
COMMENTS

A naive way to guess whether a function f:N>N is a permutation, based on just an initial subsequence (f(0),...,f(n)), is to guess "no" if (f(0),...,f(n)) contains a repeated entry or if there is some i in {0,...,n} such that i is not in {f(0),...,f(n)} and 2 i<=n; and guess "yes" otherwise. a(n) thwarts that method, causing it to change its mind infinitely often as n>infinity.
a(0)=0. Suppose a(0),...,a(n) have been defined.
1.If the above method guesses that (a(0),...,a(n)) is NOT an initial subsequence of a permutation, then unmark any "marked" numbers.
2.If the above method guesses that (a(0),...,a(n)) IS an initial subsequence of a permutation, then "mark" the smallest number not in {a(0),...,a(n)}.
3.Let a(n+1) be the least unmarked number not in {a(0),...,a(n)}.
A030301 can be derived by a similar method, where instead of trying to guess whether sequences are permutations, the naive victim is trying to guess whether sequences contain infinitely many 0s.
