A158478 Number of colors needed to paint n sectors of a circle.

0, 1, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2, 3, 2

Offset: 0

Author

Jaume Oliver Lafont, Mar 20 2009

Keywords

• nonn

Comments

No pair of adjacent sectors can have the same color.

For n > 0: smallest prime factor of A098548(n).

Decimal expansion of 61/4950. - Elmo R. Oliveira, May 05 2024

Links

• Stefano Spezia, Table of n, a(n) for n = 0..10000
• Index entries for linear recurrences with constant coefficients, signature (0,1).

Crossrefs

Cf. A010693, A020639, A098548, A158411.

Programs

Haskell

a158478 n = if n < 4 then n else 2 + mod n 2
a158478_list = [0..3] ++ cycle [2,3]
-- Reinhard Zumkeller, Nov 30 2014

Mathematica

Join[Range[0, 1], ConstantArray[{2, 3}, 20]] // Flatten (* Robert Price, Jun 15 2020 *)

PARI

a(n)=if(n<4,n,2+n%2)

Formula

G.f.: x*(1+2*x+2*x^2)/(1-x^2).

E.g.f.: 2*cosh(x) + 3*sinh(x) - 2*(1 + x). - Stefano Spezia, Mar 24 2020

a(n) = a(n-2) for n > 3. - Elmo R. Oliveira, May 05 2024

a(n) = (n mod 2) + (2 mod (n+1)). - Aaron J Grech, Sep 02 2024