A008310 Triangle of coefficients of Chebyshev polynomials T_n(x).
1, 1, -1, 2, -3, 4, 1, -8, 8, 5, -20, 16, -1, 18, -48, 32, -7, 56, -112, 64, 1, -32, 160, -256, 128, 9, -120, 432, -576, 256, -1, 50, -400, 1120, -1280, 512, -11, 220, -1232, 2816, -2816, 1024, 1, -72, 840, -3584, 6912, -6144, 2048, 13, -364, 2912, -9984, 16640, -13312, 4096
Offset: 0
Examples
Rows are: (1), (1), (-1,2), (-3,4), (1,-8,8), (5,-20,16) etc., since if c = cos(x): cos(0x) = 1, cos(1x) = 1c; cos(2x) = -1+2c^2; cos(3x) = -3c+4c^3, cos(4x) = 1-8c^2+8c^4, cos(5x) = 5c-20c^3+16c^5, etc. From _Wolfdieter Lang_, Aug 02 2014: (Start) This irregular triangle a(n,k) begins: n\k 0 1 2 3 4 5 6 7 ... 0: 1 1: 1 2: -1 2 3: -3 4 4: 1 -8 8 5: 5 -20 16 6: -1 18 -48 32 7: -7 56 -112 64 8: 1 -32 160 -256 128 9: 9 -120 432 -576 256 10: -1 50 -400 1120 -1280 512 11: -11 220 -1232 2816 -2816 1024 12: 1 -72 840 -3584 6912 -6144 2048 13: 13 -364 2912 -9984 16640 -13312 4096 14: -1 98 -1568 9408 -26880 39424 -28672 8192 15: -15 560 -6048 28800 -70400 92160 -61440 16384 ... T(4,x) = 1 - 8*x^2 + 8*x^4, T(5,x) = 5*x - 20*x^3 +16*x^5. (End)
References
- M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
- E. A. Guilleman, Synthesis of Passive Networks, Wiley, 1957, p. 593.
- Yaroslav Zolotaryuk, J. Chris Eilbeck, "Analytical approach to the Davydov-Scott theory with on-site potential", Physical Review B 63, p543402, Jan. 2001. The authors write, "Since the algebra of these is 'hyperbolic', contrary to the usual Chebyshev polynomials defined on the interval 0 <= x <= 1, we call the set of functions (21) the hyperbolic Chebyshev polynomials." (This refers to the triangle T* described in Comments.)
Links
- R. J. Mathar, Table of n, a(n) for n = 0..2600
- 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].
- Renato Ferreira Pinto Jr. and Nathaniel Harms, Testing Support Size More Efficiently Than Learning Histograms, arXiv:2410.18915 [cs.DS], 2024. See p. 40.
- D. Foata and G.-N. Han, Nombres de Fibonacci et polynomes orthogonaux
- C. Lanczos, Applied Analysis (Annotated scans of selected pages)
- I. Rivin, Growth in free groups (and other stories), arXiv:math/9911076 [math.CO], 1999.
- Eric Weisstein's World of Mathematics, Chebyshev Polynomial of the First Kind.
- Wikipedia, Chebyshev polynomials.
- Index entries for sequences related to Chebyshev polynomials.
Crossrefs
A039991 is a row reversed version, but has zeros which enable the triangle to be seen. Columns/diagonals are A011782, A001792, A001793, A001794, A006974, A006975, A006976 etc.
Row sums are one. Polynomial evaluations include A001075 (x=2), A001541 (x=3), A001091, A001079, A023038, A011943, A001081, A023039, A001085, A077422, A077424, A097308, A097310, A068203.
Cf. A053120.
Programs
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Maple
A008310 := proc(n,m) local x ; coeftayl(simplify(ChebyshevT(n,x),'ChebyshevT'),x=0,m) ; end: i := 0 : for n from 0 to 100 do for m from n mod 2 to n by 2 do printf("%d %d ",i,A008310(n,m)) ; i := i+1 ; od ; od ; # R. J. Mathar, Apr 20 2007 # second Maple program: b:= proc(n) b(n):= `if`(n<2, 1, expand(2*b(n-1)-x*b(n-2))) end: T:= n-> (p-> (d-> seq(coeff(p, x, d-i), i=0..d))(degree(p)))(b(n)): seq(T(n), n=0..15); # Alois P. Heinz, Sep 04 2019
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Mathematica
Flatten[{1, Table[CoefficientList[ChebyshevT[n, x], x], {n, 1, 13}]}]//DeleteCases[#, 0, Infinity]& (* or *) Flatten[{1, Table[Table[((-1)^k*2^(n-2 k-1)*n*Binomial[n-k, k])/(n-k), {k, Floor[n/2], 0, -1}], {n, 1, 13}]}] (* Eugeniy Sokol, Sep 04 2019 *)
Formula
a(n,m) = 2^(m-1) * n * (-1)^((n-m)/2) * ((n+m)/2-1)! / (((n-m)/2)! * m!) if n>0. - R. J. Mathar, Apr 20 2007
From Paul Weisenhorn, Oct 02 2019: (Start)
T_n(x) = 2*x*T_(n-1)(x) - T_(n-2)(x), T_0(x) = 1, T_1(x) = x.
T_n(x) = ((x+sqrt(x^2-1))^n + (x-sqrt(x^2-1))^n)/2. (End)
From Peter Bala, Aug 15 2022: (Start)
T(n,x) = [z^n] ( z*x + sqrt(1 + z^2*(x^2 - 1)) )^n.
Sum_{k = 0..2*n} binomial(2*n,k)*T(k,x) = (2^n)*(1 + x)^n*T(n,x).
exp( Sum_{n >= 1} T(n,x)*t^n/n ) = Sum_{n >= 0} P(n,x)*t^n, where P(n,x) denotes the n-th Legendre polynomial. (End)
Extensions
Additional comments and more terms from Henry Bottomley, Dec 13 2000
Comments