Counting the normalized coefficient vectors

The number of normalized coefficient vectors — six nonnegative integers summing to $t$ whose odd part $(x,y,z)$ has minimum $0$ — is

Proof. Partition the vectors according to the first zero among $(x,y,z)$. If $x=0$ the remaining five coordinates are arbitrary nonnegative integers summing to $t$, giving $\binom{t+4}{4}$ vectors. If $x>0$ and $y=0$, subtract $1$ from $x$ and the remaining five coordinates sum to $t-1$, giving $\binom{t+3}{4}$. If $x>0$, $y>0$ and $z=0$, subtract $1$ from each of $x$ and $y$ and the remaining five sum to $t-2$, giving $\binom{t+2}{4}$. Hence

Two applications of Pascal's identity collapse this to MacMahon's form: $\binom{t+4}{4}=\binom{t+3}{4}+\binom{t+3}{3}$ and $\binom{t+3}{4}=\binom{t+2}{4}+\binom{t+2}{3}$, so the sum equals $3\binom{t+3}{4}+\bigl(\binom{t+3}{3}-\binom{t+2}{3}\bigr)$, and one more instance of Pascal gives $\binom{t+3}{3}-\binom{t+2}{3}=\binom{t+2}{2}$.

Formalization Note The three counts of five-part compositions come from comps_card. Because all arithmetic stays in $\mathbb{N}$, the Pascal steps must be arranged so that no subtraction is truncated.