A112259 Expansion of x*(1+8*x)/((1-8*x)*(1+11*x+64*x^2)).
1, 5, 9, 605, 961, 16245, 284089, 29645, 15046641, 101025125, 73222249, 9908816445, 23755748641, 204034140245, 5031349566489, 1965713970605, 219320727489361, 1965265930868805, 374345220088009, 158335559155140125
Offset: 1
Examples
t = (-0.618...,1,1.618...); t^2 = (3.618...,1.381...,-1). Hence -f(t^2) = 5
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (-3,24,512).
Programs
-
Mathematica
s = {-1/GoldenRatio, 1, GoldenRatio}; trit[lst_] := Block[{a, b, c, d, e, f}, {a, b, c} = lst[[1]]; {d, e, f} = lst[[2]]; {{a, b, c}, FullSimplify[{a*d + b*f + c*e, a*e + b*d + c*f, a*f + b*e + c*d}]}]; f[n_] := FullSimplify[ -Times @@ Nest[trit, {s, s}, n][[2]]]; Table[ f[n], {n, 0, 20}] (* Robert G. Wilson v, May 16 2006 *) CoefficientList[Series[(1 + 8 x) / ((1 - 8 x) (1 + 11 x + 64 x^2)), {x, 0, 20}], x] (* Vincenzo Librandi, Aug 04 2013 *) LinearRecurrence[{-3,24,512},{1,5,9},20] (* Harvey P. Dale, Feb 28 2024 *)
Formula
t = (-1/p, 1, p). f(t) is the product 1/p*1*p. For t1 = (a, b, c) and t2 = (x, y, z), t1 - t2 = a(x, y, z) + b(z, x, y) + c(y, z, x) = (ax+bz+cy, ay+bx+cz, az+by+cx). -f(t^n) = the sequence.
G.f.: x*(1+8*x)/((1-8*x)*(1+11*x+64*x^2)). [Joerg Arndt, Aug 03 2013]
From G. C. Greubel, Sep 21 2020: (Start)
a(n) = 2^(3*n+1) * (1 - (-1)^n * T_{n}(11/16))/27, where T_{n}(x) is the Chebyshev polynomial.
a(n) = -3*a(n-1) + 24*a(n-2) + 512*a(n-3). (End)
Extensions
More terms from Robert G. Wilson v, May 16 2006
New name using g.f. from Joerg Arndt, Sep 20 2020
Comments