Alternate derivation of the Ginocchio-Haxton relation $[(2j-3)/6]$

We want the number of states with total angular momentum $J=j$ for 3 identical particles (e.g. neutrons) in a $j$ shell. We form states $M_1>M_2>M_3$ with total $M=M_1+M_2+M_3$. Consider first all states with $M=j+1$. Next form states by lowering $M_3$ by one. All such states exist because the lowest value of $M_3$ is $(j+1)-j-(j-1)=-j+2$. So far we have the total number of states with $J > j$ and $M=j$. The additional states with $M=j$ are the states with $J=j$. These additional states have the structure $M_1,M_2,M_2-1$ because if we try to raise $M_3$ we get a state not allowed by the Pauli principle, namely $M_1,M_2,M_2$. The possible values of $M_1,M_2$ are respectively $j-2n$ and $1/2+n$, where $n=0,1,2\cdots$. The total number of $J=j$ states is $N=\bar{n}+1$ (with $\bar{n}=n_{\rm max}$), while $\bar{n}$ itself is the number of seniority 3 states. The condition $M_1>M_2$ leads to $\bar{n}<(2j-1)/6$ or $N<(2j+5)/6$. This is our main result. It is easy to show that this is the same as the G-H relation\footnote{J.N.~Ginocchio and W.C.~Haxton, {\em Symmetries in Science VI}, ed. by B.~Gruber and M~Ramek, Plenum, New York (1993)} (see also Talmi's 1993 book) $\bar{n}=[(2j-3)/6]$, where $[]$ means the largest integer. Since $2j$ is an odd integer, $(2j-1)/6$ is either $I, I-1/3$ or $I-2/3$, where $I$ is an integer. If the value is $I$, then $\bar{n}=[(2j-3)/6]=[I-1/3]=I-1$. It is easy to show agreement in the other 2 cases as well. The number of $J=j$ states for the 3-particle system is equal to the number of $J=0$ states for a 4-particle system.