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

Author

Roman Witula, Sep 15 2012

Keywords

• sign
• easy

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).

Examples

			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)).
		

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).

Crossrefs

Cf. A215634.

Programs

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 *)

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

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