A215831 a(n) = 3^(-1-floor(n/3))*A215829(n).

1, -1, 9, -11, 59, -267, 433, -2041, 9753, -15443, 73547, -349875, 555073, -2641297, 12569673, -19938491, 94883195, -451526331, 716237425, -3408408073, 16219834713, -25728821507, 122437560587, -582652240611, 924236100865, -4398227463841, 20930155058697, -33200601349355

Offset: 0

Author

Roman Witula, Aug 24 2012

Keywords

• sign
• easy

Comments

The Berndt-type sequence number 8a for the argument 2Pi/9 (see A215829 for details).

From the recurrence relation for A215829 it can be proved that a(3*n+2) is divisible by 3, a(3*n) is congruent to 1 modulo 3, and a(3*n+1) is congruent to 2 modulo 3, which implies that a(3*n)+a(3*n+1) is divisible by 3.

Examples

			We have a(6)+3*a(3)=400, while a(30)+3*a(3) is divisible by 1000.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (0,0,-33,0,0,105,0,0,1).

Programs

Magma

i:=28; I:=[3,-3,27]; A215829:=[m le 3 select I[m] else -3*Self(m-1)+9*Self(m-2)+3*Self(m-3): m in [1..i]]; [3^(-1-Floor((n-1)/3))*A215829[n]: n in [1..i]]; // Bruno Berselli, Oct 02 2012

Mathematica

LinearRecurrence[{0,0,-33,0,0,105,0,0,1},{1,-1,9,-11,59,-267,433,-2041,9753},30] (* Harvey P. Dale, Nov 29 2013 *)

Formula

G.f.: (3*x^8-11*x^7+35*x^6-30*x^5-26*x^4-22*x^3-9*x^2+x-1)/(x^9+105*x^6-33*x^3-1). [Colin Barker, Oct 28 2012]