Hexagonal number

Test for hexagonal numbers

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

: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.

Congruence relations

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

Other properties

Expression using sigma notation

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

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

where the empty sum is taken to be 0.

Sum of the reciprocal hexagonal numbers

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

Multiplying the index

Using rearrangement, the next set of formulas is given:

h_{2n} = 4h_n+2n

h_{3n} = 9h_n+6n

...

h_{m*n} = m^{2}h_n+(m^{2}-m)n

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:

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

Numbers of divisors of powers of certain natural numbers

12^{n-1} for n0 has h_n divisors.

Likewise, for any natural number of the form r = p^2 q where p and q are distinct prime numbers, r^{n-1} for n0 has h_n divisors.

Proof. r^{n-1} = (p^2 q)^{n-1} = p^{2(n-1)} q^{n-1} has divisors of the form 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 r^{n-1} has [2(n - 1) + 1] [(n - 1) + 1] divisors, i.e. (2n - 1) n = h_n divisors. ∎

Hexagonal square numbers

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