cp's OEIS Frontend

a(0)=-1, a(1)=4, a(2)=-13, a(3)=49, a(n)=-4*a(n-1)+4*a(n-3)+a(n-4). - Harvey P. Dale, Sep 06 2014

a(n) = (1 - (-1)^n + 2*cos(arccos(-2)*(n+1)))/4. - Eric Simon Jacob, Aug 17 2023

Properties (from Creighton Dement, Sep 06 2004):

I: jes(n) + les(n) + tes(n) = ves(n)

II: All of the following are perfect squares: {les(2n+1); tes(2n+1); ves(2n+1); ves(2n+1) - jes(2n+1) - 1 = les(2n+1) + tes(2n+1) - 1; 3*les(2n+1) + 1 = 3*jes(n)^2 + 1}.

III: les(2n+1) divides ves(2n+1) - jes(2n+1) - 1 = les(2n+1) + tes(2n+1) - 1

IV: (jes(n))^2 = les(2n+1)

V: tes(2n) = A001570(n), sqrt( tes(2n+1) ) = A001075(n)

VI: sqrt( ves(2n+1) ) = A001835(n)

VII: sqrt( les(2n+1) ) = A001353(n)

VIII: les(n) + tes(n) = ves(2+n) + jes(n)

IX: lim n |jes(n+1)/jes(n)| = lim n |les(n+1)/les(n)| = lim n |tes(n+1)/tes(n)| = lim n |ves(n+1)/ves(n)| = 2 + sqrt(3)

Comment from Roland Bacher, Sep 07 2004: These 4 sequences satisfy jes(n+1)=-4*jes(n)-jes(n-1), les(n+1)=les(n-1)+jes(n), ves(n+1)=les(n-1)-jes(n-1)+tes(n-1), tes(n+1)=les(n-1)+3*jes(n), plus initial conditions for n=0, 1.

12*a(n) = -11 -9*(-1)^n -2*(-1)^n*A001075(n+1). - R. J. Mathar, May 21 2019

From Eric Simon Jacob, Aug 26 2023: (Start)

a(n) = ( ( sqrt(3) - 2 )^(n+1) + ( -sqrt(3) - 2 )^(n+1) + 9*(-1)^(n+1) - 11 )/12.

a(n) = ( 2*cosh( (n+1)*log(sqrt(3) - 2) ) + 9*(-1)^(n+1) - 11 )/12. (End)