A120325 Period 6: repeat [0, 0, 1, 0, 1, 0].

0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0

Offset: 0

Author

Paolo P. Lava and Giorgio Balzarotti, Jun 21 2006

Keywords

• easy
• nonn

Comments

Dirichlet series for the principal character mod 6: L(s,chi) = Sum_{n>=1} a(n+3)/n^s = (1 + 1/6^s - 1/2^s - 1/3^s) Riemann-zeta(s), e.g., L(2,chi) = A100044, L(4,chi) = 5*Pi^4/486, L(6,chi) = 91*Pi^6/87480. See Jolley eq (313) and arXiv:1008.2547 L(m=6,r=1,s). - R. J. Mathar, Jul 31 2010

Examples

			a(0) = (1/3)*(sin(0) + sin(0))^2 = 0.
a(1) = (1/3)*(sin(Pi/6) + sin(7*Pi/6))^2 = (1/3)*(1/2 - 1/2)^2 = 0.
a(2) = (1/3)*(sin(Pi/3) + sin(7*Pi/3))^2 = (1/3)*((sqrt(3))/2 + (sqrt(3))/2)^2 = 1.
a(3) = (1/3)*(sin(Pi/2) + sin(7*Pi/2))^2 = (1/3)*(1 - 1)^2 = 0.
a(4) = (1/3)*(sin(2*Pi/3) + sin(14*Pi/3))^2 = (1/3)*((sqrt(3))/2 + (sqrt(3))/2)^2 = 1.
a(5) = (1/3)*(sin(5*Pi/6) + sin(35*Pi/6))^2 = (1/3)*(1/2 - 1/2)^2 = 0.
		

References

• L. B. W. Jolley, Summation of Series, Dover (1961).

Links

• Antti Karttunen, Table of n, a(n) for n = 0..100000
• R. J. Mathar, Table of Dirichlet L-Series and Prime Zeta Modulo Functions for Small Moduli, arXiv:1008.2547 [math.NT], 2010-2015.
• Index entries for characteristic functions
• Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,1).

Crossrefs

Characteristic function of A047235. One's complement of A093719.

Cf. A100044, A059841, A276086, A358841, A358842, A358754, A358755.

Programs

Magma

[(n+3)^2 mod (2+((n+1) mod 2)) : n in [0..100]]; // Wesley Ivan Hurt, Oct 31 2014

Maple

P:=proc(n)local i,j; for i from 0 by 1 to n do j:=1/3*(sin(i*Pi/6)+sin(7*i*Pi/6))^2; print(j); od; end: P(20);
seq(abs(numtheory[jacobi](n,6)),n=3..150) ; # R. J. Mathar, Jul 31 2010

Mathematica

Table[Mod[(n + 3)^2, (5 + (-1)^n)/2], {n, 0, 100}] (* Wesley Ivan Hurt, Oct 31 2014 *)
PadRight[{},120,{0,0,1,0,1,0}] (* Harvey P. Dale, Oct 05 2016 *)

PARI

A120325(n) = ((n%3)&&!(n%2)); \\ Antti Karttunen, Dec 03 2022

Python

def A120325(n): return int(not (n+3) % 6 & 3 ^ 1) # Chai Wah Wu, May 25 2022

Formula

a(n) = (1/3)*(sin(n*Pi/6) + sin(7*n*Pi/6))^2.

From R. J. Mathar, Nov 22 2008: (Start)

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

a(n+6) = a(n). (End)

a(n) = ((n+3)*Fibonacci(n+3)) mod 2. - Gary Detlefs, Dec 13 2010

a(n) = 0 if n mod 6 = 0, otherwise a(n) = n mod 2 + (-1)^n. - Gary Detlefs, Dec 13 2010

a(n) = (n+3)^2 mod (5+(-1)^n)/2. - Wesley Ivan Hurt, Oct 31 2014

a(n) = sin(n*Pi/3)^2*(2-4*cos(n*Pi/3))/3. - Wesley Ivan Hurt, Jun 19 2016

E.g.f.: 2*(cosh(x) - cos(sqrt(3)*x/2)*cosh(x/2))/3. - Ilya Gutkovskiy, Jun 20 2016

a(n) = sign((n-3) mod 2) * sign((n-3) mod 3). - Wesley Ivan Hurt, Feb 04 2022

From Antti Karttunen, Dec 03 2022: (Start)

a(n) = 1 - A093719(n).

a(n) = [A276086(n) == 3 (mod 6)], where [ ] is the Iverson bracket.

a(n) = A059841(n) - A358841(n) - A358842(n).

For n >= 1, a(n) = A059841(n) - A358754(n) - A358755(n).

(End)

Extensions

Data section extended up to a(120) by Antti Karttunen, Dec 03 2022