A141430 - OEIS

%I #18 Apr 17 2023 10:50:47

%S 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,

%T 4,2,2,7,1,2,5,7,7,2,8,7,4,2,2,7

%N a(n) = A000111(n) mod 9.

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

%C 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).

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,-1,1).

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

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

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

%F 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.

%F 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

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

%F 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

%o (Python)

%o 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

%Y Cf. A000111, A004099, A010686, A014021.

%K nonn,easy,less

%O 0,4

%A _Paul Curtz_, Aug 06 2008

%E Edited by _R. J. Mathar_, Dec 05 2008