Editing Pronic number (section)

{{Short description|Number, product of consecutive integers}}
A '''pronic number''' is a number that is the product of two consecutive [[integer]]s, that is, a number of the form <math>n(n+1)</math>.<ref name="bon">{{citation |first1=J. H. |last1=Conway |author1-link=John H. Conway |first2=R. K. |last2=Guy |author2-link=Richard K. Guy |title=The Book of Numbers |location=New York |publisher=Copernicus |at=Figure 2.15, p.&nbsp;34 |year=1996}}.</ref> The study of these numbers dates back to [[Aristotle]]. They are also called '''oblong numbers''', '''heteromecic numbers''',<ref name="knorr">{{citation
 | last = Knorr | first = Wilbur Richard | author-link = Wilbur Knorr
 | isbn = 90-277-0509-7
 | location = Dordrecht-Boston, Mass.
 | mr = 0472300
 | pages = 144–150
 | publisher = D. Reidel Publishing Co.
 | title = The evolution of the Euclidean elements
 | url = https://books.google.com/books?id=_1H6BwAAQBAJ&pg=PA144
 | year = 1975}}.</ref> or '''rectangular numbers''';<ref name="hist"/> however, the term "rectangular number"  has also been applied to the [[composite number]]s.<ref>{{citation|url=https://www.perseus.tufts.edu/hopper/text?doc=Perseus:text:2008.01.0238:section=42|title=Plutarch, De Iside et Osiride, section 42|website=www.perseus.tufts.edu|access-date=16 April 2018}}</ref><ref>{{citation|title=Number Story: From Counting to Cryptography|first=Peter Michael|last=Higgins|publisher=Copernicus Books|year=2008|isbn=9781848000018|page=9|url=https://books.google.com/books?id=HcIwkWXy3CwC&pg=PA9}}.</ref>

The first 60 pronic numbers are:

:[[0 (number)|0]], [[2 (number)|2]], [[6 (number)|6]], [[12 (number)|12]], [[20 (number)|20]], [[30 (number)|30]], [[42 (number)|42]], [[56 (number)|56]], [[72 (number)|72]], [[90 (number)|90]], [[110 (number)|110]], [[132 (number)|132]], 156, 182, 210, 240, 272, 306, 342, 380, [[420 (number)|420]], 462, 506, 552, 600, 650, 702, 756, 812, 870, 930, 992, 1056, 1122, 1190, 1260, 1332, 1406, 1482, 1560, 1640, 1722, 1806, 1892, 1980, 2070, 2162, 2256, 2352, 2450, 2550, 2652, 2756, 2862, 2970, 3080, 3192, 3306, 3422, 3540, 3660... {{OEIS|id=A002378}}.

Letting <math>P_n</math> denote the pronic number <math>n(n+1)</math>, we have <math>P_{{-}n} = P_{n{-}1}</math>. Therefore, in discussing pronic numbers, we may assume that <math>n\geq 0</math> [[without loss of generality]], a convention that is adopted in the following sections.

==As figurate numbers==
[[File:Illustration of Triangular Number Leading to a Rectangle.svg|120px|thumb|Twice a triangular number is a pronic number]]
[[File:Illustration of Pronic Number n^2 and nplus1^2.svg|120px|thumb|The {{mvar|n}}th pronic number is {{mvar|n}} more than the {{mvar|n}}th [[square number]] and {{mvar|n}}+1 less than the ({{mvar|n}}+1)st square]]

The pronic numbers were studied as [[figurate number]]s alongside the [[triangular number]]s and [[square number]]s in [[Aristotle]]'s ''[[Metaphysics (Aristotle)|Metaphysics]]'',<ref name="knorr"/> and their discovery has been attributed much earlier to the [[Pythagoreans]].<ref name="hist">{{citation|title=Historical Encyclopedia of Natural and Mathematical Sciences, Volume 1|series=Springer reference|first=Ari|last=Ben-Menahem|publisher=Springer-Verlag|year=2009|isbn=9783540688310|page=161|url=https://books.google.com/books?id=9tUrarQYhKMC&pg=PA161}}.</ref>
As a kind of figurate number, the pronic numbers are sometimes called ''oblong''<ref name="knorr"/> because they are analogous to [[polygonal number]]s in this way:<ref name="bon"/>

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The {{mvar|n}}th pronic number is the sum of the first {{mvar|n}} [[Even and odd numbers|even]] integers, and as such is twice the {{Mvar|n}}th triangular number<ref name="bon"/><ref name="knorr"/> and {{mvar|n}} more than the {{mvar|n}}th [[square number]], as given by the alternative formula {{math|''n''<sup>2</sup> + ''n''}} for pronic numbers. Hence the {{mvar|n}}th pronic number and the {{mvar|n}}th square number (the sum of the [[Square_number#Properties|first {{mvar|n}} odd integers]]) form a [[superparticular ratio]]: 

:<math>
\frac{n(n+1)}{n^2} = \frac{n + 1}{n}
</math>

Due to this ratio, the {{mvar|n}}th pronic number is at a [[#Additional_properties|radius]] of {{mvar|n}} and {{mvar|n}} + 1 from a perfect square, and the {{mvar|n}}th perfect square is at a radius of {{mvar|n}} from a pronic number. The {{mvar|n}}th pronic number is also the difference between the [[Even and odd numbers|odd square]] {{math|(2''n'' + 1)<sup>2</sup>}} and the {{math|(''n''+1)}}st [[centered hexagonal number]].

Since the number of off-diagonal entries in a [[square matrix]] is twice a triangular number, it is a pronic number.<ref>{{citation|title=Applied Factor Analysis|first=Rudolf J.|last=Rummel|publisher=Northwestern University Press|year=1988|isbn=9780810108240|page=319|url=https://books.google.com/books?id=g_eNa_XzyEIC&pg=PA319}}.</ref>

==Sum of pronic numbers==
The partial sum of the first {{mvar|n}} positive pronic numbers is twice the value of the {{mvar|n}}th [[tetrahedral number]]:
:<math>\sum_{k=1}^{n} k(k+1) =\frac{n(n+1)(n+2)}{3}= 2T_n </math>.
The sum of the reciprocals of the positive pronic numbers (excluding 0) is a [[telescoping series]] that sums to 1:<ref name="telescope">{{citation |last=Frantz |first=Marc |title=The Calculus Collection: A Resource for AP and Beyond |pages=467–468 |year=2010 |editor1-last=Diefenderfer |editor1-first=Caren L. |series=Classroom Resource Materials |contribution=The telescoping series in perspective |contribution-url=https://books.google.com/books?id=SHJ39945R1kC&pg=PA467 |publisher=Mathematical Association of America |isbn=9780883857618 |editor2-last=Nelsen |editor2-first=Roger B. |editor1-link=Caren Diefenderfer}}.</ref>
:<math>\sum_{i=1}^{\infty} \frac{1}{i(i+1)}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\cdots=1</math>.
The [[partial sum]] of the first {{mvar|n}} terms in this series is<ref name="telescope" />
:<math>\sum_{i=1}^{n} \frac{1}{i(i+1)} =\frac{n}{n+1}</math>.
The alternating sum of the reciprocals of the positive pronic numbers (excluding 0) is a [[convergent series]]:
:<math>\sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i(i+1)}=\frac{1}{2}-\frac{1}{6}+\frac{1}{12}-\frac{1}{20}\cdots=\log(4)-1</math>.