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%I #62 Jun 23 2023 17:00:11
%S 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,
%T -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,
%U 1
%N 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.
%C The entry a(n,k), n >= 0, k = 0,1,...,g, where g=n, of this irregular triangle is the coefficient of (1+q^2)^k*q^(g-k) in the representation of the Gaussian polynomial [2n+1,1]_q = Sum_{k=0..g) a(n,k)*(1+q^2)^k*q^(g-k).
%C The sequence arises in the formal derivation of the stability polynomial B(x) = Sum_{i=0..N} d_i T(iM,x) of rank N, and degree L, where T(iM,x) denotes the Chebyshev polynomial of the first kind of degree iM. The coefficients d_i are determined by order conditions on the stability polynomial.
%C Conjecture: More generally, the Gaussian polynomial [2*n+m+1-(m mod 2),m]_q = Sum_{k=0..g(m;n)} a(m;n,k)*(1+q^2)^k*q^(g(m;n)-k), for m >= 0, n >= 0, where g(m;n) = m*n if m is odd and (2*n+1)*m/2 if m is even, and the tabf array entries a(m;n,k) are the coefficients of the g.f. for the row n polynomials G(m;n,x) = (d^m/dt^m)G(m;n,t,x)/m!|_{t=0}, with G(m;n,t,x) = (1+t)*Product_{k=1..n+(m - m (mod 2))/2}(1 + t^2 + 2*t*T(k,x/2) (Chebyshev's T-polynomials). Hence a(m;n,k) = [x^k]G(m;n,x), for k=0..g(m;n). The present entry is the instance m = 2. (Thanks to _Wolfdieter Lang_ for clarifying the text on the general prescription of a(m;n,k).)
%C Signed version of A046854, A130777.
%C Conjecture: row n is U(n, x/2) + U(n-1, x/2) where U is the sequence of Chebyshev polynomials of the second kind. - _Thomas Baruchel_, Jun 03 2018 [For a proof see the following comment.]
%C From _Wolfdieter Lang_, Oct 19 2019: (Start)
%C The row polynomial R(n, x) = Sum_{k=0..n} a(n, k)*x^k = [2*n+1]_q / q^n with the q-number [2*n+1]_q := (1 - q^n)/(1 - q), which for q = 1 becomes 2*n+1, and x = x(q) = q + q^(-1). See the simplified Name and the first comment. In terms of Chebyshev S polynomials (A049310) this q-number is written as [2*n+1]_q = q^n*S(2*n, q^(1/2) + q^(-1/2)), hence R(n, x) = S(2*n, sqrt(2+x)) = S(n, x) + S(n-1, x) (which proves the conjecture of the previous comment).
%C For the o.g.f. of R(n, x) see the formula section.
%C My motivation for looking at this sequence came from the Brändli and Beyne paper's recurrence for the polynomial P_m(s) which coincides with R(n, x), with m -> n and s -> x. (End)
%C A294099(n, k) = Sum_{j=0..k} n^j * T(n, j) for all n, k in Z. - _Michael Somos_, Jun 19 2023
%H Stephen O'Sullivan, <a href="/A267482/b267482.txt">Table of n, a(n) for n = 0..495</a>
%H Gerold Brändli and Tim Beyne, <a href="https://arxiv.org/abs/1504.02757">Modified Congruence Modulo n with Half the Amount of Residues</a>, arXiv:1504.02757 [math.NT], 2016-2017. Definition 6.for polynomials P_m(s).
%H Stephen O'Sullivan, <a href="http://dx.doi.org/10.1016/j.jcp.2015.07.050">A class of high-order Runge-Kutta-Chebyshev stability polynomials</a>, Journal of Computational Physics, 300 (2015), 665-678.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Gaussian_binomial_coefficient">Gaussian binomial coefficients</a>.
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials</a>.
%F 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}.
%F From _Wolfdieter Lang_, Oct 19 2019: (Start)
%F 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.
%F 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.
%F G.f. for row polynomials R(n, x) (that is of the triangle): G(x,z) = (1 + z)/(1 - x*z + z^2).
%F 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.)
%F (End)
%F T(n,k) = (-1)^(floor((n-k)/2))*binomial(floor((n+k)/2), k). - _François Marques_, Sep 28 2021
%e Triangle begins:
%e 1;
%e 1, 1;
%e -1, 1, 1;
%e -1, -2, 1, 1;
%e 1, -2, -3, 1, 1;
%e 1, 3, -3, -4, 1, 1;
%e -1, 3, 6, -4, -5, 1, 1;
%e -1, -4, 6, 10, -5, -6, 1, 1;
%e 1, -4, -10, 10, 15, -6, -7, 1, 1;
%e 1, 5, -10, -20, 15, 21, -7, -8, 1, 1;
%p 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);
%t 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 *)
%o (PARI) T(n,k) = (-1)^((n-k)\2)*binomial((n+k)\2, k); \\_François Marques_, Sep 28 2021
%Y Cf. A049310, A267120, A267483, A267484, A267485, A267486, A294099.
%Y Cf. A046854, A066170, A130777, A187660 all signed versions.
%K sign,tabl,easy
%O 0,8
%A _Stephen O'Sullivan_, Jan 15 2016
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