login
A267120
Triangle of coefficients of Gaussian polynomials [2n+3,3]_q represented as finite sum of terms (1+q^2)^k*q^(g-k), where k = 0,1,...,g with g=3n.
6
1, 0, -1, 1, 1, -1, 0, 5, -2, -4, 1, 1, 0, 2, -2, -15, 7, 17, -5, -7, 1, 1, 1, 0, -15, 6, 53, -23, -67, 22, 38, -8, -10, 1, 1, 0, -3, 3, 55, -28, -189, 81, 261, -90, -182, 46, 68, -11, -13, 1, 1, -1, 0, 30, -12, -229, 106, 691, -292, -1010, 359, 817, -229, -387, 79, 107, -14, -16, 1, 1
OFFSET
0,8
COMMENTS
The entry a(n,k), n >= 0, k = 0,1,...,g, where g=3n, of this irregular triangle is the coefficient of (1+q^2)^k*q^(g-k) in the representation of the Gaussian polynomial [2n+3,3]_q = Sum_{k=0..g) a(n,k)*(1+q^2)^k*q^(g-k).
Row n is of length 3n+1.
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 (A053120). 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 = 3. (Thanks to Wolfdieter Lang for clarifying the text on the general prescription of a(m;n,k).)
From Robert Israel, Jan 15 2016: (Start)
a(n,0) = A056594(n).
a(n,1) = (-1)^((n+1)/2) * A142150(n+1).
a(2n,2) = 5*(-1)^(n+1)*A000217(n), a(2n+1,2) =(-1)^n*(n+1).
It appears that Sum_{j=0..k+1} C(k+1,j)*a(n+2*j,k) = 0.
(End)
LINKS
S. O'Sullivan, A class of high-order Runge-Kutta-Chebyshev stability polynomials, Journal of Computational Physics, 300 (2015), 665-678.
FORMULA
G.f. for row polynomial: G(n,x) = (d^3/dt^3)((1+t)*Product_{i=1..n+1}(1+t^2+2t*T(i,x/2))/3!)|_{t=0}.
EXAMPLE
The irregular triangle a(n, k) begins:
n/k 0 1 2 3 4 5 6 7 8 9 10 11 12
0: 1
1: 0 -1 1 1
2: -1 0 5 -2 -4 1 1
3: 0 2 -2 -15 7 17 -5 -7 1 1
4: 1 0 -15 6 53 -23 -67 22 38 -8 -10 1 1
...
Row n=5: 0 -3 3 55 -28 -189 81 261 -90 -182 46 68 -11 -13 1 1;
Row n=6: -1 0 30 -12 -229 106 691 -292 -1010 359 817 -229 -387 79 107 -14 -16 1 1.
Row n=7: 0 4 -4 -134 70 896 -416 -2561 1073 3903 -1415 -3529 1057 1991 -467 -709 121 155 -17 -19 1 1.
... Reformatted and extended. - Wolfdieter Lang, Feb 13 2016
MAPLE
A267120 := 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+1), t$3)/3!)): if k = 0 then subs(x = 0, y) else subs(x = 0, diff(y, x$k)/k!) end if: end proc: seq(seq(A267120(n, k), k = 0 .. 3*n), n = 0 .. 20);
# More efficient:
N:= 20: # to get rows 0 to N
P[0]:= (1+t)*(t^2 + t*x + 1):
B[0]:= 1:
for n from 1 to N do
P[n]:= expand(series(P[n-1]*(1+t^2+2*t*orthopoly[T](n+1, x/2)), t, 4));
B[n]:= coeff(P[n], t, 3);
od:
seq(seq(coeff(B[n], x, j), j=0..3*n), n=0..N); # Robert Israel, Jan 15 2016
MATHEMATICA
row[n_] := 1/3! D[(1+t)*Product[1+t^2+2*t*ChebyshevT[i, x/2], {i, 1, n+1}], {t, 3}] /. t -> 0 // CoefficientList[#, x]&; Table[row[n], {n, 0, 6}] // Flatten (* Jean-François Alcover, Jan 16 2016 *)
KEYWORD
sign,tabf
AUTHOR
Stephen O'Sullivan, Jan 10 2016
STATUS
approved