仔细看了一下，还真有...b_1(n)=a_n(n)=n\uparrow\uparrow n\\ b_2(n)=(n\uparrow\uparrow n)\uparrow\uparrow (n\uparrow\uparrow n)>n\uparrow\uparrow\uparrow3\\ b_3(n)>(n\uparrow\uparrow n\uparrow\uparrow n)\uparrow\uparrow(n\uparrow\uparrow n\uparrow\uparrow n)>n\uparrow\uparrow\uparrow4\\ \dots\\ b_n(n)>n\uparrow\uparrow\uparrow(n+1)>n\uparrow\uparrow\uparrow n\\ c_1(n)>(n\uparrow\uparrow\uparrow n)\uparrow\uparrow\uparrow(n\uparrow\uparrow\uparrow n)>n\uparrow^{4}3\\ \dots\\ m_k>n\uparrow^{m_{k-1}}n

X(n,1)=a_1(n)=1\\ X(n,2)=a_n(n)=n\uparrow\uparrow n\\ X(n,3)>n\uparrow^{n\uparrow\uparrow n}n\\ X(3,3)>G(1)\\ X(3,4)>3\uparrow^{G(1)}3=G(2)\\ \dots\\ X(3,66)>G(64)