A114723 G.f.: x*(1 - 2*x^2)/(1 - x - 3*x^2 - 3*x^3 - x^4).

0, 1, 1, 2, 8, 18, 49, 129, 338, 890, 2340, 6153, 16181, 42550, 111892, 294238, 773745, 2034685, 5350526, 14070054, 36999432, 97295857, 255854841, 672810762, 1769262288, 4652554954

Offset

0,4

Comments

The first three of the sequence of polynomials: x^n-(x+1)^n are Pisots, suggesting the name Pascal-Pisots as the (x+1)^n polynomials give the Pascal triangular sequence. r = Abs[Table[x /. NSolve[Det[M - IdentityMatrix[4]*x] == 0, x][[n]], {n, 1, 4}]] gives:{0.5497, 0.831739, 0.831739, 2.62966}

Links

Table of n, a(n) for n=0..25.

Index entries for linear recurrences with constant coefficients, signature (1,3,3,1).

Formula

M = {{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {1, 3, 3, 1}}; w[0] = {0, 1, 1, 2}; w[n_] := w[n] = M.w[n - 1] a(n) = w[n][[1]].

a(n) = a(n-1)+3*a(n-2)+3*a(n-3)+a(n-4). G.f.: x*(2*x^2 -1)/(x^4 +3*x^3 +3*x^2 +x -1). [Colin Barker, Oct 11 2012]

Mathematica

M = {{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {1, 3, 3, 1}}; w[0] = {0, 1, 1, 2}; w[n_] := w[n] = M.w[n - 1] a = Flatten[Table[w[n][[1]], {n, 0, 25}]]
LinearRecurrence[{1, 3, 3, 1}, {0, 1, 1, 2}, 30] (* Harvey P. Dale, Jan 13 2015 *)

Prog

(PARI) concat(0, Vec((1 - 2*x^2)/(1 - x - 3*x^2 - 3*x^3 - x^4)+O(x^99))) \\ Charles R Greathouse IV, Oct 11 2012

Crossrefs

Sequence in context: A102713 A332217 A249763 * A267638 A153335 A119853

Adjacent sequences: A114720 A114721 A114722 * A114724 A114725 A114726

Keyword

nonn,easy

Author

Roger L. Bagula, Feb 18 2006

Status

approved