OFFSET
0,1
COMMENTS
(Start) Let A be the unit-primitive matrix (see [Jeffery])
A=A_(10,2)=
(0 0 1 0 0)
(0 1 0 1 0)
(1 0 1 0 1)
(0 1 0 2 0)
(0 0 2 0 1).
Then a(n)=Trace(A^n). For m=1,2,..., A^(m) can also be written
A^(m)=
[ F(m-1)^2 0 F(m)^2 0 F(m-1)*F(m) ]
[ 0 F(2*m-1) 0 F(2*m) 0 ]
[ F(m)^2 0 F(m+1)^2 0 F(m)*F(m+1) ]
[ 0 F(2*m) 0 F(2*m+1) 0 ]
[ 2*F(m-1)*F(m) 0 2*F(m)*F(m+1) 0 F(2*m+1)-F(m)*F(m+1) ],
where F(m-1)=A000045(n) are the Fibonacci numbers and m=n+1. Hence also a(n+1)=Trace(A^(n+1))=F(m-1)^2+F(2*m-1)+F(m+1)^2+2*F(2*m+1)-F(m)*F(m+1). (End)
Evidently one of a class of accelerator sequences for Catalan's constant based on traces of successive powers of a unit-primitive matrix A_(N,r), 0<r<Floor(N/2), and for which the closed-form expression for a(n) is derived from the eigenvalues of A_(N,r).
LINKS
L. E. Jeffery, Unit-primitive matrices.
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).
FORMULA
G.f.: 5*(1-x-x^2)/((1+x)*(1-3*x+x^2)).
a(n) = 2*a(n-1)+2*a(n-2)-a(n-3), n>2, a(0)=5, a(1)=5, a(2)=15.
a(n) = Sum_{k=1..5} ((w_k)^2-1)^n, w_k = 2*cos((2*k-1)*Pi/10).
a(n) = (-1)^n+2*(1/tau^(2*n)+tau^(2*n)), tau = (1+sqrt(5))/2=1.618033....
E.g.f.: cosh(x) + 4*exp(3*x/2)*cosh(sqrt(5)*x/2) - sinh(x). - Stefano Spezia, Jul 09 2024
MATHEMATICA
CoefficientList[Series[5 (1-x-x^2)/((1+x)(1-3x+x^2)), {x, 0, 40}], x] (* or *) LinearRecurrence[{2, 2, -1}, {5, 5, 15}, 40] (* Harvey P. Dale, Nov 26 2016 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
L. Edson Jeffery, Apr 20 2011
STATUS
approved