A215665 a(n) = 3*a(n-2) - a(n-3), with a(0)=0, a(1)=a(2)=-3.

0, -3, -3, -9, -6, -24, -9, -66, -3, -189, 57, -564, 360, -1749, 1644, -5607, 6681, -18465, 25650, -62076, 95415, -211878, 348321, -731049, 1256841, -2541468, 4501572, -8881245, 16046184, -31145307, 57019797, -109482105, 202204698, -385466112, 716096199

Offset: 0

Author

Roman Witula, Aug 20 2012

Keywords

• sign
• easy

Comments

The Berndt-type sequence number 6 for the argument 2Pi/9 defined by the first relation from the section "Formula" below. Two sequences connected with a(n) (possessing the respective numbers 5 and 7) are discussed in A215664 and A215666 - for more details see comments to A215664 and Witula's reference. We have a(n) - a(n+1) = A215664(n).

From initial values and the recurrence formula we deduce that a(n)/3 are all integers.

We note that a(10) is the first element of a(n) which is positive integer and all (-1)^n*a(n+10) are positive integer, which can be obtained from the title recurrence relation.

The following decomposition holds (X - c(1)*c(2)^n)*(X - c(2)*c(4)^n)*(X - c(4)*c(1)^n) = X^3 - a(n)*X^2 - A215917(n-1)*X + (-1)^n.

If X(n) = 3*X(n-2) - X(n-3), n in Z, with X(n) = a(n) for every n=0,1,..., then X(-n) = abs(A215919(n)) = (-1)^n*A215919(n) for every n=0,1,...

Examples

			We have a(1)=a(2)=a(8)=-3, a(3)=a(6)=-9, a(4)+a(11)=-10*a(10), and 47*a(5)=2*a(11).
		

References

• R. Witula, Ramanujan type formulas for arguments 2Pi/7 and 2Pi/9, Demonstratio Math., (in press, 2012).
• D. Chmiela and R. Witula, Two parametric quasi-Fibonacci numbers of the nine order, (submitted, 2012).

Links

• Index entries for linear recurrences with constant coefficients, signature (0,3,-1).

Crossrefs

Cf. A215455, A215634, A215635, A215636, A215664, A214699, A215007, A214683.

Programs

Mathematica

LinearRecurrence[{0,3,-1}, {0,-3,-3}, 50]

PARI

concat(0,Vec(-3*(1+x)/(1-3*x^2+x^3)+O(x^99))) \\ Charles R Greathouse IV, Oct 01 2012

Formula

a(n) = c(1)*c(2)^n + c(2)*c(4)^n + c(4)*c(1)^n, where c(j) := 2*cos(2*Pi*j/9).

G.f.: -3*x*(1+x)/(1-3*x^2+x^3).