Hexagonal number

One can efficiently test whether a positive integer x is a hexagonal number by computing

${\displaystyle n={\frac {{\sqrt {8x+1}}+1}{4}}.}$ 

If n is an integer, then x is the nth hexagonal number. If n is not an integer, then x is not hexagonal.

• ${\displaystyle h_{n}\equiv n{\pmod {4}}}$ 
• ${\displaystyle h_{3n}+h_{2n}+h_{n}\equiv 0{\pmod {2}}}$ 

The nth number of the hexagonal sequence can also be expressed by using sigma notation as

${\displaystyle h_{n}=\sum _{k=0}^{n-1}{(4k+1)}}$ 

where the empty sum is taken to be 0.

The sum of the reciprocal hexagonal numbers is 2ln(2), where ln denotes natural logarithm.

${\displaystyle {\begin{aligned}\sum _{k=1}^{\infty }{\frac {1}{k(2k-1)}}&=\lim _{n\to \infty }2\sum _{k=1}^{n}\left({\frac {1}{2k-1}}-{\frac {1}{2k}}\right)\\&=\lim _{n\to \infty }2\sum _{k=1}^{n}\left({\frac {1}{2k-1}}+{\frac {1}{2k}}-{\frac {1}{k}}\right)\\&=2\lim _{n\to \infty }\left(\sum _{k=1}^{2n}{\frac {1}{k}}-\sum _{k=1}^{n}{\frac {1}{k}}\right)\\&=2\lim _{n\to \infty }\sum _{k=1}^{n}{\frac {1}{n+k}}\\&=2\lim _{n\to \infty }{\frac {1}{n}}\sum _{k=1}^{n}{\frac {1}{1+{\frac {k}{n}}}}\\&=2\int _{0}^{1}{\frac {1}{1+x}}dx\\&=2[\ln(1+x)]_{0}^{1}\\&=2\ln {2}\\&\approx {1.386294}\end{aligned}}}$ 

Using rearrangement, the next set of formulas is given:

${\displaystyle h_{2n}=4h_{n}+2n}$ 

${\displaystyle h_{3n}=9h_{n}+6n}$ 

${\displaystyle ...}$ 

${\displaystyle h_{m*n}=m^{2}h_{n}+(m^{2}-m)n}$ 

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:

${\displaystyle {\frac {h_{m}+m}{h_{n}+n}}=\left({\frac {m}{n}}\right)^{2}}$ 

${\displaystyle 12^{n-1}}$  for n>0 has ${\displaystyle h_{n}}$  divisors.

Likewise, for any natural number of the form ${\displaystyle r=p^{2}q}$  where p and q are distinct prime numbers, ${\displaystyle r^{n-1}}$  for n>0 has ${\displaystyle h_{n}}$  divisors.

Proof. ${\displaystyle r^{n-1}=(p^{2}q)^{n-1}=p^{2(n-1)}q^{n-1}}$  has divisors of the form ${\displaystyle p^{k}q^{l}}$ , for k = 0 ... 2(n − 1), l = 0 ... n − 1. Each combination of k and l yields a distinct divisor, so ${\displaystyle r^{n-1}}$  has ${\displaystyle [2(n-1)+1][(n-1)+1]}$  divisors, i.e. ${\displaystyle (2n-1)n=h_{n}}$  divisors. ∎

The sequence of numbers that are both hexagonal and perfect squares starts 1, 1225, 1413721,... OEIS: A046177.

• Centered hexagonal number

• Weisstein, Eric W. "Hexagonal Number". MathWorld.