Search: id:A140883 Results 1-1 of 1 results found. %I A140883 %S A140883 2,1,1,1,0,1,4,3,3,4,9,0,16,0,9,16,5,20,20,5,16,31,0,30,0,30,0,31,64,7, %T A140883 112,56,56,112,7,64,129,0,288,0,320,0,288,0,129,256,9,576,120,432,432, %U A140883 120,576,9,256,511,0,1230,0,720,0,720,0,1230,0,511 %V A140883 2,1,1,1,0,1,4,-3,-3,4,9,0,-16,0,9,16,5,-20,-20,5,16,31,0,-30,0,-30,0, 31,64,-7,-112,56, %W A140883 56,-112,-7,64,129,0,-288,0,320,0,-288,0,129,256,9,-576,-120,432,432,-120, -576,9,256, %X A140883 511,0,-1230,0,720,0,720,0,-1230,0,511 %N A140883 A coefficient triangular sequence made from the ChebyshevT polynomials, T(x,n) and their toral inverse (or reversed coefficient) polynomial x^n*T(1/x,n): p(x,n)=T[x,n]+x^n*T(1/x,n). %C A140883 Row sums are all two. %C A140883 All this does is make a symmetrical coefficient triangle %C A140883 since the double integration is no where zero, they aren't orthogonal; %C A140883 Table[Integrate[p[x, n]*p[x, m]/ %C A140883 Sqrt[1 - x^2], {x, -1, 1}], {n, 0, 10}, {m, 0, 10}] %F A140883 p(x,n)=T[x,n]+x^n*T(1/x,n); Out_n,m=Coefficients(p(x,n)). %e A140883 {2}, %e A140883 {1, 1}, %e A140883 {1, 0, 1}, %e A140883 {4, -3, -3, 4}, %e A140883 {9, 0, -16, 0, 9}, %e A140883 {16, 5, -20, -20, 5, 16}, %e A140883 {31, 0, -30, 0, -30, 0, 31}, %e A140883 {64, -7, -112, 56, 56, -112, -7, 64}, %e A140883 {129, 0, -288, 0, 320, 0, -288, 0, 129}, %e A140883 {256, 9, -576, -120, 432, 432, -120, -576, 9, 256}, %e A140883 {511, 0, -1230, 0, 720, 0, 720, 0, -1230, 0, 511} %t A140883 Clear[p, x, n, m, a]; p[x_, n_] := ChebyshevT[n, x] + ExpandAll[x^n*ChebyshevT[n, 1/x]]; Table[p[x, n], {n, 0, 10}]; a = Table[CoefficientList[p[x, n], x], {n, 0, 10}]; Flatten[a] %Y A140883 Cf. A053120. %Y A140883 Sequence in context: A068029 A158208 A117274 this_sequence A064744 A135997 A026609 %Y A140883 Adjacent sequences: A140880 A140881 A140882 this_sequence A140884 A140885 A140886 %K A140883 uned,tabl,sign %O A140883 1,1 %A A140883 Roger L. Bagula and Gary W. Adamson (rlbagulatftn(AT)yahoo.com), Jul 22 2008 Search completed in 0.001 seconds