Problem 6

For a feasible sequence of length $4m+2$ with $m\ge2$, keep $a'_j=a_j$ except for the following replacements (all omitted exponents stay unchanged): $a'_{m+2}=a_{m+2}+2$; for $j=2,\ldots,6$, set $a'_{2m+j}=a_{2m+j}+1$ and $a'_{4m+2j}=a_{2m+j}+1$; and set $a'_{4m+2j+1}=a_{m+2}+3$ for $j=1,\ldots,6$. The first replacement preserves both sums because $2^{-a}=2^{-(a+2)}+6\,2^{-(a+3)}$ and $(m+2)3^{-a}=(m+2)3^{-(a+2)}+\sum_{j=1}^6(4m+2j+1)3^{-(a+3)}$. Each pair with index $2m+j$ preserves both sums since $(2m+j)+(4m+2j)=3(2m+j)$. The new sequence therefore has length $4m+13$ and remains feasible.