login
A189234
Expansion of (5-4*x-12*x^2+6*x^3+3*x^4)/(1-x-4*x^2+3*x^3+3*x^4-x^5).
7
5, 1, 9, 4, 25, 16, 78, 64, 257, 256, 874, 1013, 3034, 3953, 10684, 15229, 38017, 58056, 136338, 219508, 491870, 824737, 1782735, 3083887, 6484514, 11489516, 23652443, 42688039, 86459608, 158270401, 316576903, 585868009, 1160673633, 2166063365, 4259693562
OFFSET
0,1
COMMENTS
(Start) Let U be the unit-primitive matrix (see [Jeffery])
U=U_(11,1)=
(0 1 0 0 0)
(1 0 1 0 0)
(0 1 0 1 0)
(0 0 1 0 1)
(0 0 0 1 1).
Then a(n)=Trace(U^n). (End)
Evidently one of a class of accelerator sequences for Catalan's constant based on traces of successive powers of a unit-primitive matrix U_(N,r) (0<r<floor(N/2)) and for which the closed-form expression for a(n) is derived from the eigenvalues of U_(N,r).
LINKS
A. Akbary, Q. Wang, A generalized Lucas sequence and permutation binomials, Proc. Amer. Math. Soc. 134 (2006) 15-22, sequence a(n) with l=11.
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=5).
FORMULA
G.f.: (5-4*x-12*x^2+6*x^3+3*x^4)/(1-x-4*x^2+3*x^3+3*x^4-x^5).
a(n)=a(n-1)+4*a(n-2)-3*a(n-3)-3*a(n-4)+a(n-5), {a(m)}={5,1,9,4,25}, m=0..4.
a(n)=Sum_{k=1..5} (x_k)^n; x_k=2*(-1)^(k-1)*cos(k*Pi/11).
MATHEMATICA
CoefficientList[Series[(5-4x-12x^2+6x^3+3x^4)/(1-x-4x^2+3x^3+ 3x^4-x^5), {x, 0, 40}], x] (* or *) LinearRecurrence[{1, 4, -3, -3, 1}, {5, 1, 9, 4, 25}, 40] (* Harvey P. Dale, Jan 18 2012 *)
PROG
(PARI) Vec((5-4*x-12*x^2+6*x^3+3*x^4)/(1-x-4*x^2+3*x^3+3*x^4-x^5)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012
CROSSREFS
Unsigned version of A094650.
Sequence in context: A293198 A193029 A094650 * A199736 A207873 A198671
KEYWORD
nonn,easy
AUTHOR
L. Edson Jeffery, Apr 18 2011
STATUS
approved