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 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. 7
 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 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,8 COMMENTS 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). 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. 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).) Signed version of A046854, A130777. 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.] From Wolfdieter Lang, Oct 19 2019: (Start) 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). For the o.g.f. of R(n, x) see the formula section. 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) 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. Definition 6.for polynomials P_m(s). S. O'Sullivan, A class of high-order Runge-Kutta-Chebyshev stability polynomials, Journal of Computational Physics, 300 (2015), 665-678. Wikipedia, Gaussian binomial coefficients. 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) EXAMPLE 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; 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); 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 *) CROSSREFS Cf. A049310, A267120, A267483, A267484, A267485, A267486. Sequence in context: A130461 A225631 A306209 * A130777 A187660 A066170 Adjacent sequences:  A267479 A267480 A267481 * A267483 A267484 A267485 KEYWORD sign,tabl,easy AUTHOR Stephen O'Sullivan, Jan 15 2016 STATUS approved

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Last modified December 9 16:42 EST 2019. Contains 329879 sequences. (Running on oeis4.)