Pronic number

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A pronic number is a number that is the product of two consecutive integers, that is, a number of the form n ( n + 1 ) {\displaystyle n(n+1)} {\displaystyle 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 P n {\displaystyle P_{n}} {\displaystyle P_{n}} denote the pronic number n ( n + 1 ) {\displaystyle n(n+1)} {\displaystyle n(n+1)}, we have P − n = P n − 1 {\displaystyle P_{{-}n}=P_{n{-}1}} {\displaystyle P_{{-}n}=P_{n{-}1}}. Therefore, in discussing pronic numbers, we may assume that n ≥ 0 {\displaystyle n\geq 0} {\displaystyle n\geq 0} 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 × 22 × 33 × 44 × 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 ) n 2 = n + 1 n {\displaystyle {\frac {n(n+1)}{n^{2}}}={\frac {n+1}{n}}} {\displaystyle {\frac {n(n+1)}{n^{2}}}={\frac {n+1}{n}}}

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 = 1 n k ( k + 1 ) = n ( n + 1 ) ( n + 2 ) 3 = 2 T n {\displaystyle \sum _{k=1}^{n}k(k+1)={\frac {n(n+1)(n+2)}{3}}=2T_{n}} {\displaystyle \sum _{k=1}^{n}k(k+1)={\frac {n(n+1)(n+2)}{3}}=2T_{n}}.

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

∑ i = 1 ∞ 1 i ( i + 1 ) = 1 2 + 1 6 + 1 12 + 1 20 ⋯ = 1 {\displaystyle \sum _{i=1}^{\infty }{\frac {1}{i(i+1)}}={\frac {1}{2}}+{\frac {1}{6}}+{\frac {1}{12}}+{\frac {1}{20}}\cdots =1} {\displaystyle \sum _{i=1}^{\infty }{\frac {1}{i(i+1)}}={\frac {1}{2}}+{\frac {1}{6}}+{\frac {1}{12}}+{\frac {1}{20}}\cdots =1}.

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

∑ i = 1 n 1 i ( i + 1 ) = n n + 1 {\displaystyle \sum _{i=1}^{n}{\frac {1}{i(i+1)}}={\frac {n}{n+1}}} {\displaystyle \sum _{i=1}^{n}{\frac {1}{i(i+1)}}={\frac {n}{n+1}}}.

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

∑ i = 1 ∞ ( − 1 ) i + 1 i ( i + 1 ) = 1 2 − 1 6 + 1 12 − 1 20 ⋯ = log ⁡ ( 4 ) − 1 {\displaystyle \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} {\displaystyle \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}.

In quantum mechanics

In quantum mechanics, a wave-particle's spin angular momentum L {\displaystyle L} {\displaystyle L}, specifically its total spin magnitude, is defined by the reduced Planck constant ℏ = h 2 π {\displaystyle \hbar ={\frac {h}{2\pi }}} {\displaystyle \hbar ={\frac {h}{2\pi }}} times the square root of a half-integer pronic number n 2 {\displaystyle {\frac {n}{2}}} {\displaystyle {\frac {n}{2}}}:[8]

L = ℏ n 2 ( n 2 + 1 ) {\displaystyle L=\hbar {\sqrt {{\frac {n}{2}}({\frac {n}{2}}+1)}}} {\displaystyle L=\hbar {\sqrt {{\frac {n}{2}}({\frac {n}{2}}+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.[9][10]

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

n ( n + 1 ) + ( n + 1 ) ( n + 2 ) 2 = ( n + 1 ) 2 {\displaystyle {\frac {n(n+1)+(n+1)(n+2)}{2}}=(n+1)^{2}} {\displaystyle {\frac {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

n 2 ≤ n ( n + 1 ) < ( n + 1 ) 2 < ( n + 1 ) ( n + 2 ) < ( n + 2 ) 2 . {\displaystyle n^{2}\leq n(n+1)<(n+1)^{2}<(n+1)(n+2)<(n+2)^{2}.} {\displaystyle n^{2}\leq n(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:

⌊ m ⌋ ⋅ ⌈ m ⌉ = m . {\displaystyle \lfloor {\sqrt {m}}\rfloor \cdot \lceil {\sqrt {m}}\rceil =m.} {\displaystyle \lfloor {\sqrt {m}}\rfloor \cdot \lceil {\sqrt {m}}\rceil =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

100 n ( n + 1 ) + 25 = 100 n 2 + 100 n + 25 = ( 10 n + 5 ) 2 {\displaystyle 100n(n+1)+25=100n^{2}+100n+25=(10n+5)^{2}} {\displaystyle 100n(n+1)+25=100n^{2}+100n+25=(10n+5)^{2}}.

The difference between two consecutive unit fractions is the reciprocal of a pronic number:[11]

1 n − 1 n + 1 = ( n + 1 ) − n n ( n + 1 ) = 1 n ( n + 1 ) {\displaystyle {\frac {1}{n}}-{\frac {1}{n+1}}={\frac {(n+1)-n}{n(n+1)}}={\frac {1}{n(n+1)}}} {\displaystyle {\frac {1}{n}}-{\frac {1}{n+1}}={\frac {(n+1)-n}{n(n+1)}}={\frac {1}{n(n+1)}}}

References

  1. Conway, J. H.; Guy, R. K. (1996), The Book of Numbers, New York: Copernicus, Figure 2.15, p. 34.
  2. Knorr, Wilbur Richard (1975), The evolution of the Euclidean elements, Dordrecht-Boston, Mass.: D. Reidel Publishing Co., pp. 144–150, ISBN 90-277-0509-7, MR 0472300.
  3. Ben-Menahem, Ari (2009), Historical Encyclopedia of Natural and Mathematical Sciences, Volume 1, Springer reference, Springer-Verlag, p. 161, ISBN 9783540688310.
  4. "Plutarch, De Iside et Osiride, section 42", www.perseus.tufts.edu, retrieved 16 April 2018
  5. Higgins, Peter Michael (2008), Number Story: From Counting to Cryptography, Copernicus Books, p. 9, ISBN 9781848000018.
  6. Rummel, Rudolf J. (1988), Applied Factor Analysis, Northwestern University Press, p. 319, ISBN 9780810108240.
  7. Frantz, Marc (2010), "The telescoping series in perspective", in Diefenderfer, Caren L.; Nelsen, Roger B. (eds.), The Calculus Collection: A Resource for AP and Beyond, Classroom Resource Materials, Mathematical Association of America, pp. 467–468, ISBN 9780883857618.
  8. Sebens, Charles T. (2018-05-26). "How Electrons Spin". arXiv.org. Retrieved 2026-08-07.
  9. McDaniel, Wayne L. (1998), "Pronic Lucas numbers" (PDF), Fibonacci Quarterly, 36 (1): 60–62, doi:10.1080/00150517.1998.12428962, MR 1605345, archived from the original (PDF) on 2017-07-05, retrieved 2011-05-21.
  10. McDaniel, Wayne L. (1998), "Pronic Fibonacci numbers" (PDF), Fibonacci Quarterly, 36 (1): 56–59, doi:10.1080/00150517.1998.12428961, MR 1605341.
  11. This identity is a special case ( r = 1 {\displaystyle r=1} {\displaystyle r=1}) of the more general formula: ∑ k = 0 r ( − 1 ) k ( r k ) 1 n + k = r ! ∏ j = 0 r ( n + j ) {\displaystyle \sum _{k=0}^{r}(-1)^{k}{\binom {r}{k}}{\frac {1}{n+k}}={\frac {r!}{\prod _{j=0}^{r}(n+j)}}} {\displaystyle \sum _{k=0}^{r}(-1)^{k}{\binom {r}{k}}{\frac {1}{n+k}}={\frac {r!}{\prod _{j=0}^{r}(n+j)}}}. See: Meyer, David. "A Useful Mathematical Trick, Telescoping Series, and the Infinite Sum of the Reciprocals of the Triangular Numbers" (PDF). David Meyer's GitHub. p. 1. Retrieved 2024-11-26.