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

0,4

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

Table of n, a(n) for n=0..49.

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

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

Prog

(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

Crossrefs

Cf. A000111, A004099, A010686, A014021.

Sequence in context: A078320 A306992 A325437 * A021392 A131688 A226213

Adjacent sequences: A141427 A141428 A141429 * A141431 A141432 A141433

Keyword

nonn,easy,less

Author

Paul Curtz, Aug 06 2008

Extensions

Edited by R. J. Mathar, Dec 05 2008

Status

approved