Decagonal number

Template:Short description Template:Refimprove In mathematics, a decagonal number is a figurate number that extends the concept of triangular and square numbers 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 ith 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

$d_n=4n^2-3n$.

The first few decagonal numbers are:

0, 1, 10, 27, 52, 85, 126, 175, 232, 297, 370, 451, 540, 637, 742, 855, 976, 1105, 1242, 1387, 1540, 1701, 1870, 2047, 2232, 2425, 2626, 2835, 3052, 3277, 3510, 3751, 4000, 4257, 4522, 4795, 5076, 5365, 5662, 5967, 6280, 6601, 6930, 7267, 7612, 7965, 8326 Template:OEIS.

The nth 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

$D_n=n^2+3(n^2-n)$.

Properties

• Decagonal numbers consistently alternate parity.
• $D_n$ is the sum of the first $n$ natural numbers congruent to 1 mod 8.
• $D_n$ is number of divisors of $48^{n-1}$.
• The only decagonal numbers that are square numbers are 0 and 1.
• The decagonal numbers follow the following recurrence relations:

$D_n=D_{n-1}+8n-7,D_0=0$

$D_n=2D_{n-1}-D_{n-2}+8,D_0=0,D_1=1$

$D_n=3D_{n-1}-3D_{n-2}+D_{n-3},D_0=0,D_1=1,D_2=10$

Sum of reciprocals

The sum of the reciprocals of the decagonal numbers admits a simple closed form: $$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{4n^2-3n}+\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(4n-3)}=\ln \left(2\right)+\frac{\pi }{6}.$$

Proof

This derivation rests upon the method of adding a "constructive zero": $$\begin{array}{cc}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(4n-3)}& =\frac{4}{3}\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{4n-3}-\frac{1}{4n})\\ & =\frac{2}{3}\sum _{n=1}^{\mathrm{\infty }}(\frac{2}{4n-3}-\frac{2}{4n}+(\frac{1}{4n-1}-\frac{1}{4n-2})-(\frac{1}{4n-1}-\frac{1}{4n-2}))\end{array}$$ Rearranging and considering the individual sums: $$\begin{array}{cc}& =\frac{2}{3}\sum _{n=1}^{\mathrm{\infty }}[(\frac{1}{4n-3}-\frac{1}{4n-2}+\frac{1}{4n-1}-\frac{1}{4n})+(\frac{1}{4n-2}-\frac{1}{4n})+(\frac{1}{4n-3}-\frac{1}{4n-1})]\\ & =\frac{2}{3}\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{4n-3}-\frac{1}{4n-2}+\frac{1}{4n-1}-\frac{1}{4n})+\frac{1}{3}\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{2n-1}-\frac{1}{2n})+\frac{2}{3}\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{2(2n-1)-1}-\frac{1}{2(2n)-1})\\ & =\frac{2}{3}\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{n}+\frac{1}{3}\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{n}+\frac{2}{3}\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{2n-1}\\ & =\ln \left(2\right)+\frac{\pi }{6}.\end{array}$$