Search: id:A123956
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%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
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