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A267482
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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.
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7
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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
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OFFSET
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0,8
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COMMENTS
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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).)
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.]
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)
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LINKS
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FORMULA
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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}.
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
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EXAMPLE
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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;
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MAPLE
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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
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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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PROG
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(PARI) T(n, k) = (-1)^((n-k)\2)*binomial((n+k)\2, k); \\François Marques, Sep 28 2021
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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