I'd heard the term pronic number before but not promic number. However, the number associated with my diurnal age today, 26970, is such a number according to OEIS A007531:
A007531 | \(a(n) = n \times (n-1) \times (n-2) \) |
Pronic numbers are also known as oblong (Merzbach and Boyer 1991, p. 50) or heteromecic numbers. However, "pronic" seems to be a misspelling of "promic" (from the Greek promekes, meaning rectangular, oblate, or oblong). However, no less an authority than Euler himself used the term "pronic," so attempting to "correct" it at this late date seems inadvisable.
So that clears matters up and it would seem that the terms pronic and oblong can be applied to numbers of the form \(n \times (n-1) \) with \(n \geq 1\) as well as those of the form \(n \times (n-1) \times (n-2) \) with \(n \geq 2\). Anyway, this is just nomenclature and these types of numbers might indeed be referred to as generalised factorial numbers because they can be written in the following form:$$P_{n,k}=\frac{n!}{(n-k)!} \text{ where } k \geq 2$$The notation \(P_{n,k}\) can be interpreted as the \(n\)-th pronic number of dimension \(k\). When \(k=2\) the dimensionality is two, when \(k=3\) the dimensionality is three and so on. In this notational system, we have:$$26970=P_{31,3}=\frac{31!}{(31-3)!}=31 \times 30 \times 29$$The sequences for various values of \(k\) can be generated using this SageMath code. For example, when \(k=4\), the sequence becomes OEIS A052762:
24, 120, 360, 840, 1680, 3024, 5040, 7920, 11880, 17160, 24024, 32760, 43680, 57120, 73440, 93024, 116280, 143640, 175560, 212520, 255024, 303600, 358800, 421200, 491400, 570024, 657720, 755160, 863040, 982080, 1113024, 1256640, 1413720, 1585080, 1771560, 1974024, 2193360, 2430480, 2686320, 2961840, 3258024, 3575880, 3916440, 4280760, 4669920, 5085024, 5527200
In my post titled My Yearly Pronic Number on Saturday, 11th June 2022, I noted that:$$\sum_{n=1}^{\infty} \frac{1}{n \times (n+1)}=1$$To this it can be added that:
- \( \displaystyle \sum_{n=2}^{\infty} \dfrac{1}{P_{n,2}}=1 \)
- \( \displaystyle \sum_{n=3}^{\infty} \dfrac{1}{P_{n,3}}=\dfrac{1}{4} \)
- \( \displaystyle \sum_{n=4}^{\infty} \dfrac{1}{P_{n,4}}=\dfrac{1}{18} \)
- \( \displaystyle \sum_{n=5}^{\infty} \dfrac{1}{P_{n,5}}=\dfrac{1}{96} \)
- \( \displaystyle \sum_{n=6}^{\infty} \dfrac{1}{P_{n,6}}=\dfrac{1}{600} \)
No comments:
Post a Comment