Let the maximum possible number of members of $C_{k}$ be $n_{k}$ . The maximum number means we assume that $C$ is freely generated from $B$ by $\{f,g\}$ . $n_{0}=0$ , $n_{1}=2$ . For $C_{2}$ , the sequence of length $2$ has $a$ or $b$ as the first element, so that $n_{2}=6$ , and the members with possible sequences are $C_{1}a$ , $C_{1}b$ , $ag(a)$ , $bg(b)$ , $af(a,a)$ , and $bf(b,b)$ . For $C_{3}$ , to produce a new element we either apply $g$ to a function in $C_{2}$ , or apply $f$ to 2 different elements or two functions in a sequence in $C_{2}$ , so that we get $18$ more elements, $24$ overall: $C_{2}a$ , $C_{2}b$ , $a\in C_{2}g(a)$ , $b\in C_{2}g(b)$ , $a\in C_{2}f(a,a)$ , $b\in C_{2}f(b,b)$ , $ag(a)g(g(a))$ , $bg(b)g(g(b))$ , $af(a,a)g(f(a,a))$ , $bf(b,b)g(f(b,b))$ , $[ab|ba]f(a,b)$ , $[ab|ba]f(b,a)$ , $ag(a)f(a,g(a))$ , $ag(a)f(g(a),a)$ , $ag(a)f(g(a),g(a))$ , $bg(b)f(b,g(b))$ , $bg(b)f(g(b),b)$ , $bg(b)f(g(b),g(b))$ , $af(a,a)f(a,f(a,a))$ , $af(a,a)f(f(a,a),a)$ , $af(a,a)f(f(a,a),f(a,a))$ , $bf(b,b)f(b,f(b,b))$ , $bf(b,b)f(f(b,b),b)$ , $bf(b,b)f(f(b,b),f(b,b))$ . Similarly for $n_{4}$ .