Decagonal number

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).^[1] 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

${\displaystyle d_{n}=4n^{2}-3n}${\displaystyle d_{n}=4n^{2}-3n}.^[2]

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 (sequence A001107 in the 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

${\displaystyle D_{n}=n^{2}+3\left(n^{2}-n\right)}${\displaystyle D_{n}=n^{2}+3\left(n^{2}-n\right)}.

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

${\displaystyle D_{n}=D_{n-1}+8n-7,D_{0}=0}${\displaystyle D_{n}=D_{n-1}+8n-7,D_{0}=0}

${\displaystyle D_{n}=2D_{n-1}-D_{n-2}+8,D_{0}=0,D_{1}=1}${\displaystyle D_{n}=2D_{n-1}-D_{n-2}+8,D_{0}=0,D_{1}=1}

${\displaystyle D_{n}=3D_{n-1}-3D_{n-2}+D_{n-3},D_{0}=0,D_{1}=1,D_{2}=10}${\displaystyle D_{n}=3D_{n-1}-3D_{n-2}+D_{n-3},D_{0}=0,D_{1}=1,D_{2}=10}

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

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

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