A094256 Expansion of x / ( (x-1)*(x^3 - 9*x^2 + 6*x - 1) ).

1, 7, 34, 143, 560, 2108, 7752, 28101, 100947, 360526, 1282735, 4552624, 16131656, 57099056, 201962057, 714012495, 2523515514, 8916942687, 31504028992, 111295205284, 393151913464, 1388758662221, 4905479957435, 17327203698086, 61202661233823, 216176614077600

Offset: 1

Author

Gary W. Adamson, Apr 25 2004

Keywords

• nonn
• easy

Comments

Previous name was: Let M = the 4 X 4 matrix [0 1 0 0 / 0 0 1 0 / 0 0 0 1 / -1 10 -15 7]. Perform M^n * [1 0 0 0] = [p q r s]. Then a(n-3), a(n-2), a(n-1), a(n) = -p, -q, -r, -s respectively.

a(n)/a(n-1) tends to 3.53208888624... = 4*cos^2(Pi/9), which is an eigenvalue of the matrix and a root of the polynomial x^4 - 6x^3 + 15x^2 -10x + 1 = 0 (having roots 4*cos^2(r*Pi/9), with r = 1,2,3,4).

Number of (s(0), s(1), ..., s(2n+4)) such that 0 < s(i) < 9 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+4, s(0) = 1, s(2n+4) = 7. - Herbert Kociemba, Jun 13 2004

From Wolfdieter Lang, Mar 27 2020: (Start)

This sequence, with offset -5, starting with -85, -10, -1, 0, 0, 0, 1, 7, ... appears in the formula for the n-th power of the 4 X 4 tridiagonal matrix given in A332602 as M_4 = matrix([1,1,0,0], [1,2,1,0], [0,1,2,1], [0,0,1,2]): (M_4)^n = a(n-2)*(M_4)^3 + b(n)*(M_4)^2 + c(n)*M_4 - a(n-3)*1_4, for n >= 0, with the 4 X 4 unit Matrix 1_4, b(n) = -15*a(n-3) + 10*a(n-4) - a(n-5), and c(n) = 10*a(n-3) - a(n-4). Proof from the characteristc polynomial of M_4 (see a comment in A332602) and the Cayley-Hamilton theorem.

From the proof that A094829(n+3)/A094829(n+2) -> rho(9)^2 = A332438 for n-> infinitiy, with rho(9) = 2*cos(Pi/9) = A332437 (see a comment in A094829), and a formula given below the same limit is obtained for a(n+1)/a(n) for n -> infinity, as stated in a comment above. (End)

Examples

			a(2), a(3), a(4), a(5) = 7, 34, 143, 560, since M^5 * [1 0 0 0] = [ -7 -34 -143 -560].
Cayley-Hamilton: (M_4)^5 = a(3)*(M_4)^3 + b(5)*(M_4)^2 + c(5)*M_4 - a(2)*1_4 = 34*(M_4)^3 - 95*(M_4)^2 + 69*M_4 - 7*1_4. - _Wolfdieter Lang_, Mar 27 2020
		

References

• C. V. Durell and A. Robson, "Advanced Trigonometry", Dover 2003, p. 216.

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..1824
• G. Kreweras, Sur les éventails de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #15 (1970), 3-41. [Annotated scanned copy]
• László Németh and László Szalay, Sequences Involving Square Zig-Zag Shapes, J. Int. Seq., Vol. 24 (2021), Article 21.5.2.
• Index entries for linear recurrences with constant coefficients, signature (7,-15,10,-1).

Crossrefs

a(n) = A005023(n-1), n > 1. - R. J. Mathar, Sep 05 2008

Cf. A080938, A094829, A332602, A332437, A332438.

Programs

Magma

I:=[1,7,34,143]; [n le 4 select I[n] else 7*Self(n-1) - 15*Self(n-2) + 10*Self(n-3) - Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jul 25 2015

Mathematica

Table[ (MatrixPower[{{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {-1, 10, -15, 7}}, n].{-1, 0, 0, 0})[[4]], {n, 24}] (* Robert G. Wilson v, Apr 28 2004 *)
LinearRecurrence[{7, -15, 10, -1}, {1, 7, 34, 143}, 40] (* Vincenzo Librandi, Jul 25 2015 *)

PARI

Vec(x / ( (x-1)*(x^3-9*x^2+6*x-1) ) + O(x^30)) \\ Michel Marcus, Jul 25 2015

Formula

From Herbert Kociemba, Jun 13 2004: (Start)

a(n) = (2/9)*Sum_{r=1..8} sin(r*Pi/9)*sin(7*r*Pi/9)*(2*cos(r*Pi/9))^(2n+4).

a(n) = 7*a(n-1) - 15*a(n-2) + 10*a(n-3) - a(n-4).

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

3*a(n) = 1 - A094829(n+2) + 8*A094829(n+1) - A094829(n). - R. J. Mathar, Jun 29 2012 [offset corrected, and A094829(1) = 0. - Wolfdieter Lang, Mar 27 2020]

a(n) = (1/3)*(1 + 2*A094829(n+1) + 8*A094829(n) - A094829(n-1)), for n >= 1, with A094829(1) and A094829(0) = 0. - Wolfdieter Lang, Mar 27 2020

Extensions

More terms from Robert G. Wilson v, Apr 28 2004

a(25)-a(26) from Vincenzo Librandi, Jul 25 2015

New name (using g.f. from Herbert Kociemba) from Joerg Arndt, Jul 25 2015