OFFSET
0,2
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.
LINKS
S. Alexander, On Guessing Whether A Sequence Has A Certain Property, arXiv:1011.6626 [math.LO], 2010-2012, J. Int. Seq. 14 (2011) # 11.4.4.
FORMULA
a(0) = 0; if n = A068156(k+1) = 6*2^k - 3 for some k >= 0 then a(n) = A033484(k) = (n-1)/2; otherwise, a(n) = n+1. - Andrey Zabolotskiy, Feb 27 2025
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Sam Alexander, Nov 26 2010
EXTENSIONS
a(22) corrected and further terms added by Andrey Zabolotskiy, Feb 27 2025
STATUS
approved