A215829 a(n) = -3*a(n-1) + 9*a(n-2) + 3*a(n-3), with a(0)=3, a(1)=-3, a(2)=27.

3, -3, 27, -99, 531, -2403, 11691, -55107, 263331, -1250883, 5957307, -28339875, 134882739, -641835171, 3054430539, -14535159939, 69169849155, -329162695299, 1566411248475, -7454188455651, 35472778517331, -168806797907427, 803312835011307

Offset: 0

Author

Roman Witula, Aug 24 2012

Keywords

• sign
• easy

Comments

The Berndt-type sequence number 8 for the argument 2*Pi/9 defined by the trigonometric relations from the Formula section below.

From the general recurrence relation: b(n) = -3*b(n-1) + 9*b(n-2) + 3*b(n-3), i.e., b(n) - b(n-2) = 8*b(n-2) + 3(b(n-3) - b(n-1)) the following summation formulas can be easily deduced: b(2*n+1) + 3*b(2*n) - 3*b(0) - b(1) = 8*Sum_{k=1..n} b(2*k-1) and b(2*n+2) + 3*b(2*n+1) - b(2) - 3*b(1) = 8*Sum_{k=1..n} b(2*k). Hence it follows that (a(2*n+1) + 3*a(2*n))/2 are all integers congruent to 3 modulo 4, and (a(2*n+2) + 3*a(2*n+1))/2 are all integers congruent to 1 modulo 4.

We note that all numbers 3^(-1-floor(n/3))*a(n) = A215831(n) and 3^(-n-2)*a(3*n+2) are integers.

The following decomposition holds true: (X - k(1)^n)*(X - (-k(2))^n)*(X - k(3)^n) = X^3 - sqrt(3)^(-n)*a(n)*X^2 + sqrt(3)^(-n)*T(n) - sqrt(3)^(-n), where T(2*n+1) = sqrt(3)*A215945(n) and T(2*n) = A215948(n). [Roman Witula, Aug 30 2012]

Examples

			We have k(1)^3 - k(2)^3 + k(4)^3 = -11*sqrt(3).
		

References

• D. Chmiela and R. Witula, Two parametric quasi-Fibonacci numbers of the nine order, (submitted, 2012).

Links

• Roman Witula, Ramanujan Type Trigonometric Formulas: The General Form for the Argument 2*Pi/7, Journal of Integer Sequences, Vol. 12 (2009), Article 09.8.5
• Index entries for linear recurrences with constant coefficients, signature (-3, 9, 3).

Crossrefs

Cf. A215945, A215948, A216034, A215455, A215634, A215635, A215636, A215664, A215665, A215666, A215831, A215575, A108716, A215794, A215828.

Programs

Mathematica

LinearRecurrence[{-3, 9, 3}, {3, -3, 27}, 50]

Formula

a(n) = (k(1)^n + (-k(2))^n + k(4)^n)*(sqrt(3))^n = (-1+4*c(1))^n + (-1+4*c(2))^n + (-1+4*c(4))^n, where k(j) := cot(2*Pi*j/9) and c(j) := cos(2*Pi*j/9).

G.f.: (3 + 6*x - 9*x^2)/(1 + 3*x - 9*x^2 - 3*x^3). [corrected by Georg Fischer, May 10 2019]