A201219 - OEIS

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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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,3

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.

FORMULA

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

MATHEMATICA

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