We start from $$ \sum_{r=0}^{N N_f} \ h^{2 l} \sum_{\tau \vdash r }s_{\tau}(1^{N_f})s_{\tau}(1^{N_f}) = \left ( \prod_{n=1}^{N} \ \sum_{l_n=0}^{N_f} \ \ e_{l_n} (H) \right ) \det_{1\le i,j\le N} e_{l_i-i+j} (H)= $$ where $s_{\sigma}(1^{N_f})$ is Schur function and $\sigma \vdash r$ run over partition, or alternatively we are using $e_{l_n} (H)$ - elementary symmetric function, where $H=(h,...,h)$($N_f$ times). Lets make the expansion of the sum around $h=1$ $$ =\left ( \prod_{n=1}^{N} \ \sum_{l_n=0}^{N_f} \ h^{2 l_n} \ \binom{N_f}{l_n} \right ) \det_{1\le i,j\le N} \binom{N_f}{l_i-i+j} =|h^{2 l_n} =1+2 l_n \log h + \cdots | $$ $$ = \det_{1\le i,j\le N} \binom{N_f^2}{N_f-i+j} + \log h \sum_{k=1}^N \det_{1\le i,j\le N} (N_f+ k-j) \binom{N_f^2}{N_f -i+j} + $$ $$ {} + (\log h)^2 \left \{ \sum_{k=1}^N \det_{1\le i,j\le N} \frac{(N_f+ k-j)(N_f^2+(1-N_f)(k-j))}{2 N_f -1} \binom{N_f^2}{N_f-i+j} \right. + $$ $$ \left. +\sum_{k, l=1}^N \det_{1\le i,j\le N} (N_f+ k-j) (N_f+ l-j) \binom{N_f^2}{N_f-i+j} \right \} + $$ $$ {} + (\log h)^3\left \{\sum_{k=1}^N \det_{1\le i,j\le N}\!\!\!\! \frac{ (N_f+k-j)^2 \left(N_f(N_f+1)+ (2- N_f)(k-j)\right)}{3(2 N_f-1)} \!\!\binom{N_f^2}{N_f-i+j}+ \cdots \right \}+ O( (\log h)^4) . $$ Using $$ \sum_{s=0}^{N_f} \ \binom{N_f}{s} \binom{N_f}{s-k +l} = \binom{2 N_f}{N_f-k+l} $$ we came to $$ \det_{1\le k,l \le N} \ \binom{2 N_f}{N_f-k+l} = \frac{ G[N+ 2 N_f+1] G[N+1] G[N_f+1]^2 } { G[2 N_f+1] G[N+ N_f+1]^2 }\ = Z_0 \ . $$ First correction is simple $$ \sum_{k=1}^N \det_{1\le i,j\le N} (N_f+ k-j) \binom{N_f^2}{N_f-i+j} = N_f N \, Z_0 $$ because of $$ \sum_{k=1}^N \det_{1\le i,j\le N} ( k-j) \binom{N_f^2}{N_f-i+j} = \epsilon^{j_1 .. j_N} [(1- j_1)+\cdots +(N- j_N)] $$ $$ \times\binom{N_f^2}{N_f-1+j_1} \cdots \binom{N_f^2}{N_f-N+j_N} = 0 \ . $$ But with next terms we get in trouble (looking at $- 2k\cdot j$ terms) $$ \sum_{k=1}^N \det_{1\le i,j\le N} ( k-j)^2 \binom{N_f^2}{N_f-i+j} = \epsilon^{j_1 \cdots j_N} \left \{ [(1+ j_1^2)+ \cdots +(N^2+ j_N^2)] + \right. $$ $$ \left. {} - 2( j_1+ 2 j_2+ \cdots+ N j_N) \right \} \binom{N_f^2}{N_f-1+j_1} \cdots \binom{N_f^2}{N_f-N+j_N}= 2 \frac{N(N+1)(2 N+1)}{6} Z_0 + \text{???} $$ How is to get determinants of this type?