I'd like a closed form expression $x(n)$ for the number of orbits of the symmetric group on $3$ points acting on the triples in $\{ (a,b,c) \mid a,b,c \in \Bbb{Z}, 1 \leq a,b,c \leq n, c = 2n−a−b \}$.

I feel like this should be a really basic problem, but my standard method of attack fails: look it up in OEIS and prove the known formula. My backup plan of "think about it" has failed: I don't know how to deal with the restriction on $c$. (Without the restriction, I happen to have learned this is the same thing as counting $1 \leq a \leq b \leq c \leq n$, and I happen to have learned this is Binomial($n+2$, $3$) because you need to place two bars between $n$ stars, but I have no general context for this.)

I suspect this is a pretty standard counting problem even with the restriction, but I never really learned how to count (fish, fish, fish, …, fish, fish, …, fish).

The counts $x(n)$ for $n=1$ to $30$ are: $0, 1, 2, 3, 4, 6, 7, 9, 11, 13, 15, 18, 20, 23, 26$, $29, 32, 36, 39, 43, 47, 51, 55, 60, 64, 69, 74, 79, 84, 90$.

I think there are $(n-1)\cdot(n+4)/2$ triples, but even that is a little fuzzy (increase increases by $1$ each time). I have no idea how many of them have $2$ equal components.


For each $n$ in $\mathbb{N}$, let $A_n \subset [n]^3$ be the set of triples $(a,b,c)$ such that $a+b+c=2n$. We want to count the number of orbits of $S_3$ acting on the triples of $A_n$. This is given by Burnside's lemma as $$ |A_n/S_3| = \frac{1}{6}\sum_{g \in S_3}|A_n^{g}|, $$ where for each $g \in S_3$, $A_n^g$ is the set of triples fixed by $g$. So we need to count the elements of $A_n$ (since these are preserved by the identity operation), the elements of $A_n$ that are of the form $(a,a,c)$ (since these are preserved under a two-component swap), and the elements of $A_n$ that are of the form $(a,a,a)$ (since these are preserved under the two rotations). Calling these three counts $a_n$, $b_n$, and $c_n$ respectively, the result will be $$ |A_n/S_3| = \frac{1}{6}\left(a_n + 3b_n + 2c_n\right). $$ The number of triples of the form $(a,a,a)\in A_n$ is $c_n=1$ if $n$ is a multiple of $3$, and $0$ otherwise. The number of triples of the form $(a,a,c)\in A_n$ is $b_n = \lfloor{n/2}\rfloor$, since the doubled element can take any value from $\lceil{n/2}\rceil$ to $n-1$, inclusive. Finally, we must compute the total number of triples $(a,b,c)\in A_n$. If $(a,b,c)$ were chosen from all of $\mathbb{N}^3$, the count would be $(n-1)(2n-1)$, since this is the number of ways to break a line of $2n$ elements in two distinct places. Here we must remove the cases where one of the components is greater than $n$. The number to be removed is $$ \begin{eqnarray} 3 \sum_{i=n+1}^{2n-2} (2n-i-1) &=& 3\sum_{i=1}^{n-2} (i) \\ &=& \frac{3}{2}(n-1)(n-2), \end{eqnarray} $$ giving us $$ a_n = (n-1)(2n-1) - \frac{3}{2}(n-1)(n-2) = \frac{1}{2}(n-1)(n+4). $$

So the final result is $$ \begin{eqnarray} |A_n/S_3| &=& \frac{1}{6}\left(\frac{1}{2}(n-1)(n+4) + 3\left\lfloor{\frac{1}{2}n}\right\rfloor + 2c_n\right) \\ &=& \frac{1}{12}\left(n^2 + 6n - 4\right) + \epsilon_n, \end{eqnarray} $$ where $|\epsilon_n| < 1/2$; i.e., the number of orbits is the closest integer to $(n^2+6n-4)/12$.


It is $x(n)=\lfloor \frac{n^2+4n+1}{12} \rfloor + \lfloor \frac {n+3}{6} \rfloor$

No proof, just used your data, took two differences, and played.

  • $\begingroup$ For your formula, I get: 0, 1, 0, 1, 2, 4, 5, 7, 7, 9, 11, 14, 16, 19, 20, 23, 26, 30, 33, 37, 39, 43, 47, 52, 56, 61, 64, 69, 74, 80. Was something copied down wrong? $\endgroup$ – Jack Schmidt Mar 17 '11 at 14:59
  • $\begingroup$ @Jack: Yes, the sign between the terms was - and should be +. Fixed now. $\endgroup$ – Ross Millikan Mar 17 '11 at 15:14
  • $\begingroup$ Thanks! I've split the problem into mod 6, and on each residue class it appears polynomial. I think I can get a proof there. $\endgroup$ – Jack Schmidt Mar 17 '11 at 16:18

This one can also be done using the Polya Enumeration Theorem. Recall the cycle index of the symmetric group on three elements which can be computed using pen and paper by factorizing the six constituent permutations and has the value $$Z(S_3) = \frac{1}{6}\left(a_1^3 + 3 a_1 a_2 + 2 a_3 \right).$$ Then the desired value is given by $$[z^{2n}] Z(S_3)(z+z^2+\cdots+z^n).$$ Actually doing the substitution we get $$[z^{2n}]\frac{1}{6}(z+z^2+\cdots+z^n)^3 \\+ [z^{2n}]\frac{1}{2} (z+z^2+\cdots+z^n) (z^2+z^4+\cdots+z^{2n}) \\+ [z^{2n}]\frac{1}{3} (z^3+z^6+\cdots+z^{3n}).$$ Now the contribution from the first term is $$[z^{2n}]\frac{z^3}{6}(1+z+\cdots+z^{n-1})^3 = [z^{2n-3}]\frac{1}{6} \frac{(1-z^n)^3}{(1-z)^3}.$$ This is in fact (use the Newton binomial) $$[z^{2n-3}]\frac{1}{6} \frac{1-3z^n}{(1-z)^3} =\frac{1}{6}{2n-3+2\choose 2} -\frac{1}{2}{n-3+2\choose 2} \\= \frac{1}{6}{2n-1\choose 2} -\frac{1}{2}{n-1\choose 2} \\ = \frac{1}{12} n^2 +\frac{1}{4} n - \frac{1}{3}.$$ For the second term we see by inspection that for those terms from $z+z^2+\cdots+z^n$ that are even there exists a matching term from $z^2+z^4+\cdots+z^{2n}$ that gives an exponent sum of $2n$ making for a contribution of $$\frac{1}{2}\bigg\lfloor\frac{n}{2}\bigg\rfloor = \frac{1}{4} n - \frac{1}{4} (n\bmod 2).$$ Finally the third term contributes the value $1/3$ if $n$ is divisible by three. This gives the final result $$\frac{1}{12} n^2 +\frac{1}{2} n + \begin{cases} 0 & \quad\text{if}\quad n \bmod 6 = 0,\\ -\frac{7}{12} & \quad\text{if}\quad n \bmod 6 = 1,5 \\ -\frac{1}{3} & \quad\text{if}\quad n \bmod 6 = 2,4 \\ -\frac{1}{4} & \quad\text{if}\quad n \bmod 6 = 3. \end{cases}.$$

The computation at this MSE link is somewhat similar, as is the one at this MSE link II which also uses the Newton binomial.


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