A097947 Expansion of g.f. (2+7*x+2*x^2)/((x^2-1)*(1+4*x+x^2)).
-2, 1, -6, 16, -62, 225, -842, 3136, -11706, 43681, -163022, 608400, -2270582, 8473921, -31625106, 118026496, -440480882, 1643897025, -6135107222, 22896531856, -85451020206, 318907548961, -1190179175642, 4441809153600, -16577057438762, 61866420601441, -230888624967006
Offset: 0
Links
- Index entries for linear recurrences with constant coefficients, signature (-4,0,4,1).
Programs
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Mathematica
LinearRecurrence[{-4, 0, 4, 1}, {-2, 1, -6, 16}, 27] (* Robert P. P. McKone, Aug 25 2023 *)
Formula
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)
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)
Comments