A141430 a(n) = A000111(n) mod 9.

1, 1, 1, 2, 5, 7, 7, 2, 8, 7, 4, 2, 2, 7, 1, 2, 5, 7, 7, 2, 8, 7, 4, 2, 2, 7, 1, 2, 5, 7, 7, 2, 8, 7, 4, 2, 2, 7, 1, 2, 5, 7, 7, 2, 8, 7, 4, 2, 2, 7

Offset: 0

Author

Paul Curtz, Aug 06 2008

Keywords

• nonn
• easy
• less

Comments

After the initial 1,1, the sequence is periodic with period 12.

This sequence's periodic part is a shuffled version of the two period-6 sequences A070366 and A010697. The sequence contains only the digits 1, 2, 4, 5, 7 and 8 (those of A141425).

Links

• Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,-1,1).

Crossrefs

Cf. A000111, A004099, A010686, A014021.

Programs

Python

def A141430(n): return (2, 7, 1, 2, 5, 7, 7, 2, 8, 7, 4, 2)[n%12] if n>1 else 1 # Chai Wah Wu, Apr 17 2023

Formula

a(n) = A000111(n) mod 9 = A004099(n) mod 9.

a(n+12) = a(n), n > 1.

a(n) + a(n+6) = 9, n > 1.

a(n+11-p) - a(n+p) = 6 (p=0 or 5), 0 (p=1 or 4), -3 (p=2 or 3), any n > 1.

G.f.: (6x^8-5x^7+x^6+2x^5+3x^4+x^3+1) / ((1-x)(x^2+1)(x^4-x^2+1)). - R. J. Mathar, Dec 05 2008

a(n) = 9/2 +(-1)^floor(n/2)*A010686(n)/2 - 3*A014021(n), n > 1. - R. J. Mathar, Dec 05 2008

a(n) = 9/2 - (3/2)*cos(Pi*n/6) + (1/2)*3^(1/2)*sin(Pi*n/6) - (1/2)*cos(Pi*n/2) - (5/2)*sin(Pi*n/2) - (3/2)*cos(5*Pi*n/6) - (1/2)*3^(1/2)*sin(5*Pi*n/6). - Richard Choulet, Dec 12 2008

Extensions

Edited by R. J. Mathar, Dec 05 2008