Pyramidal Numbers

Today I turned 26588 days old and one of the properties of this number is that it's a 15-gonal or pentadecagonal number and a member of OEIS A177890:

A177890

15-gonal (or pentadecagonal) pyramidal numbers:                 a(\(n\)) = \( \dfrac{n (n+1)  (13n-10)}{6}\)

The initial members of the sequence are:

0, 1, 16, 58, 140, 275, 476, 756, 1128, 1605, 2200, 2926, 3796, 4823, 6020, 7400, 8976, 10761, 12768, 15010, 17500, 20251, 23276, 26588, 30200, 34125, 38376, 42966, 47908, 53215, 58900, 64976, 71456, 78353, 85680, 93450, 101676, 110371, 119548, 129220

GENERAL FORMULA

There is a general formula for \(P(n)\), the \(n\)-th polygonal pyramidal number, that uses \(T(n)\), the \(n\)-th triangular number, and \(s\) representing the number of sides of the polygon. Here it is:$$P(n,s)=T(n) \times \frac{(s-2) \times n - (s-5)}{3}$$In the case of 15-gonal numbers, the formula becomes:$$ \begin{align} P(n)&=T(n) \times \frac{13n - 10}{3}\\&=\frac{n(n+1)(13n-10)}{6} \text{ where }T(n)=\frac{n(n+1)}{2}\end{align}$$GENERATING FUNCTION

The generating function for pyramidal numbers is given by:$$G(x,s)=x \, \frac{(s-3)x+1}{(1-x)^4}$$In the case of the pentadecagonal numbers, this results in:$$G(x)=x \, \frac{12x+1}{(1-x)^4}$$PARTICULAR EXAMPLE

Figure 1 shows how a square pyramidal number is constructed:

Figure 1: source

Note that the 3-dimensional pyramidal numbers are constructed from 2-dimensional polygons stacked one on top of the other. In Figure 1, these polygons are squares. The sequence of square pyramidal numbers is given by:$$P_n^{^ {\,4}}=\frac{n(n+1)(2n+1)}{6}$$Figure 2 shows an actual pile of cannonballs forming a square-based pyramid.

Figure 2: source

On the topic of cannonballs, the cannonball problem can be stated as follows:

The cannonball problem asks for the sizes of pyramids of cannonballs that can also be spread out to form a square array, or equivalently, which numbers are both square and square pyramidal. Besides 1, there is only one other number that has this property: 4900, which is both the 70th square number and the 24th square pyramidal number. Source.

SUMS OF RECIPROCALS

The sums of the reciprocals of the pyramidal polygonal numbers all converge. Here are some examples:

• triangular pyramidal: \( \displaystyle \sum_{n=1}^{\infty} \dfrac{6}{n(n+1)(n+2)}=\dfrac{3}{2} \)
• square pyramidal: \( \displaystyle \sum_{n=1}^{\infty} \dfrac{6}{n(n+1)(2n+1)}=6(3-4 \log(2)) \)
• pentagonal pyramidal: \( \displaystyle \sum_{n=1}^{\infty} \dfrac{6}{n^2(n+1)}=\dfrac{\pi^2}{3}-2 \)

More results are listed at this site. There is a rather formidable general formula for the sum of the reciprocals of the pyramidal numbers. I don't claim to understand it but here it is:$$ - \, \frac{6 [ s-5 +(s-2)(\psi ( \frac{3}{s-2} ) +\gamma)]}{(s-5)(2s-7)} $$where \( \psi(x) \) is the digamma function and \( \gamma \) is the  Euler-Mascheroni constant.

Alternating sums of reciprocals are also convergent. An example is the alternating sum of reciprocals of square pyramidal numbers where we have:$$6\sum_{n=1}^{\infty}  \dfrac{(-1)^{n-1}}{n(n+1)(2n+1)}=6(\pi-3)$$