A249578 List of triples (r,s,t): the matrix M = [[4,12,9][2,7,6][1,4,4]] is raised to successive powers, then (r,s,t) are the square roots of M[3,1], M[1,1], M[1,3] respectively.

0, 1, 0, 1, 2, 3, 4, 7, 12, 15, 26, 45, 56, 97, 168, 209, 362, 627, 780, 1351, 2340, 2911, 5042, 8733, 10864, 18817, 32592, 40545, 70226, 121635, 151316, 262087, 453948, 564719, 978122, 1694157, 2107560, 3650401, 6322680

Offset

0,5

Comments

M is the 'Fibonacci matrix' F = [[1,2,1][1,1,0][1,0,0]] taken to the third power and flipped on a vertical axis.

Sequence identities:

2a(3n-2) + a(3n-1) = a(3n+1)

2a(3n) + a(3n+1) = a(3n+3)

a(3n-2) + a(3n-1) + a(3n+1) = a(3n+2)

a(3n) + a(3n+1) + a(3n-3) = a(3n+2)

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

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (0,0,4,0,0,-1).

Formula

a(n) = 4*a(n-3)-a(n-6).

G.f.: x*(3*x^4-2*x^3+x^2+1) / (x^6-4*x^3+1). - Colin Barker, Nov 04 2014

Example

M^0 = the 3 X 3 identity matrix = [[1,0,0][0,1,0][0,0,1]]. M[3,1] = 0; M[1,1] = 1; M[1,3] = 0. So the first triple is r = a(0) = 0; s = a(1) = 1; t = a(2) = 0.
M^1 = [[4,12,9][2,7,6][1,4,4]], so r = a(3) = 1; s = a(4) = 2; t = a(5) = 3.

Mathematica

CoefficientList[Series[x (3 x^4 - 2 x^3 + x^2 + 1) / (x^6 - 4 x^3 + 1), {x, 0, 70}], x] (* Vincenzo Librandi, Nov 04 2014 *)
LinearRecurrence[{0, 0, 4, 0, 0, -1}, {0, 1, 0, 1, 2, 3}, 40] (* Harvey P. Dale, Jan 17 2017 *)

Prog

(PARI) concat(0, Vec(x*(3*x^4-2*x^3+x^2+1)/(x^6-4*x^3+1) + O(x^100))) \\ Colin Barker, Nov 04 2014
(Magma) I:=[0, 1, 0, 1, 2, 3]; [n le 6 select I[n] else 4*Self(n-3)-Self(n-6): n in [1..40]]; // Vincenzo Librandi, Nov 04 2014

Crossrefs

a(3n) = the n-th term of A001353.

a(3n+1) = n-th term of A001075.

a(3n+2) = n-th term of A005320.

Sequence in context: A140463 A223578 A073147 * A325244 A376987 A217786

Adjacent sequences: A249575 A249576 A249577 * A249579 A249580 A249581

Keyword

nonn,easy

Author

Russell Walsmith, Nov 03 2014

Status

approved