Search: id:A089069 Results 1-1 of 1 results found. %I A089069 %S A089069 0,0,2,2,1,0,0,0,1,1,1,0,0,0,1,1,1,0,0,0,1,1,1,0,0,0,1,1,1,0,0,0,1,1,1, 0,0, %T A089069 0,1,1,1,0,0,0,1,1,0,1,0,0,2,2,1,0,0,0,1,2,1,0,0,0,1,1,0,1,3,2,1,0,0,1, 3,2,1, %U A089069 0,0,1,3,2,1,0,0,1,3,2,1,0,0,1,3,2,1,0,0,1,3,2,1,0,0,1,3,2,1,0,0,1,3,2, 1,0,0,1,3 %V A089069 0,0,2,-2,-1,0,0,0,1,-1,-1,0,0,0,1,-1,-1,0,0,0,1,-1,-1,0,0,0,1,-1,-1,0, 0,0,1,-1,-1,0,0, %W A089069 0,1,-1,-1,0,0,0,1,-1,0,1,0,0,2,-2,-1,0,0,0,1,-2,-1,0,0,0,1,-1,0,1,3,-2, 1,0,0,1,3,-2,1, %X A089069 0,0,1,3,-2,1,0,0,1,3,-2,1,0,0,1,3,-2,1,0,0,1,3,-2,1,0,0,1,3,-2,1,0,0, 1,3,-2,1,0,0,1,3 %N A089069 Let m0 be the matrix {{0,1,0},{0,1,1},{1,0,-q}}, m1={{0,1,0},{0,0,1}, {1,1,q1}}; if n even a(n) = (3,3)-element of m[n-1]*m0, if n odd a(n) = (3,3)-element of m[n-1]*m1. %C A089069 Alternating even-odd matrix recursive sequence using the x^3-x^2-x-1=0 Pisot balanced by the x^3-x-1=0 minimal Pisot. %C A089069 A new minimal Pisot recursive matrix developed with eigenvalue equation x^3-x-1=0 using the Blackmore-Kappraff "bonacci" matrix as the pattern. %t A089069 (* Adamson's matrix functions alternating x^3 Pisot and minimal Pisot*) digits=200 Solve[x^3-x-1==0, x] k=positive root of minimal Pisot 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}} m[n_Integer?Positive] := If[Mod[n, 2]==0, m[n-1].m0, m[n-1].m1] m[0] = {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}} a=Table[Floor[m[n][[3, 3]]], {n, 1, digits}] %Y A089069 Sequence in context: A163534 A030619 A016370 this_sequence A143535 A016270 A049783 %Y A089069 Adjacent sequences: A089066 A089067 A089068 this_sequence A089070 A089071 A089072 %K A089069 sign %O A089069 1,3 %A A089069 Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Dec 03 2003 Search completed in 0.002 seconds