(\frac{1}{4n-3}-\frac{1}{4n}\right) \\
&=\frac{2}{3}\sum_{n=1}^{\infty}\left(\frac{2}{4n-3}-\frac{2}{4n}+\left(\frac{1}{4n-1}-\frac{1}{4n-2}\right)-\left(\frac{1}{4n-1}-\frac{1}{4n-2}\right)\right)
\end{align}
</math>
Rearranging and considering the individual sums:
<math display=block>
\begin{align}
&= \frac{2}{3} \sum_{n=1}^{\infty} \left[ \left(\frac{1}{4n-3} - \frac{1}{4n-2} + \frac{1}{4n-1} - \frac{1}{4n} \right) + \left(\frac{1}{4n-2} - \frac{1}{4n} \right) + \left(\frac{1}{4n-3} - \frac{1}{4n-1} \right) \right] \\ 
&= \frac{2}{3} \sum_{n=1}^{\infty} \left( \frac{1}{4n-3} - \frac{1}{4n-2} + \frac{1}{4n-1} - \frac{1}{4n} \right) + \frac{1}{3} \sum_{n=1}^{\infty} \left( \frac{1}{2n-1} - \frac{1}{2n} \right) + \frac{2}{3} \sum_{n=1}^{\infty} \left( \frac{1}{2(2n-1)-1} - \frac{1}{2(2n)-1} \right) \\ 
&= \frac{2}{3} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} + \frac{1}{3} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} + \frac{2}{3} \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{2n-1} \\
&= \ln\left(2\right)+\frac{\pi}{6}.
\end{align}
</math>

{{Figurate numbers}}
{{Classes of natural numbers |state=collapsed}}
{{num-stub}}

[[Category:Figurate numbers]]