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Test for hexagonal numbers

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

n = 8 x + 1 + 1 4 . {\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.

Congruence relations

• h n ≡ n ( mod 4 ) {\displaystyle h_{n}\equiv n{\pmod {4}}}
• h 3 n + h 2 n + h n ≡ 0 ( mod 2 ) {\displaystyle 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 = ∑ k = 0 n − 1 ( 4 k + 1 ) {\displaystyle 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.

∑ k = 1 ∞ 1 k ( 2 k − 1 ) = lim n → ∞ 2 ∑ k = 1 n ( 1 2 k − 1 − 1 2 k ) = lim n → ∞ 2 ∑ k = 1 n ( 1 2 k − 1 + 1 2 k − 1 k ) = 2 lim n → ∞ ( ∑ k = 1 2 n 1 k − ∑ k = 1 n 1 k ) = 2 lim n → ∞ ∑ k = 1 n 1 n + k = 2 lim n → ∞ 1 n ∑ k = 1 n 1 1 + k n = 2 ∫ 0 1 1 1 + x d x = 2 [ ln ⁡ ( 1 + x ) ] 0 1 = 2 ln ⁡ 2 ≈ 1.386294 {\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}}}

Multiplying the index

Using rearrangement, the next set of formulas is given:

h 2 n = 4 h n + 2 n {\displaystyle h_{2n}=4h_{n}+2n}

h 3 n = 9 h n + 6 n {\displaystyle h_{3n}=9h_{n}+6n}

. . . {\displaystyle ...}

h m ∗ n = m 2 h n + ( m 2 − m ) n {\displaystyle 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:

h m + m h n + n = ( m n ) 2 {\displaystyle {\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 {\displaystyle 12^{n-1}} for n>0 has h n {\displaystyle h_{n}} divisors.

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

Proof. r n − 1 = ( p 2 q ) n − 1 = p 2 ( n − 1 ) q n − 1 {\displaystyle r^{n-1}=(p^{2}q)^{n-1}=p^{2(n-1)}q^{n-1}} has divisors of the form p k q l {\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 r n − 1 {\displaystyle r^{n-1}} has [ 2 ( n − 1 ) + 1 ] [ ( n − 1 ) + 1 ] {\displaystyle [2(n-1)+1][(n-1)+1]} divisors, i.e. ( 2 n − 1 ) n = h n {\displaystyle (2n-1)n=h_{n}} divisors. ∎

Hexagonal square numbers

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

See also

• Centered hexagonal number

External links

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