Heptagonal number

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A heptagonal number is a figurate number that represents a heptagon. The n-th heptagonal number is given by the formula

\frac{5n^2 - 3n}{2}.
The first five heptagonal numbers.

The first few heptagonal numbers are:

1, 7, 18, 34, 55, 81, 112, 148, 189, 235, 286, 342, 403, 469, 540, 616, 697, 783, 874, 970, 1071, 1177, 1288, 1404, 1525, 1651, 1782, … (sequence A000566 in OEIS)

Parity[edit]

The parity of heptagonal numbers follows the pattern odd-odd-even-even. Like square numbers, the digital root in base 10 of a heptagonal number can only be 1, 4, 7 or 9. Five times a heptagonal number, plus 1 equals a triangular number.

Generalized heptagonal numbers[edit]

A generalized heptagonal number is obtained by the formula

T_n + T_{\lfloor \frac{n}{2} \rfloor},

where Tn is the nth triangular number. The first few generalized heptagonal numbers are:

1, 4, 7, 13, 18, 27, 34, 46, 55, 70, 81, 99, 112, … (sequence A085787 in OEIS)

Every other generalized heptagonal number is a regular heptagonal number. Besides 1 and 70, no generalized heptagonal numbers are also Pell numbers.[1]

Sum of reciprocals[edit]

A formula for the sum of the reciprocals of the heptagonal numbers is given by:[2]


\sum_{n=1}^\infty \frac{2}{n(5n-3)} = \frac{1}{15}{\pi}{\sqrt{25-10\sqrt{5}}}+\frac{2}{3}\ln(5)+\frac{{1}+\sqrt{5}}{3}\ln\left(\frac{1}{2}\sqrt{10-2\sqrt{5}}\right)+\frac{{1}-\sqrt{5}}{3}\ln\left(\frac{1}{2}\sqrt{10+2\sqrt{5}}\right)

Test for heptagonal numbers[edit]

Let y be a positive integer.

\sqrt{40n +9} +3\over10

Heptagonal roots[edit]

In analogy to the square root of x, one can calculate the heptagonal roots of x.

The positive heptagonal root of x are given by the formula:

n = \frac{\sqrt{40x + 9} + 3}{10}

Derivation of heptagonal root formula[edit]

The heptagonal roots n of x are derived by:

x = \frac{5n^2 - 3n}{2}
2x = 5n^2 - 3n
5n^2 - 3n - 2x = 0
n = \frac{-(-3) \pm \sqrt{(-3)^2 - (4 \times 5 \times -2x)}}{2 \times 5} (use quadratic formula to solve for n)
n = \frac{3 \pm \sqrt{9 - (-40x)}}{10}
n = \frac{3 \pm \sqrt{9 + 40x}}{10}

Rearrange this to:

n = \frac{\pm \sqrt{40x + 9} + 3}{10}

Properties regarding the heptagonal roots[edit]

If n_{1} = \frac{\sqrt{40x + 9} + 3}{10}, and n_2 = \frac{-\sqrt{40x + 9} + 3}{10},

n_2 = -(n_1 - \frac{3}{5})

Derivation[edit]

n_2 = \frac{-\sqrt{40x + 9} + 3}{10}
n_2 = -\frac{\sqrt{40x + 9} - 3}{10}
n_2 = -\left(\frac{\sqrt{40x + 9} + 3}{10} - \frac{6}{10}\right)

Now substitute \frac{\sqrt{40x + 9} + 3}{10} with n_1.

n_2 = n_1 - \frac{6}{10} = n_1 - \frac{3}{5}.

References[edit]

  1. ^ B. Srinivasa Rao, "Heptagonal Numbers in the Pell Sequence and Diophantine equations 2x^2 = y^2(5y - 3)^2 \pm 2" Fib. Quart. 43 3: 194
  2. ^ Beyond the Basel Problem: Sums of Reciprocals of Figurate Numbers