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 A089154 A palindromic matrix version of the alternating matrix sequence at the X^3 level 3 X 3: BAAB. 0
 4, 30, 190, 1176, 7252, 44694, 275422, 1697232, 10458820, 64450158, 397159774, 2447408808, 15081612628, 92937084582, 572704120126, 3529161805344, 21747674952196, 134015211518526, 825838944063358, 5089048875898680 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS This result is the best behaved of the ABBA and BAAB combinations of Pisots with n-Bonacci's to n=5 that I tried with even odd versions of this recursive method. LINKS FORMULA A=m0={{0, 1, 0}, {0, 1, 1}, {1, 0, -1}} B=m1={{0, 1, 0}, {0, 0, 1}, {1, 1, 1}} AB=m0.m1 BA=m1.m0 a(n) = Floor[m[n][[3, 3]]] element of If[Mod[n, 2]==0, m[n-1].mab, m[n-1].mba] Empirical G.f.: 2*x*(x+2) / ((x-1)*(x^2+6*x-1)). [Colin Barker, Dec 02 2012] MATHEMATICA (* Adamson's matrix functions alternating x^3-x^2-x-1 Pisot and minimal Pisot:BAAB Palindromic *) digits=100 Solve[x^3-x-1==0, x] k=1.32471795724474605' q=N[k^2-1/k, 20] m0={{0, 1, 0}, {0, 1, 1}, {1, 0, -q}} NSolve[x^3-x^2-x-1==0, x] k1=1.83928675521416113 q1=k1^2-k1-1/k1 m1={{0, 1, 0}, {0, 0, 1}, {1, 1, q1}} mab=m0.m1 mba=m1.m0 m[n_Integer?Positive] := If[Mod[n, 2]==0, m[n-1].mab, m[n-1].mba] m[0] = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}} a=Delete[Union[Table[Floor[m[n][[3, 3]]], {n, 1, digits}]], 1] CROSSREFS Sequence in context: A132849 A115867 A057416 * A113450 A268218 A272493 Adjacent sequences:  A089151 A089152 A089153 * A089155 A089156 A089157 KEYWORD nonn,uned AUTHOR Roger L. Bagula, Dec 06 2003 STATUS approved

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Last modified February 29 08:26 EST 2020. Contains 332355 sequences. (Running on oeis4.)