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 A114723 G.f.: x*(1 - 2*x^2)/(1 - x - 3*x^2 - 3*x^3 - x^4). 0

%I

%S 0,1,1,2,8,18,49,129,338,890,2340,6153,16181,42550,111892,294238,

%T 773745,2034685,5350526,14070054,36999432,97295857,255854841,

%U 672810762,1769262288,4652554954

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

%C 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}

%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,3,1).

%F 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]].

%F 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]

%t 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}]]

%t LinearRecurrence[{1,3,3,1},{0,1,1,2},30] (* _Harvey P. Dale_, Jan 13 2015 *)

%o (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

%K nonn,easy

%O 0,4

%A _Roger L. Bagula_, Feb 18 2006

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Last modified September 25 11:39 EDT 2021. Contains 347654 sequences. (Running on oeis4.)