An estimate for the number of terms in the Hilbert–Kamke problem. II

It is proved that there exist integers $A_1,\dots,A_n$ such that the system of congruences
$$ \sum^s_{i=1}\binom{x_i}j=A_j(\bmod 2^{\alpha(n,j)}),\qquad j=1,\dots,n, $$
where $\alpha(n,j)$ denotes the exponent of the highest power of 2 dividing $(n!/(j-1)!)2^{[(n-j+1)/2]+1}$, is solvable in integers $x_1,\dots,x_s$ only if the necessary condition $s\geqslant H(n)$ holds, where
$$ H(n)=\sum_{0\leqslant k\leqslant[\ln n/\ln 2]}2^k(2^{[n/2^k]}-1). $$
From this the estimate $r(n)\geqslant H(n)$ is derived for the number $r(n)$ of terms in the Hilbert–Kamke problem. Combined with a result from the previous paper, this gives the formula $r(n)=H(n)$ for $n\geqslant12$.
Bibliography: 4 titles.