

A182660


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


2



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
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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 wouldbe 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], 20102012.
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, k1, 0)); \\ Antti Karttunen, Jul 22 2018


CROSSREFS

Cf. A082691, A135416, A209229.
Sequence in context: A104261 A028702 A083929 * A286100 A338210 A122698
Adjacent sequences: A182657 A182658 A182659 * A182661 A182662 A182663


KEYWORD

nonn


AUTHOR

Sam Alexander, Nov 27 2010


STATUS

approved



