Earlier this month, I watched La Vuelta, one of the three grand tours of cycling (a trio that includes the Tour de France). I noticed the peloton—the main group of riders—took on an aerodynamic profile that sometimes looked like a triangle, sometimes looked like a rhombus, and sometimes looked somewhere in between.
For example, the figure below shows the four possible formations between a triangle and a rhombus when the peloton’s maximum width is four riders:
For certain numbers of riders, multiple formations like these are possible. In particular, there are two formations with 15 riders: a triangle that’s five riders wide at the base and an almost-rhombus that’s four riders wide in the middle, but missing the bottommost rider, as shown below.
After 15, what is the next smallest number of riders that similarly has two distinct formations between a triangle and a rhombus?
Taking a hint from the formations for $N=15$ riders, let's try to find the formula for all number of riders that have two formations between a triangle and a rhombus where one formation is a triangle and the other has one more row. Let's denote these numbers as $N_k^{(1)}.$ Let's start with a triangle with $n$ rows, then we could get one triangle of $n+k$ rows, and another formation by adding $k+1$ rows of descending length, from $n-1$ through $n-k-1.$ In this case, in the first formation, there are $\sum_{i=1}^k (n+i) = kn + \frac{k(k+1)}{2}$ riders added beyond the initial triangle. In the second formation, there are $\sum_{i=1}^{k+1} (n-i) = (k+1)n - \frac{(k+1)(k+2)}{2}$ additional riders. In this case we can then solve for $n$ in terms of $k$ to see that $$kn + \frac{k(k+1)}{2} = (k+1)n - \frac{(k+1)(k+2)}{2} \Rightarrow n = \frac{k(k+1)}{2} + \frac{(k+1)(k+2)}{2} = (k+1)^2.$$ Thus, for any $k \in \mathbb{N},$ $N_k^{(1)}$ is number of riders in a triangle with $(k+1)^2 + k$ rows, that is $$N_k^{(1)} = \frac{(k+1)(k+2)(k^2+3k+1)}{2}.$$
We can continue to define more configurations, e.g., let's say all number of riders that have two formations between a triangle and a rhombus where one formation is a triangle and the other has $\ell$ more rows, that is, $N^{(\ell)}_k.$ For the case of $\ell = 2,$ we would have $$\sum_{i=1}^k (n+i) = kn+\frac{k(k+1)}{2} = (k+2)n - \frac{(k+2)(k+3)}{2} = \sum_{i=1}^{k+2} (n-i),$$ which would imply that $$2n = \frac{k(k+1)}{2} + \frac{(k+2)(k+3)}{2} = k^2 + 3k + 3 = (k+1) (k+2) + 1;$$ however, since $(k+1)(k+2)$ is even, $(k+1) (k+2) + 1$ is always odd for all k, so there can be no solutions and there are no $k$ for which $N^{(2)}_k$ is defined. Luckily, all is not lost since as soon as we reach $\ell = 3,$ things start getting somewhat better. For $\ell = 3$, we have $$\sum_{i=1}^k (n+i) = kn+\frac{k(k+1)}{2} = (k+3)n - \frac{(k+3)(k+4)}{2} = \sum_{i=1}^{k+3} (n-i),$$ which would imply that $$3n = \frac{k(k+1)}{2} + \frac{(k+3)(k+4)}{2} = k^2 + 4k + 6 = (k+3)(k+1) + 3,$$ that is, $$n = \frac{(k+1)(k+3)}{3} + 1, \,\,\,\text{for $k \equiv 0, 2 \!\! \mod 3.$}$$ So since the total number of riders is the same as a triangle of $n+k$ rows, we have $$N^{(3)}_k = \frac{(k+1)(k+6)(k^2+7k+9)}{18},$$ whenever $k \equiv 0, 2 \mod \!\! 3.$ Similarly, we can find $$N_k^{(4)} = \frac{(k+2)(k+7)(k^2+9k+10)}{32},$$ whenever $k \equiv 1,2 \mod \!\! 4$ and $k \gt 2$ and $$N_k^{(5)} = \frac{(k^2+11k+15)(k^2+11k+20)}{100},$$ whenever $k \equiv 0, 4 \mod\!\!5$ and $k \gt 3.$
Though it is certainly possible that there might be others, I am somewhat confident that the next smallest number of riders with two or more formations is $N_2^{(3)} = 36.$ I am less confident, but will put it out there regardless, that $1275$ is the smallest number of riders that has three different formations, since we have both $N_9^{(3)} = N_{10}^{(4)} = 1275.$ In particular, the three formations for $1275$ riders are (a) a triangle of $50$ rows; (b) a triangle with $40$ rows, followed by $14$ rows of descending length; or (c) a triangle of $41$ rows, followed by $13$ rows of descending length.
For completeness, my working version of the sequence for number of riders that can form multiple triambuses or rhombangles or whatever we want to call them is: $$15, 36, 60, 66, 78, 95, 190, 210, 253, 325, 390, 406, 435, 518, 861, 903, 946, 1275, 1351, 1540, 1661, 2346, 2556, 2775, 3081, 3452, 3486, 4005, 4064, ....$$ However, more accurately, we can just throw on the working conditions, that it is the sequence of all numbers of riders that can form multiple triambuses where the overall number of rows in the different triambus formations differs by less than the index where I gave up, er, I mean less than $6$.
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