A193680 Period 6 sequence 0,1,2,0,2,1.

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

Offset: 0

Author

Wolfdieter Lang, Sep 30 2011

Keywords

• nonn
• easy
• mult

Comments

This sequence can be extended periodically to negative values of n.

See a comment on A203571 where a k-family of 2k-periodic sequences P_k has been defined. The present sequence is P_3. - Wolfdieter Lang, Feb 02 2012

Examples

			a(8) = 8(mod 3) = 2 because (-1)^floor(8/3)= +1; 8\3 = 2 is even.
a(4) = (3-4)(mod 3) = 2, because (-1)^floor(4/3) is -1; 4\3 = 1 is odd.
		

Links

• Rémi Guillaume, Formulas for A193680, and some proofs.
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,1).

Crossrefs

Cf. signed versions A112300, A186809.

Programs

Mathematica

Table[{0, 1, 2, 0, 2, 1}, {15}] // Flatten (* Jean-François Alcover, Jun 21 2013 *)
PadRight[{},120,{0,1,2,0,2,1}] (* Harvey P. Dale, Jul 25 2020 *)

PARI

a(n)=[0,1,2,0,2,1][n%6+1] \\ Charles R Greathouse IV, Oct 07 2011

Formula

a(n) = n (mod 3) if (-1)^floor(n/3)=+1 else (3 - n)(mod 3), n>=0. (-1)^floor(n/3) is the parity of the quotient floor(n/3), sometimes denoted by n\3.

O.g.f.: x*(1+2*x+2*x^3+x^4)/(1-x^6).

a(n) = 1-((-1)^(n+1)+cos(Pi*n/3)+3*cos(2*Pi*n/3))/3. - R. J. Mathar, Oct 07 2011, corrected by Vaclav Kotesovec, Feb 19 2023

a(n) = floor((71/364)*3^(n+1)) mod 3. - Hieronymus Fischer, Jan 04 2013

a(n) = floor((4007/333333)*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 04 2013

From Amiram Eldar, Jan 01 2023: (Start)

Multiplicative with a(2^e) = 2, a(3^e) = 0, and a(p^e) = 1 for p >= 5.

Dirichlet g.f.: zeta(s)*(1+1/2^s-1/3^s-1/6^s). (End)

Extensions

Keyword:mult added by Andrew Howroyd, Jul 31 2018