A182660 a(2^(k+1)) = k; 0 everywhere else.

0, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset

0,9

Comments

A surjection N->N designed to spite a guesser who is trying to guess whether it's a surjection, using the following naive guessing method: Guess that (n0,...,nk) is a subsequence of a surjection iff it contains every natural less than log_2(k+1).

This sequence causes the would-be guesser to change his mind infinitely often.

a(0)=0. Assume a(0),...,a(n) have been defined.

If the above guesser guesses that (a(0),...,a(n)) IS the beginning of a surjective sequence, then let a(n+1)=0. Otherwise let a(n+1) be the least number not in (a(0),...,a(n)).

Links

Antti Karttunen, Table of n, a(n) for n = 0..65537

S. Alexander, On Guessing Whether A Sequence Has A Certain Property, arXiv:1011.6626 [math.LO], 2010-2012.

S. Alexander, On Guessing Whether A Sequence Has A Certain Property, J. Int. Seq. 14 (2011) # 11.4.4.

Prog

(Magma) [ exists(t){ k: k in [1..Ceiling(Log(n+1))] | n eq 2^(k+1) } select t else 0: n in [0..100] ];
(PARI) A182660(n) = if(n<2, 0, my(p = 0, k = isprimepower(n, &p)); if(2==p, k-1, 0)); \\ Antti Karttunen, Jul 22 2018

Crossrefs

Cf. A082691, A135416, A209229.

Sequence in context: A104261 A028702 A083929 * A286100 A359431 A338210

Adjacent sequences: A182657 A182658 A182659 * A182661 A182662 A182663

Keyword

nonn

Author

Sam Alexander, Nov 27 2010

Status

approved