A267482 Triangle of coefficients of Gaussian polynomials [2n+1,1]_q represented as finite sum of terms (1+q^2)^k*q^(g-k), where k = 0,1,...,g with g=n.
1, 1, 1, -1, 1, 1, -1, -2, 1, 1, 1, -2, -3, 1, 1, 1, 3, -3, -4, 1, 1, -1, 3, 6, -4, -5, 1, 1, -1, -4, 6, 10, -5, -6, 1, 1, 1, -4, -10, 10, 15, -6, -7, 1, 1, 1, 5, -10, -20, 15, 21, -7, -8, 1, 1
Offset: 0
Examples
Triangle begins: 1; 1, 1; -1, 1, 1; -1, -2, 1, 1; 1, -2, -3, 1, 1; 1, 3, -3, -4, 1, 1; -1, 3, 6, -4, -5, 1, 1; -1, -4, 6, 10, -5, -6, 1, 1; 1, -4, -10, 10, 15, -6, -7, 1, 1; 1, 5, -10, -20, 15, 21, -7, -8, 1, 1;
Links
- Stephen O'Sullivan, Table of n, a(n) for n = 0..495
- Gerold Brändli and Tim Beyne, Modified Congruence Modulo n with Half the Amount of Residues, arXiv:1504.02757 [math.NT], 2016-2017. Definition 6.for polynomials P_m(s).
- Stephen O'Sullivan, A class of high-order Runge-Kutta-Chebyshev stability polynomials, Journal of Computational Physics, 300 (2015), 665-678.
- Wikipedia, Gaussian binomial coefficients.
- Index entries for sequences related to Chebyshev polynomials.
Crossrefs
Programs
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Maple
A267482 := proc (n, k) local y: y := expand(subs(t = 0, diff((1+t)*product(1+t^2+2*t*ChebyshevT(i, x/2), i = 1 .. n),t))): if k = 0 then subs(x = 0, y) else subs(x = 0, diff(y, x$k)/k!) end if: end proc: seq(seq(A267482(n, k), k = 0 .. n), n = 0 .. 20);
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Mathematica
row[n_] := D[(1+t)*Product[1+t^2+2*t*ChebyshevT[i, x/2], {i, 1, n}], t] /. t -> 0 // CoefficientList[#, x]&; Table[row[n], {n, 0, 20}] // Flatten (* Jean-François Alcover, Jan 16 2016 *)
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PARI
T(n,k) = (-1)^((n-k)\2)*binomial((n+k)\2, k); \\François Marques, Sep 28 2021
Formula
G.f. for row polynomial: G(n,x) = (d^2/dt^2)((1+t)*Product_{i=1..n+1}(1+t^2+2t*T(i,x/2)))|_{t=0}.
From Wolfdieter Lang, Oct 19 2019: (Start)
Row polynomial R(n, x) = S(2*n, sqrt(2+x)) = S(n, x) + S(n-1, x) = Sum_{k=0..n} (-1)^k*binomial(2*n-k, k)*(2 + x)^(n-k), for n >= 0. (See the Thomas Baruchel conjecture and the proof above.) For the S(n, x) coefficients see A049310.
R(n, x) = Sum_{j=0} (-1)^e(n,j)*binomial(e(n,j) + j, j)*x^j*, with e(n,j) := floor((n-j)/2). See eq. (12) of the Brändli and Beyne paper.
G.f. for row polynomials R(n, x) (that is of the triangle): G(x,z) = (1 + z)/(1 - x*z + z^2).
Recurrence for R(n, x): R(-1, x) = -1, R(0, x) = 1, R(n, x) = x*R(n-1, x) - R(n-2, x), for n >= 1. (See the Brändli and Beyne link, polynomials P_m(s) in Definition 6.)
(End)
T(n,k) = (-1)^(floor((n-k)/2))*binomial(floor((n+k)/2), k). - François Marques, Sep 28 2021
Comments