Decagonal number

• 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}^{\infty}\frac{1}{4n^{2}-3n}+\sum_{n=1}^{\infty}\frac{1}{n\left(4n-3\right)}=\ln\left(2\right)+\frac{\pi}{6}. Proof This derivation rests upon the method of adding a "constructive zero": \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} Rearranging and considering the individual sums: \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} ==References==