OFFSET
0,1
COMMENTS
Let A be the adjacency matrix of the graph P_3 with a loop added at the end. Then a(n) = trace(A^n). A is a 'reverse Jordan matrix' [0,0,1;0,1,1;1,1,0]. a(n) = abs(A094648(n)).
From L. Edson Jeffery, Mar 22 2011: (Start)
Let A be the unit-primitive matrix (see [Jeffery])
A = A_(7,1) =
(0 1 0)
(1 0 1)
(0 1 1).
Then a(n) = Trace(A^n). (End)
LINKS
Michael De Vlieger, Table of n, a(n) for n = 0..3910
A. Akbary, Q. Wang, A generalized Lucas sequence and permutations binomials, Proc. Am. Math. Soc. 134 (2006) 15-22, sequence a(n) with l=7.
Robin Chapman and Nicholas C. Singer, Eigenvalues of a bidiagonal matrix, Amer. Math. Monthly, 111 (2004), p. 441.
Tomislav Došlić, Mate Puljiz, Stjepan Šebek, and Josip Žubrinić, On a variant of Flory model, arXiv:2210.12411 [math.CO], 2022.
L. E. Jeffery, Unit-primitive matrix
Genki Shibukawa, New identities for some symmetric polynomials and their applications, arXiv:1907.00334 [math.CA], 2019.
Q. Wang, On generalized Lucas sequences, Contemp. Math. 531 (2010) 127-141, Table 1 (k=3).
Index entries for linear recurrences with constant coefficients, signature (1,2,-1).
FORMULA
G.f.: (3-2*x-2*x^2)/(1-x-2*x^2+x^3);
a(n) = a(n-1) + 2*a(n-2) - a(n-3);
a(n) = (2*sqrt(7)*sin(atan(sqrt(3)/9)/3)/3+1/3)^n + (1/3-2*sqrt(7)*sin(atan(sqrt(3)/9)/3+Pi/3)/3)^n + (2*sqrt(7)*cos(acot(-sqrt(3)/9)/3)/3+1/3)^n.
a(n) = 2^n*((cos(Pi/7))^n+(cos(3*Pi/7))^n+(cos(5*Pi/7))^n). - Vladimir Shevelev, Aug 25 2010
a(n) = (-1)^n*A094648(n). - R. J. Mathar, Nov 05 2024
MATHEMATICA
CoefficientList[Series[(3 - 2 x - 2 x^2)/(1 - x - 2 x^2 + x^3), {x, 0, 33}], x] (* Michael De Vlieger, Aug 21 2019 *)
PROG
(PARI) {a(n)=if(n>=0, n+=1; polsym(x^3-x^2-2*x+1, n-1)[n], n=1-n; polsym(1-x-2*x^2+x^3, n-1)[n])} /* Michael Somos, Aug 03 2006 */
(PARI) a(n)=trace([0, 1, 0; 1, 0, 1; 0, 1, 1]^n); /* Joerg Arndt, Apr 30 2011 */
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Jul 16 2004
STATUS
approved