Search: id:A123956 Results 1-1 of 1 results found. %I A123956 %S A123956 1,1,1,1,2,2,1,3,4,4,1,4,8,8,8,1,5,12,20,16,16,1,6,18,32,48,32,32,1,7, %T A123956 24,56,80,112,64,64,1,8,32,80,160,192,256,128,128,1,9,40,120,240,432, %U A123956 448,576,256,256,1,10,50,160,400,672,1120,1024,1280,512,512 %V A123956 -1,1,1,-1,-2,-2,1,-3,4,4,-1,4,8,-8,-8,1,5,-12,-20,16,16,-1,-6,-18,32, 48,-32,-32,1,-7, %W A123956 24,56,-80,-112,64,64,-1,8,32,-80,-160,192,256,-128,-128,1,9,-40,-120, 240,432,-448, %X A123956 -576,256,256,-1,-10,-50,160,400,-672,-1120,1024,1280,-512,-512 %N A123956 Chebyshev polynomial recursion into matrices back into triangular sequence: p(k, x) = 2*x*p(k - 1, x) - p(k - 2, x) The absolute values of the coefficients of the two triangular sequences match exactly. %C A123956 My reasoning: The Steinbach matrices produce Chebyshev polynomials, which are orthogonal Polynomials. %C A123956 They are determinant-one matrices and if we include Legendre polynomials that give Matrices such that they are elements of SL(n,Q) special linear groups with Q->rational elements that give characteristic polynomials that are orthogonal to each other. %C A123956 Since the Bonacci and Steinbach are both Fibonacci like, they are associated with tori. The SL[n,C] ( complex domain) are associated with the permutation groups A[n]. %C A123956 So the Steinbach matrices are behaving like special kinds of permutation groups ( a subset of the larger group: Q < C). %C A123956 So as long as the determinants are one or negative one and the matrix elements are Integers or fractions, they are a special kind of n-th level permutation group associated with toral polynomials. %C A123956 I also figure the connection of the Bessel type polynomials which are Laurent expansion types to the Hermite Taylor expansion : Laurent outside the unit disk:--> Bessel-> Sum[a[n]/x^n,{n,1,Infinity}] Taylor inside the unit disk:--> Hermite ( vibrational)-> Sum[a[n]*x^n,{n, 0,Infinity}] %C A123956 If we associate the unit disk to determinant one, that even begins to make some sense! I figure since the Chebyshev SL(2,Z) integer group is poster boy for these matrices I'd do that : Polynomial recursion for Chebyshev's is: p[k_, x_] := p[k, x] = 2*x*p[k - 1, x] - p[k - 2, x] Matrices my program gives: 1 X 1 {{0}}, 2 X 2 {{0, 1}, {-1, 0}}, 3 X 3 {{0,1, 0}, {0, 0, 1}, {-2, -2, 0}}, 4 X 4 {{0, 1, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}, {-4, -4, 3, 0}}, 5 X 5 {{0, 1, 0, 0, 0}, {0, 0, 1, 0, 0}, {0, 0, 0, 1, 0}, {0, 0, 0, 0, 1}, {-8, -8, 8, 4, 0}}, 6 X 6 {{0, 1, 0, 0, 0, 0}, {0, 0, 1, 0, 0, 0}, {0, 0, 0, 1, 0, 0}, {0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 1}, {-16, -16, 20, 12, -5, 0}} %D A123956 CRC Standard Mathematical Tables and Formulae, 16th ed. 1996, p. 484. %D A123956 M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 799. %H A123956 M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy]. %F A123956 p(k, x) = 2*x*p(k - 1, x) - p(k - 2, x) p(k,x)->a(n,m) a(n,m)->t(n,m) %e A123956 Triangle begins: %e A123956 {-1}, %e A123956 {1, 1}, %e A123956 {-1, -2, -2}, %e A123956 {1, -3, 4, 4}, %e A123956 {-1, 4, 8, -8, -8}, %e A123956 {1, 5, -12, -20, 16, 16}, %e A123956 {-1, -6, -18, 32, 48, -32, -32}, %e A123956 {1, -7, 24, 56, -80, -112,64, 64}, %e A123956 {-1, 8, 32, -80, -160, 192, 256, -128, -128}, %e A123956 {1, 9, -40, -120, 240, 432, -448, -576, 256, 256}, %e A123956 {-1, -10, -50, 160, 400, -672, -1120, 1024, 1280, -512, -512} %e A123956 The absolute values of the coefficients of the two triangular sequences match exactly. %e A123956 Table[Abs[Flatten[b][[n]]] - Abs[Flatten[w][[n]]], {n, 1, Min[Length[Flatten[b]], Length[Flatten[w]]]}] %e A123956 {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, %e A123956 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, %e A123956 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0} %t A123956 p[ -1, x] = 0; p[0, x] = 1; p[1, x] = x + 1; p[k_, x_] := p[k, x] = 2*x*p[k - 1, x] - p[k - 2, x]; w = Table[CoefficientList[p[n, x], x], {n, 0, 10}] An[d_] := Table[If[n == d && m 1/y)] /. 1/y -> 1, y], {d, 1, 11}]; Flatten[%] %Y A123956 Cf. A123235. %Y A123956 Sequence in context: A118032 A089692 A066201 this_sequence A113594 A102563 A121496 %Y A123956 Adjacent sequences: A123953 A123954 A123955 this_sequence A123957 A123958 A123959 %K A123956 uned,probation,tabl,sign %O A123956 1,5 %A A123956 Roger Bagula and Gary Adamson (rlbagulatftn(AT)yahoo.com), Oct 27 2006 Search completed in 0.001 seconds