Suppose $\mathbb{F}$ be a frame corresponding to a partition $m_1 \geqslant m_2\geqslant...\geqslant m_r>0$ and $f(\mathbb{F})$ represents the number of standard young tableaux. Then i want to prove that
$(n+1)f(\mathbb{F})=\sum_{j=1}^{r+1}f(m_i,m_j+1)$
where we assume $m_{r+1}=0$ and $ n= \sum_{i=1}^{r}m_i$ and use $f(\mathbb{F})=\sum_{j=1}^{r}f(m_i,m_j-1)$
Notation: $f(m_i,m_j-1)$ denotes number of standard tableau corresponding to the partition obtained from mother partition by reducing $1$ in the $j$th place and all other remains same, we assume $f(m_1,...,m_k,0,0,0....)=f(m_1,...,m_k)$
My try: I tried to do it inductively and tried to go from right to left as follows.
$f(m_1+1,m_2,...,m_r)= f(m_i) + f(m_1+1,m_2-1,...,m_r)+...+f(m_1+1,m_2,...,m_r-1)$ $f(m_1,m_2+1,...,m_r)=.............$
and then adding side by side i got the proof. but suddenly i realised for the second equation if $m_1=m_2$ then quantity on the left side is always 0 while the quantity on the right side is always greater than zero as $f(m_i)>0$ so,
where am i wrong?