Editing Hexagonal number (section)

==Other properties==
===Expression using sigma notation===
The ''n''th number of the hexagonal sequence can also be expressed by using [[sigma notation]] as

:<math> h_n = \sum_{k=0}^{n-1}{(4k+1)} </math>

where the [[empty sum]] is taken to be 0.

===Sum of the reciprocal hexagonal numbers===
The sum of the reciprocal hexagonal numbers is {{math|2ln(2)}}, where {{math|ln}} denotes [[natural logarithm]].
:<math>\begin{align} \sum_{k=1}^{\infty} \frac{1}{k(2k-1)} &= \lim_{n \to \infty}2\sum_{k=1}^{n} \left(\frac{1}{2k-1} - \frac{1}{2k} \right)\\                            &= \lim_{n \to \infty}2\sum_{k=1}^{n} \left(\frac{1}{2k-1} +  \frac{1}{2k} - \frac{1}{k} \right)\\ &= 2 \lim_{n \to \infty}\left(\sum_{k=1}^{2n}\frac{1}{k} - \sum_{k=1}^{n}\frac{1}{k}\right)\\                                                         &= 2 \lim_{n \to \infty}\sum_{k=1}^{n}\frac{1}{n+k} \\ &= 2 \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}\\ &= 2 \int_{0}^{1}\frac{1}{1+x}dx \\ &= 2 [ \ln(1+x) ]_{0}^{1} \\ &= 2 \ln{2}\\
                     & \approx{1.386294}
\end{align}</math>

===Multiplying the index===
Using rearrangement, the next set of formulas is given:
 
<math> h_{2n} = 4h_n+2n </math>

<math> h_{3n} = 9h_n+6n </math>

<math>...</math>

<math> h_{m*n} = m^{2}h_n+(m^{2}-m)n</math>

===Ratio relation===
Using the final formula from before with respect to ''m'' and then ''n'', and then some reducing and moving, one can get to the following equation:

<math>\frac{h_{m}+m}{h_{n}+n}=\left(\frac{m}{n}\right)^2</math>

===Numbers of divisors of powers of certain natural numbers===
<math>12^{n-1}</math> for ''n''>0 has <math>h_n</math> divisors.

Likewise, for any natural number of the form <math>r = p^2 q</math> where ''p'' and ''q'' are distinct prime numbers, <math> r^{n-1}</math> for ''n''>0 has <math>h_n</math> divisors.

''Proof.'' <math>r^{n-1} = (p^2 q)^{n-1} = p^{2(n-1)} q^{n-1} </math>
has divisors of the form <math>p^k q^l</math>, for ''k'' = 0 ... 2(''n'' &minus; 1), ''l'' = 0 ... n &minus; 1. Each combination of ''k'' and ''l'' yields a distinct divisor, so <math>r^{n-1}</math> has <math>[2(n - 1) + 1] [(n - 1) + 1]</math> divisors, i.e. <math> (2n - 1) n = h_n</math> divisors. ∎