Pronic number

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Short description: Number, product of consecutive integers

A pronic number is a number that is the product of two consecutive integers, that is, a number of the form n(n+1).[1] The study of these numbers dates back to Aristotle. They are also called oblong numbers, heteromecic numbers,[2] or rectangular numbers;[3] however, the term "rectangular number" has also been applied to the composite numbers.[4][5]

The first 60 pronic numbers are:

0, 2, 6, 12, 20, 30, 42, 56, 72, 90, 110, 132, 156, 182, 210, 240, 272, 306, 342, 380, 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... (sequence A002378 in the OEIS).

Letting Pn denote the pronic number n(n+1), we have Pn=Pn1. Therefore, in discussing pronic numbers, we may assume that n0 without loss of generality, a convention that is adopted in the following sections.

As figurate numbers

Twice a triangular number is a pronic number
The nth pronic number is n more than the nth square number and n+1 less than the (n+1)st square

The pronic numbers were studied as figurate numbers alongside the triangular numbers and square numbers in Aristotle's Metaphysics,[2] and their discovery has been attributed much earlier to the Pythagoreans.[3] As a kind of figurate number, the pronic numbers are sometimes called oblong[2] because they are analogous to polygonal numbers in this way:[1]

* * * * *
* * *
* * * *
* * * *
* * * *
* * * * *
* * * * *
* * * * *
* * * * *
1 × 2 2 × 3 3 × 4 4 × 5

The nth pronic number is the sum of the first n even integers, and as such is twice the nth triangular number[1][2] and n more than the nth square number, as given by the alternative formula n2 + n for pronic numbers. Hence the nth pronic number and the nth square number (the sum of the first n odd integers) form a superparticular ratio:

n(n+1)n2=n+1n

Due to this ratio, the nth pronic number is at a radius of n and n + 1 from a perfect square, and the nth perfect square is at a radius of n from a pronic number. The nth pronic number is also the difference between the odd square (2n + 1)2 and the (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.[6]

Sum of pronic numbers

The partial sum of the first n positive pronic numbers is twice the value of the nth tetrahedral number:

k=1nk(k+1)=n(n+1)(n+2)3=2Tn.

The sum of the reciprocals of the positive pronic numbers (excluding 0) is a telescoping series that sums to 1:[7]

i=11i(i+1)=12+16+112+120=1.

The partial sum of the first n terms in this series is[7]

i=1n1i(i+1)=nn+1.

The alternating sum of the reciprocals of the positive pronic numbers (excluding 0) is a convergent series:

i=1(1)i+1i(i+1)=1216+112120=log(4)1.

Additional properties

Pronic numbers are even, and 2 is the only prime pronic number. It is also the only pronic number in the Fibonacci sequence and the only pronic Lucas number.[8][9]

The arithmetic mean of two consecutive pronic numbers is a square number:

n(n+1)+(n+1)(n+2)2=(n+1)2

So there is a square between any two consecutive pronic numbers. It is unique, since

n2n(n+1)<(n+1)2<(n+1)(n+2)<(n+2)2.

Another consequence of this chain of inequalities is the following property. If m is a pronic number, then the following holds:

mm=m.

The fact that consecutive integers are coprime and that a pronic number is the product of two consecutive integers leads to a number of properties. Each distinct prime factor of a pronic number is present in only one of the factors n or n + 1. Thus a pronic number is squarefree if and only if n and n + 1 are also squarefree. The number of distinct prime factors of a pronic number is the sum of the number of distinct prime factors of n and n + 1.

If 25 is appended to the decimal representation of any pronic number, the result is a square number, the square of a number ending on 5; for example, 625 = 252 and 1225 = 352. This is so because

100n(n+1)+25=100n2+100n+25=(10n+5)2.

The difference between two consecutive unit fractions is the reciprocal of a pronic number:[10]

1n1n+1=(n+1)nn(n+1)=1n(n+1)

References

  1. 1.0 1.1 1.2 Conway, J. H.; Guy, R. K. (1996), The Book of Numbers, New York: Copernicus, Figure 2.15, p. 34 .
  2. 2.0 2.1 2.2 2.3 The evolution of the Euclidean elements, Dordrecht-Boston, Mass.: D. Reidel Publishing Co., 1975, pp. 144–150, ISBN 90-277-0509-7, https://books.google.com/books?id=_1H6BwAAQBAJ&pg=PA144 .
  3. 3.0 3.1 Ben-Menahem, Ari (2009), Historical Encyclopedia of Natural and Mathematical Sciences, Volume 1, Springer reference, Springer-Verlag, p. 161, ISBN 9783540688310, https://books.google.com/books?id=9tUrarQYhKMC&pg=PA161 .
  4. Plutarch, De Iside et Osiride, section 42, https://www.perseus.tufts.edu/hopper/text?doc=Perseus:text:2008.01.0238:section=42, retrieved 16 April 2018 
  5. Higgins, Peter Michael (2008), Number Story: From Counting to Cryptography, Copernicus Books, p. 9, ISBN 9781848000018, https://books.google.com/books?id=HcIwkWXy3CwC&pg=PA9 .
  6. Rummel, Rudolf J. (1988), Applied Factor Analysis, Northwestern University Press, p. 319, ISBN 9780810108240, https://books.google.com/books?id=g_eNa_XzyEIC&pg=PA319 .
  7. 7.0 7.1 Frantz, Marc (2010), "The telescoping series in perspective", in Diefenderfer, Caren L.; Nelsen, Roger B., The Calculus Collection: A Resource for AP and Beyond, Classroom Resource Materials, Mathematical Association of America, pp. 467–468, ISBN 9780883857618, https://books.google.com/books?id=SHJ39945R1kC&pg=PA467 .
  8. McDaniel, Wayne L. (1998), "Pronic Lucas numbers", Fibonacci Quarterly 36 (1): 60–62, doi:10.1080/00150517.1998.12428962, http://www.mathstat.dal.ca/FQ/Scanned/36-1/mcdaniel2.pdf, retrieved 2011-05-21 .
  9. McDaniel, Wayne L. (1998), "Pronic Fibonacci numbers", Fibonacci Quarterly 36 (1): 56–59, doi:10.1080/00150517.1998.12428961, http://www.fq.math.ca/Scanned/36-1/mcdaniel1.pdf .
  10. This identity is a special case (r=1) of the more general formula: k=0r(1)k(rk)1n+k=r!j=0r(n+j). See: Meyer, David. "A Useful Mathematical Trick, Telescoping Series, and the Infinite Sum of the Reciprocals of the Triangular Numbers". p. 1. https://davidmeyer.github.io/qc/tricks.pdf.