A201219 a(1) = 0; for n>1, a(n) = 1 if n is a power of 2, otherwise a(n) = 2.

0, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2

Offset: 1

Author

N. J. A. Sloane, Nov 28 2011

Keywords

• nonn

Comments

This solves the Emissary puzzle.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..16384
• Puzzles Column, Emissary, Fall 2011, page 9, puzzle 6.

Crossrefs

Cf. A043529, A209229.

Programs

Mathematica

Join[{0},Table[If[IntegerQ[Log2[n]],1,2],{n,2,200}]] (* Harvey P. Dale, Dec 03 2021 *)

Formula

a(1) = 0, for n > 1, a(n) = 2 - A209229(n). - Antti Karttunen, Nov 19 2017