Patterns of 1-gon numbers

Question

The general pattern of the n-gonal numbers is that the mth n-gonal number is equal to $$\left(\frac{n}{2}-1\right)m^2-\left(\frac{n}{2}-2\right)m .$$ For instance, the formula of the triangular numbers is $$\frac{m(m+1)}{2}$$ which can be expanded to $$\frac{1}{2}m^2+\frac{1}{2}m.$$ However, if we take $n = 1$, the pattern specializes to $$-\frac{1}{2}m^2+\frac{3}{2}m \text{.}$$ For $m = 1 ,\ldots, 10$, the values are $$1, 1, 0, 2, -5, -9, -14, -20, -27, -35 .$$

Here are my two questions.

1. Is there a general pattern between consecutive $1$-gon numbers? So far, for $m \geq 5$, the consecutive numbers have a difference of $m - 1$, e.g. the fifth and sixth $1$-gons differ by $4$, the sixth and seventh 1-gons differ by $5$, etc. Of course, they differ where the fifth is larger than the sixth, which is larger than the seventh, but the absolute difference seems to be $m-1$.
2. What would the geometric representation of these $1$-gon numbers be? Again, as a negative representation is visually impossible, assume the absolute values. In the same way that the triangular numbers are layers of dots of consecutive numbers, what would these absolute values of the $1$-gons be?

Answer 1

Answers by calculation

If we were to define the "$1$-gon numbers" algebraically, then using $n=1$, we obtain $a_m=\left(\dfrac {(1)}2-1\right)m^2-\left(\dfrac {(1)}2-2\right)m=-\dfrac{1}{2}m^2+\dfrac{3}{2}m$

As mentioned in the question, this leads to the sequence (starting from $m=0$): $0,1,1,0,-2,-5,-9,-14,-20,\ldots$. Note that the negative numbers are very close to the triangular numbers, except they are off by $1$ and have the wrong sign. In fact, for $m\ge0$, we have $a_{m+2}=-\dfrac{1}{2}(m+2)^2+\dfrac{3}{2}(m+2)=-\left(\dfrac{m(m+1)}{2}-1\right)$, which confirms that that pattern continues forever.

Differences

This explains the pattern of differences of consecutive monogonal numbers: the pattern is essentially just the same as that for the triangular numbers, we have $a_{m+3}-a_{m+2}=-\left(\dfrac{(m+1)(m+2)}{2}-1\right) - \left(-\left(\dfrac{m(m+1)}{2}-1\right)\right)=-(m+1)$

Geometric picture

Putting aside the first two entries in the sequence, this also gives you a geometric picture for the absolute values of the negative entries: draw triangular numbers and remove one point.

Better Intuition

Warmup: $n=2$

I prefer to imagine polygonal numbers as having a fixed vertex, as suggested by this public domain illustration of the hexagonal numbers:

As you decrease the number of sides for a polygon, the angles decrease in measure, and the two sides incident to the fixed vertex move closer together. If we let the angle decrease from ${60}^\circ$ for the triangular numbers ($n=3$) to $0^\circ$ for $n=2$, then the two sides would coincide, and it makes geometric sense that the numbers are forced to count dots in a segment: $\bullet$, $\bullet-\bullet$, $\bullet-\bullet-\bullet$, etc.

And algebraically, when we substitute $n=2$ into the formula for polygonal numbers, we obtain $\left(\dfrac {(2)}2-1\right)m^2-\left(\dfrac {(2)}2-2\right)m=m$, which matches the observed pattern perfectly.

$n=1$

Following the pattern of decreasing the angle, we could decrease it from $0$ to $-{60}^\circ$, passing the sides with the fixed vertex through each other, and using the signed/directed angle as a clue that we should count some area negatively. (For those who have not encountered directed area but are familiar with the cross/vector product of three-dimensional vectors, we swapped the vectors for the sides and so negated the cross product that represented twice the area.)

This is suggestive of a triangle whose area counts negatively. But that would give the sequence of negatives of the triangular numbers, which is not what we want. However, because we're in a discrete setting, with dots on the boundary of the a shape, we have some options.

We can think of the $m$th number as starting with a side of length $m$ that then "sweeps out" or "grows" the regular polygon with the desired angle. And all new points may be counted negatively if the angle was negative, but the initial side is positive.

This gives a sequence of triangular diagrams where one side counts positively and the rest count negatively. For instance for $m=4$, we would have something like this:

In general, this geometric idea of one positive edge and the rest of the triangle negative suggests a formula like $a_m=m-\dfrac{(m-1)m}{2}$, which does indeed equal the formula we started with: $-\dfrac12m^2+\dfrac32m$.

$n<1$

We can continue this further, making more shapes that mostly count negatively. For example, with $n=0$, we have things like the following diagram suggesting that $m=5$ should yield $5-4*5=-15$: $$\begin{matrix}+&+&+&+&+\\\times&\times&\times&\times&\times\\\times&\times&\times&\times&\times\\\times&\times&\times&\times&\times\\\times&\times&\times&\times&\times\end{matrix}$$

In general, for these $n\le2$ we expect a negative-counting $(4-n)$-gon shape, except we need to add $m$ to cancel the one edge that shouldn't be negative, and add another $m$ to make the one edge positive. These pictures give the correct answers in general, as we have $$\left(\dfrac {n}2-1\right)m^2-\left(\dfrac {n}2-2\right)m=2m-\left(\left(\dfrac {(4-n)}2-1\right)m^2-\left(\dfrac {(4-n)}2-2\right)m\right)$$