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<div>A '''heptagonal number''' is a [[figurate number]] that represents a [[heptagon]]. The ''n''-th heptagonal number is given by the formula <br />
:<math>\frac{5n^2 - 3n}{2}</math>.<br />
[[File:Heptagonal numbers.svg|thumbnail|right|The first five heptagonal numbers.]]<br />
The first few heptagonal numbers are:<br />
:[[1 (number)|1]], [[7 (number)|7]], [[18 (number)|18]], [[34 (number)|34]], [[55 (number)|55]], [[81 (number)|81]], [[112 (number)|112]], [[148 (number)|148]], [[189 (number)|189]], [[235 (number)|235]], 286, 342, 403, 469, 540, [[616 (number)|616]], 697, 783, 874, 970, 1071, 1177, 1288, 1404, 1525, 1651, 1782, … {{OEIS|id=A000566}}<br />
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==Parity==<br />
The parity of heptagonal numbers follows the pattern odd-odd-even-even. Like [[square number]]s, 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]].<br />
<br />
==Generalized heptagonal numbers==<br />
A '''generalized heptagonal number''' is obtained by the formula<br />
:<math>T_n + T_{\lfloor \frac{n}{2} \rfloor},</math><br />
where ''T''<sub>''n''</sub> is the ''n''th triangular number. The first few generalized heptagonal numbers are:<br />
:1, [[4 (number)|4]], 7, [[13 (number)|13]], 18, [[27 (number)|27]], 34, [[46 (number)|46]], 55, [[70 (number)|70]], 81, [[99 (number)|99]], 112, … {{OEIS|id=A085787}}<br />
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Every other generalized heptagonal number is a regular heptagonal number. Besides 1 and 70, no generalized heptagonal numbers are also [[Pell number]]s.<ref>B. Srinivasa Rao, "Heptagonal Numbers in the Pell Sequence and [[Diophantine equation]]s <math>2x^2 = y^2(5y - 3)^2 \pm 2</math>" ''[[Fibonacci Quarterly|Fib. Quart.]]'' '''43''' 3: 194</ref><br />
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==Sum of reciprocals==<br />
A formula for the sum of the reciprocals of the heptagonal numbers is given by:<ref>[http://www.math.psu.edu/sellersj/downey_ong_sellers_cmj_preprint.pdf Beyond the Basel Problem: Sums of Reciprocals of Figurate Numbers]</ref><br />
<br />
:<math><br />
\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) <br />
</math><br />
<br />
== Test for heptagonal numbers ==<br />
<math>\sqrt{40n +9} +3\over10</math><br />
==References==<br />
<br />
<references/><br />
<br />
{{Classes of natural numbers}}<br />
[[Category:Figurate numbers]]</div>24.59.64.214