Given an admissible indexing φ of the countable atomless Boolean algebra B, an automorphism F of B is said to be recursively presented (relative to φ) if there exists a recursive function $p \in \operatorname{Sym}(\omega)$ such that F ⚬ φ = φ ⚬ p. Our key result on recursiveness: Both the subset of $\operatorname{Aut}(\mathscr{B})$ consisting of all those automorphisms which are recursively presented relative to some indexing, and its complement, the set of all "totally nonrecursive" automorphisms, are uncountable. This arises (...) as a consequence of the following combinatorial investigations: (1) A comparison of the cycle structures of f and f̄, where f is a permutation of some free basis of B and f̄ is the automorphism of B induced by f. (2) An explicit description of the permutations of ω whose conjugacy classes in $\operatorname{Sym}(\omega)$ are (a) uncountable, (b) countably infinite, and (c) finite. (shrink)

Our results concern the existence of a countable extension U of the countable atomless Boolean algebra B such that U is a "nonconstructive" extension of B. It is known that for any fixed admissible indexing φ of B there is a countable nonconstructive extension U of B (relative to φ). The main theorem here shows that there exists an extension U of B such that for any admissible indexing φ of B, U is nonconstructive (relative to φ). Thus, in this (...) sense U is a countable totally nonconstructive extension of B. (shrink)