0,3

The Floor[N[]] is necessary in Mathenmatica to get from a radial expression to an integer one. Without the Floor they all have decimal places: \!\({0.`, 0.9999999999999999`, 1.9999999999999998`, 5.999999999999999`, \ 11.999999999999998`, 24.999999999999996`, 48.00000000000001`, \ 91.00000000000001`, 168.00000000000003`, 306.`, 550.0000000000001`, \ 978.9999999999999`, 1728.0000000000007`, 3029.0000000000005`, \ 5278.000000000001`, 9150.`, 15791.999999999998`, 27148.999999999996`, \ 46511.99999999999`, 79439.`, 135300.`, 229866.0000000001`, 389642.0000000002`, 659111.0000000002`, \ 1.1128320000000005`*^6}\) Limit[a[[n + 1]]/a[[n]],n->Infinity]=(1+Sqrt[5])/2 but in a "slow" logrithmic limit.

f[n]=Binet[n] g[x]=ztransform[f[n]] h[x]=D[g[x],{x,1}] w[n]=Inverseztransform[h[x]] a(n) = Abs[w[n]]

Mathematica code for the function is:

\!\(\*

RowBox[{"\[Piecewise]", GridBox[{

{\(1\/5\ 2\^\(\(-1\) -

n\)\ \((\((\(-5\) + \(AT)5)\)\ \((

1 + \(AT)5)\)\^n - \((1 - \(AT)5)\)\^n\ \((5 + \(AT)5)\))\)\ \((\(-1\) +

n)\)\), \(n >= 1\)},

{\(1\/25\ 2\^\(-n\)\ \((\((1 + \(AT)5)\)\^n\ \((\(AT)5 - 5\ n)\) - \((1 - \

\(AT)5)\)\^n\ \((\(AT)5 + 5\ n)\))\)\),

TagBox["True",

"PiecewiseDefault",

AutoDelete->False,

DeletionWarning->True]}

},

ColumnSpacings->1.2,

ColumnAlignments->{Left}]}]\)

b0 = x /. Solve[x^2 - x - 1 == 0, x][[2]] b1 = x /. Solve[x^2 - x - 1 == 0, x][[1]] a0[n_] := (b0^n - b1^n)/(b0 - b1); f[x_] = ZTransform[a0[n], n, x] dg[x_] = D[f[x], {x, 1}] w[n_] = InverseZTransform[dg[x], x, n] a=Table[Abs[Floor[N[w[n]]], {n, 1, 25}]

Sequence in context: A137829 A045925 A128020 this_sequence A140659 A099495 A034875

Adjacent sequences: A116559 A116560 A116561 this_sequence A116563 A116564 A116565

nonn,uned,probation,obsc

Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Mar 17 2006