User blog comment:All Things Logical/ABACABADABACABA numbers/@comment-5529393-20171128045632

ABACABADABACABA(n) = $$n^{2^n - 3} + 2 \cdot n^{2^n - 5} + n^{2^n - 7} + \cdots$$, so

$$ n^{2^n - 3} \le ABACABADABACABA(n) < n^{2^n - 3} + 3 \cdot n^{2^n - 5}$$ for all n.