A216757 a(n) = 6*a(n-1) - 9*a(n-2) + 3*a(n-3).

0, -3, -15, -63, -252, -990, -3861, -15012, -58293, -226233, -877797, -3405564, -13211910, -51254775, -198838152, -771371667, -2992450959, -11608875207, -45035307612, -174709321686, -677764787229, -2629310751036, -10200109386213, -39570153919641, -153507871295037

Offset

1,2

Comments

a(n) = X(2*n-1)/sqrt(3), where X(n) = 3*X(n-2) - sqrt(3)*X(n-3), with X(0)=3, X(1)=0, and X(2)=6.

The Berndt-type sequence number 13 for the argument 2*Pi/9 defined by the relation a(n)*sqrt(3) = s(1)^(2*n-1) - s(2)^(2*n-1) + s(4)^(2*n-1), where s(j) := 2*sin(2*Pi*j/9). For the respective sums with the even powers of sines - see A215634.

We note that X(n) = s(1)^n + (-s(2))^n + s(4)^n -- see Witula's book for details. Moreover the numbers of the form a(n)*3^(-1-floor((n-1)/3)) are integers.

The following summation formulas hold: Sum_{k=3..n} a(k) = 3*(2*a(n-1) - a(n-2) + 1), X(2*n+1) - X(1)*3^n = X(2*n+1) = -sqrt(3)*Sum_{k=1..n} X(2*(n-k))*3^(k-1), and X(2*n) - X(0)*3^n = X(2*n) - 3^(n+1) = -sqrt(3)*Sum{k=1..n} X(2*(n-k))*3^(k-1).

References

R. Witula, On some applications of formulas for sums of the unimodular complex numbers, Wyd. Pracowni Komputerowej Jacka Skalmierskiego, Gliwice 2011 (in Polish).

Links

G. C. Greubel, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (6,-9,3).

Formula

a(n) = 6*a(n-1) - 9*a(n-2) + 3*a(n-3).

G.f.: -3*x^2*(1 - x)/(1 - 6*x + 9*x^2 - 3*x^3).

a(n) = Sum_{k=0..n} 3*(-1)^k*(binomial(2*n-1, n+9*k+7) - binomial(2*n-1, n+9*k+1)). - Greg Dresden, Jan 28 2023

Example

We have s(1)^5 - s(2)^5 + s(4)^5 = 5*(s(1)^3 - s(2)^3 + s(4)^3) = -15*sqrt(3), s(1)^9 - s(2)^9 + s(4)^9 = 4*(s(1)^7 - s(2)^7 + s(4)^7) = -252*sqrt(3),
39*(s(1)^11 - s(2)^11 + s(4)^11) = 10*(s(1)^13 - s(2)^13 + s(4)^13) = -38610*sqrt(3),
s(1)^7 - s(2)^7 + s(4)^7 = 4*(s(1)^5 - s(2)^5 + s(4)^5) + (s(1)^3 - s(2)^3 + s(4)^3) = -63*sqrt(3), and  s(1)^15 - s(2)^15 + s(4)^15 = 1000*(s(1)^5 - s(2)^5 + s(4)^5) + 4*(s(1)^3 - s(2)^3 + s(4)^3) = -15012*sqrt(3).
We note that a(6) = 3*(a(5) + a(4) + a(3)).

Mathematica

LinearRecurrence[{6, -9, 3}, {0, -3, -15}, 30]
CoefficientList[Series[-3*x^2*(1 - x)/(1 - 6*x + 9*x^2 - 3*x^3), {x, 0, 5 0}], x] (* G. C. Greubel, Apr 17 2017 *)

Prog

(PARI) concat(0, Vec(-3*(1-x)/(1-6*x+9*x^2-3*x^3)+O(x^99))) \\ Charles R Greathouse IV, Oct 01 2012

Crossrefs

Cf. A215634.

Sequence in context: A229277 A218313 A218190 * A218366 A359082 A359083

Adjacent sequences: A216754 A216755 A216756 * A216758 A216759 A216760

Keyword

sign,easy

Author

Roman Witula, Sep 15 2012

Status

approved