A228565 -id:A228565 - cp's OEIS frontend

A244419 Coefficient triangle of polynomials related to the Dirichlet kernel. Rising powers. Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)).

1, 1, 2, -1, 2, 4, -1, -4, 4, 8, 1, -4, -12, 8, 16, 1, 6, -12, -32, 16, 32, -1, 6, 24, -32, -80, 32, 64, -1, -8, 24, 80, -80, -192, 64, 128, 1, -8, -40, 80, 240, -192, -448, 128, 256, 1, 10, -40, -160, 240, 672, -448, -1024, 256, 512, -1, 10, 60, -160, -560, 672, 1792, -1024, -2304, 512, 1024

Offset: 0

Wolfdieter Lang, Jul 29 2014

This is the row reversed version of A180870. See also A157751 and A228565.
The Dirichlet kernel is D(n,x) = Sum_{k=-n..n} exp(i*k*x) = 1 + 2*Sum_{k=1..n} T(n,x) = S(n, 2*y) + S(n-1, 2*y) = S(2*n, sqrt(2*(1+y))) with y = cos(x), n >= 0, with the Chebyshev polynomials T (A053120) and S (A049310). This triangle T(n, k) gives in row n the coefficients of the polynomial Dir(n,y) = D(n,x=arccos(y)) = Sum_{m=0..n} T(n,m)*y^m. See A180870, especially the Peter Bala comments and formulas.
This is the Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)) due to the o.g.f. for Dir(n,y) given by (1+z)/(1 - 2*y*z + z^2) = G(z)/(1 - y*F(z)) with G(z) = (1+z)/(1+z^2) and F(z) = 2*z/(1+z^2) (see the Peter Bala formula under A180870). For Riordan triangles and references see the W. Lang link 'Sheffer a- and z- sequences' under A006232.
The A- and Z- sequences of this Riordan triangle are (see the mentioned W. Lang link in the preceding comment also for the references): The A-sequence has o.g.f. 1+sqrt(1-x^2) and is given by A(2*k+1) = 0 and A(2*k) [2, -1/2, -1/8, -1/16, -5/128, -7/256, -21/1024, -33/2048, -429/32768, -715/65536, ...], k >= 0. (See A098597 and A046161.)
  The Z-sequence has o.g.f. sqrt((1-x)/(1+x)) and is given by
  [1, -1, 1/2, -1/2, 3/8, -3/8, 5/16, -5/16, 35/128, -35/128, ...]. (See A001790 and A046161.)
The column sequences are A057077, 2*(A004526 with even numbers signed), 4*A008805 (signed), 8*A058187 (signed), 16*A189976 (signed), 32*A189980 (signed) for m = 0, 1, ..., 5.
The row sums give A005408 (from the o.g.f. due to the Riordan property), and the alternating row sums give A033999.
The row polynomials Dir(n, x), n >= 0, give solutions to the diophantine equation (a + 1)*X^2 - (a - 1)*Y^2 = 2 by virtue of the identity (a + 1)*Dir(n, -a)^2 - (a - 1)*Dir(n, a)^2 = 2, which is easily proved inductively using the recurrence Dir(n, a) = (1 + a)*(-1)^(n-1)*Dir(n-1, -a) + a*Dir(n-1, a) given below by Wolfdieter Lang. - Peter Bala, May 08 2025

			The triangle T(n,m) begins:
  n\m  0   1   2    3    4    5    6     7     8    9    10 ...
  0:   1
  1:   1   2
  2:  -1   2   4
  3:  -1  -4   4    8
  4:   1  -4 -12    8   16
  5:   1   6 -12  -32   16   32
  6:  -1   6  24  -32  -80   32   64
  7:  -1  -8  24   80  -80 -192   64   128
  8:   1  -8 -40   80  240 -192 -448   128   256
  9:   1  10 -40 -160  240  672 -448 -1024   256  512
  10: -1  10  60 -160 -560  672 1792 -1024 -2304  512  1024
  ...
Example for A-sequence recurrence: T(3,1) = Sum_{j=0..2} A(j)*T(2,j) = 2*(-1) + 0*2 + (-1/2)*4 = -4. Example for Z-sequence recurrence: T(4,0) = Sum_{j=0..3} Z(j)*T(3,j) = 1*(-1) + (-1)*(-4) + (1/2)*4 + (-1/2)*8 = +1. (For the A- and Z-sequences see a comment above.)
Example for the alternate recurrence: T(4,2) = 2*T(3,1) - T(3,2) = 2*(-4) - 4 = -12. T(4,3) = 0*T(3,2) + T(3,3) = T(3,3) = 8. - _Wolfdieter Lang_, Jul 30 2014

Wikipedia, Dirichlet kernel.

Dir(n, x) : A005408 (x = 1), A002878 (x = 3/2), A001834 (x = 2), A030221 (x = 5/2), A002315 (x = 3), A033890 (x = 7/2), A057080 (x = 4), A057081 (x = 9/2), A054320 (x = 5), A077416 (x = 6), A028230 (x = 7), A159678 (x = 8), A049629 (x = 9), A083043 (x = 10),
(-1)^n * Dir(n, x): A122367 (x = -3/2); A079935 (x = -2), A004253 (x = -5/2), A001653 (x = -3), A049685 (x = -7/2), A070997 (x = -4), A070998 (x = -9/2), A072256(n+1) (x = -5).
Cf. A180870 (row reversed), A157751, A228565, A049310, A053120, A057077, A004526, A008805, A058187, A189976, A189980, A005408, A033999.

T[n_, k_] := T[n, k] = Which[k == 0, (-1)^Quotient[n, 2], (0 <= n && n < k) || (n == -1 && k == 1), 0, True, 2 T[n-1, k-1] - T[n-2, k]];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 28 2019, from Sage *)

def T(n, k):
    if k == 0: return (-1)^(n//2)
    if (0 <= n and n < k) or (n == -1 and k == 1): return 0
    return 2*T(n-1, k-1) - T(n-2, k)
for n in range(11): [T(n,k) for k in (0..n)] # Peter Luschny, Jul 29 2014

T(n, m) = [y^m] Dir(n,y) for n >= m >= 0 and 0 otherwise, with the polynomials Dir(y) defined in a comment above.
T(n, m) = 2^m*(S(n,m) + S(n-1,m)) with the entries S(n,m) of A049310 given there explicitly.
O.g.f. for polynomials Dir(y) see a comment above (Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2))).
O.g.f. for column m: ((1 + x)/(1 + x^2))*(2*x/(1 + x^2))^m, m >= 0, (Riordan property).
Recurrence for the polynomials: Dir(n, y) = 2*y*Dir(n-1, y) - Dir(n-2, y), n >= 1, with input D(-1, y) = -1 and D(0, y) = 1.
Triangle three-term recurrence: T(n,m) = 2*T(n-1,m-1) - T(n-2,m) for n >= m >= 1 with T(n,m) = 0 if 0 <= n < m, T(0,0) = 1, T(-1,1) = 0 and T(n,0) = A057077(n) = (-1)^(floor(n/2)).
From Wolfdieter Lang, Jul 30 2014: (Start)
In analogy to A157751 one can derive a recurrence for the row polynomials Dir(n, y) = Sum_{m=0..n} T(n,m)*y^m also using a negative argument but only one recursive step: Dir(n,y) = (1+y)*(-1)^(n-1)*Dir(n-1,-y) + y*Dir(n-1,y), n >= 1, Dir(0,y) = 1 (Dir(-1,y) = -1). See also A180870 from where this formula can be derived by row reversion.
This entails another triangle recurrence T(n,m) = (1 + (-1)^(n-m))*T(n-1,m-1) - (-1)^(n-m)*T(n-1,m), for n >= m >= 1 with T(n,m) = 0 if n < m and T(n,0) = (-1)^floor(n/2). (End)
From Peter Bala, Aug 14 2022: (Start)
The row polynomials Dir(n,x), n >= 0, are related to the Chebyshev polynomials of the first kind T(n,x) by the binomial transform as follows:
(2^n)*(x - 1)^(n+1)*Dir(n,x) = (-1) * Sum_{k = 0..2*n+1} binomial(2*n+1,k)*T(k,-x).
Note that Sum_{k = 0..2*n} binomial(2*n,k)*T(k,x) = (2^n)*(1 + x)^n*T(n,x). (End)
From Peter Bala, May 04 2025: (Start)
For n >= 1, the n-th row polynomial Dir(n, x) = (-1)^n * (U(n, -x) - U(n-1, -x)) = U(2*n, sqrt((1+x)/2)), where U(n, x) denotes the n-th Chebyshev polynomial of the second kind.
For n >= 1 and x < 1, Dir(n, x) = (-1)^n * sqrt(2/(1 - x )) * T(2*n+1, sqrt((1 - x)/2)), where T(n, x) denotes the n-th Chebyshev polynomial of the first kind.
Dir(n, x)^2 - 2*x*Dir(n, x)*Dir(n+1, x) + Dir(n+1, x)^2 = 2*(1 + x).
Dir(n, x) = (-1)^n * R(n, -2*(x+1)), where R(n, x) is the n-th row polynomial of the triangle A085478.
Dir(n, x) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n+k, 2*k) * (2*x + 2)^k. (End)

A157751 Triangle of coefficients of polynomials F(n,x) in descending powers of x generated by F(n,x)=(x+1)*F(n-1,x)+F(n-1,-x), with initial F(0,x)=1.

Original entry on oeis.org

1, 1, 2, 1, 2, 4, 1, 4, 4, 8, 1, 4, 12, 8, 16, 1, 6, 12, 32, 16, 32, 1, 6, 24, 32, 80, 32, 64, 1, 8, 24, 80, 80, 192, 64, 128, 1, 8, 40, 80, 240, 192, 448, 128, 256, 1, 10, 40, 160, 240, 672, 448, 1024, 256, 512, 1, 10, 60, 160, 560, 672, 1792, 1024, 2304, 512, 1024, 1, 12, 60, 280, 560, 1792, 1792, 4608, 2304, 5120, 1024, 2048

Offset: 0

Clark Kimberling, Mar 05 2009

Conjecture 1. If n>1 is even then F(n,x) has no real roots.
Conjecture 2. If n>0 is odd then F(n,x) has exactly one real root, r,
and if n>4 then 0 < -r < n.
Conjectures 1 and 2 are true. [From Alain Thiery (Alain.Thiery(AT)math.u-bordeaux1.fr), May 14 2010]
Cayley (1876) states "We, in fact, find  1 + sin u = 1 + x,  1 - sin 3u = (1 + x)(1 - 2x)^2,  1 + sin 5u = (1 + x)(1 + 2x - 4x^2)^2,  1 - sin 7u = (1 + x)(1 - 4x - 4x^2 + 8x^3)^2, &c.". - Michael Somos, Jun 19 2012
Appears to be the unsigned row reverse of A180870 and A228565. - Peter Bala, Feb 17 2014
From Wolfdieter Lang, Jul 29 2014: (Start)
This triangle is the Riordan triangle ((1+z)/(1-z^2), 2*z/(1-z^2)). For Riordan triangles see the W. Lang link 'Sheffer a-and z-sequences' under A006232, also for references. The o.g.f. given by Peter Bala in the formula section refers to the row reversed triangle. The usual information on this triangle, like o.g.f. for the columns, the row sums, the alternating row sums, the recurrences using A- and Z-sequences, etc. follows from this Riordan property. The Riordan proof follows from the given o.g.f. by Peter Bala, call it Grev(x,z), by row reversion: G(x,z) = Grev(1/x,x*z) = (1+z)/(1- 2*x*z - z^2) = G(z)*(1/(1 - x*F(z))) with G(z) = (1+z)/(1-z^2) and F(z) = z*2/(1-z^2). See A244419 for the discussion for a signed version of this triangle.
(End)

			Rows 0 to 8:
1
1 2
1 2 4
1 4 4 8
1 4 12 8 16
1 6 12 32 16 32
1 6 24 32 80 32 64
1 8 24 80 80 192 64 128
1 8 40 80 240 192 448 128 256
(Row 8) = (1, 4*2, 10*4, 10*8, 15*16, 6*32, 7*64, 1*128, 1*256).
First few polynomials:
F(0,x)=1, F(1,x)=x+2, F(2,x)=x^2+2*x+4, F(3,x)=x^3+4*x^2+4*x+8.
The row polynomials R(n,x) start: 1, 1 + 2*x = x*F(1,1/x), 1 + 2*x + 4^x^2 = x^2*F(2,1/x), ...  - _Wolfdieter Lang_, Jul 29 2014

A. Cayley, On an Expression for 1 +- sin(2p+1)u in Terms of sin u, Messenger of Mathematics, 5 (1876), pp. 7-8 = Mathematical Papers Vol. 10, n. 630, pp. 1-2.

G. C. Greubel, Rows n=0..100 of triangle, flattened

Cf. A140070, A140071, A180870, A228565, A078057, A123335, A000012, 2*A004526, 4*A008805, 8*A058187, 16*A189976, 32*A189980.

T[n_, 0]:= 1; T[n_, n_]:= 2^n; T[n_, k_]:= T[n, k] = T[n-1, k] + (1 + (-1)^(n-k))*T[n-1, k-1]; Table[T[n, k], {n, 0, 15}, {k, 0, n}] (* G. C. Greubel, Sep 24 2018 *)

t(n,k) = if(k==0, 1, if(k==n, 2^n, t(n-1,k) + (1+(-1)^(n-k))*t(n-1,k-1)));
for(n=0,15, for(k=0,n, print1(t(n,k), ", "))) \\ G. C. Greubel, Sep 24 2018

Count the top row as row 0 and let C(n,k) denote the usual binomial
coefficient. For row 2n, define p(0)=C(n,0), p(1)=C(n,1), p(2)=C(n+1,2),
p(3)=C(n+1,3), p(4)=C(n+2,4), p(5)=c(n+2,5),..., until reaching two final
1's: p(2n-1)=C(2n-1,2n-1) and p(2n)=C(2n,2n). Then the k-th number in row
2n is p(k)*2^k. For row 2n+1, define q(0)=C(n,0), q(1)=C(n+1,1),
q(2)=C(n+1,2), q(3)=C(n+2,3),..., until reaching q(2n+1)=C(2n+1,2n+1).
Then the k-th number in row 2n+1 is q(k)*2^k.
From Peter Bala, Jan 17 2014: (Start)
Working with an offset of 0, the o.g.f. is (1 + x*z)/(1 - 2*z - x^2*z^2) = 1 + (x + 2)*z + (x^2 + 2*x + 4)*z^2 + ....
Recurrence equation: T(n,k) = 2*T(n-1,k-1) + T(n-2,k-2) with T(n,0) = 1.
The polynomials G(n,x) defined by G(0,x) = 1 and G(n,x) = x*F(n-1,x) for n >= 1 satisfy G(n,x) = (x + 1)*G(n-1,x) - G(n-1,-x). Cf. A140070 and A140071. (End)
From Wolfdieter Lang, Jul 29 2014: (Start)
O.g.f. for the row polynomials (rising powers of x) R(n,x) = x^n*F(n,1/x): (1+z)/(1 - 2*x*z - z^2). Riordan triangle ((1+z)/(1-z^2), 2*z/(1-z^2)). See a comment above.
Recurrence for the row polynomials R(n,x) = (1+x)*R(n-1,x) - (-1)^n*x*R(n-1,-x), n >= 1, R(0,x) = 1.
R(n,x) = Ftilde(n,2*x) + Ftilde(n-1,2*x) with the monic Fibonacci polynomials Ftilde(n,x) given in A168561.
Recurrence for the triangle: R(n,m) = R(n-1,m) + (1 + (-1)^(n-m))*R(n-1,m-1), n >= m >= 1, R(n,m) = 0 if n < m, R(n,0) = 1.
O.g.f. column sequences ((1+x)/(1-x^2))*(2*x/(1-x^2))^m, m >= 0. See A000012, 2*A004526, 4*A008805, 8*A058187, 16*A189976, 32*A189980, ...
Row sums A078057. Alternating row sums A123335.
(End)

Offset corrected to 0. Cf.s added, keyword easy added by Wolfdieter Lang, Jul 29 2014

A180870 D(n, x) is the Dirichlet kernel sin((n+1/2)x)/sin(x/2). The triangle gives in row n the coefficients of descending powers of x of the polynomial D(n, arccos(x)).

Original entry on oeis.org

1, 2, 1, 4, 2, -1, 8, 4, -4, -1, 16, 8, -12, -4, 1, 32, 16, -32, -12, 6, 1, 64, 32, -80, -32, 24, 6, -1, 128, 64, -192, -80, 80, 24, -8, -1, 256, 128, -448, -192, 240, 80, -40, -8, 1, 512, 256, -1024, -448, 672, 240, -160, -40, 10, 1

Offset: 0

Jonny Griffiths, Sep 21 2010

D(n, arccos(x)) = U(n, x) + U(n-1, x) where U(n, x) are the Chebyshev polynomials of the second kind. These polynomials arise naturally in the investigation of the integer triples (p, q, (p*q + 1)/(p + q)).

Chebyshev polynomials of the fourth kind, usually denoted by W(n, x) (see, for example, Mason and Handscomb, Chapter 1, Definition 1.3). See A228565 for Chebyshev polynomials of the third kind. Cf. A157751. - Peter Bala, Jan 17 2014

			The triangle T(n,m) begins:
n\m    0   1     2     3    4   5    6    7   8  9  10 ...
0:     1
1:     2   1
2:     4   2    -1
3:     8   4    -4    -1
4:    16   8   -12    -4    1
5:    32  16   -32   -12    6   1
6:    64  32   -80   -32   24   6   -1
7:   128  64  -192   -80   80  24   -8   -1
8:   256 128  -448  -192  240  80  -40   -8   1
9:   512 256 -1024  -448  672 240 -160  -40  10  1
10: 1024 512 -2304 -1024 1792 672 -560 -160  60 10  -1
... reformatted - _Wolfdieter Lang_, Jul 26 2014
Recurrence: T(4,2) = (1 + 1)*T(3,2) - T(3,1) = 2*(-4) - 4 = -12. T(4,3) = 0*T(3,3) - (-1)*T(3,2) = T(3,2) = -4. - _Wolfdieter Lang_, Jul 30 2014

J. C. Mason and D. C. Handscomb, Chebyshev polynomials, Chapman and Hall/CRC 2002.

Paul Barry, On the Group of Almost-Riordan Arrays, arXiv preprint arXiv:1606.05077 [math.CO], 2016.
Wikipedia, Dirichlet kernel.

Cf. A008312, A028297, A157751, A228565, A049310, A244419 (row reversed triangle).

ogf := (1 + t)/(1 - 2*x*t + t^2):
ser := simplify(series(ogf, t, 12)): tc := n -> coeff(ser, t, n):
Trow := n -> local k; seq(coeff(tc(n), x, n-k), k = 0..n):
seq(print(Trow(n)), n = 0..9);  # Peter Luschny, Oct 07 2024

row(n) = {if (n==0, return([1])); f = 2*x+1; for (k = 2, n, for (i = 1, (k-1)\2 + 1, f += (-1)^(i+1)*(binomial(k-i, i-1)*(2*x)^(k-2*i+2) - 2*binomial(k-1-i, i-1)*(2*x)^(k-2*i)););); Vec(f);} \\ Michel Marcus, Jul 18 2014

From Peter Bala, Jan 17 2014: (Start)
O.g.f. (1 + t)/(1 - 2*x*t + t^2) = 1 + (2*x + 1)*t + (4*x^2 + 2*x - 1)*t^2 + ....
Recurrence equation: W(0,x) = 1, W(1,x) = 2*x + 1 and W(n,x) = 2*x*W(n-1,x) - W(n-2,x) for n >= 2.
In terms of U(n,x), the Chebyshev polynomials of the second kind, we have W(n,x) = U(2*n,u) with u = sqrt((1 + x)/2). Also binomial(2*n,n)*W(n,x) = 2^(2*n)*Jacobi_P(n,1/2,-1/2,x). (End)
Row sums: 2*n+1. - Michel Marcus, Jul 16 2014
T(n,m) = [x^(n-m)](U(n, x) + U(n-1, x)) = [x^(n-m)] S(2*n, sqrt(2*(1+x))), n >= m >= 0, with U(n, x) = S(n, 2*x). The coefficient triangle of the Chebyshev S-polynomials is given in A049310. See the Peter Bala comments above. - Wolfdieter Lang, Jul 26 2014
From Wolfdieter Lang, Jul 30 2014: (Start)
O.g.f. for the row polynomials R(n,x) = Sum_{m=0..n} T(n,m)*x^m, obtained from the one given by Peter Bala above by row reversion: (1 + x*t)/(1 - 2*t + (x*t)^2).
In analogy to A157751 one can derive a recurrence for the row polynomials R(n, x) = x^n*Dir(n,1/x) with Dir(n,x) = U(n,x) + U(n-1,x) using also negative arguments but only one recursive step: R(n,x) = (1+x)*R(n-1,-x) + R(n-1,x), n >= 1, R(0,x) = 1 (R(-1,x) = -1/x). Proof: derive the o.g.f. and compare it with the known one.
This entails the triangle recurrence T(n,m) = (1 + (-1)^m)* T(n-1,m) - (-1)^m*T(n-1,m-1), for n >= m >= 1 with T(n,m) = 0 if n < m and T(n,0) = 2^n. (End)

Missing term in sequence corrected by Paul Curtz, Dec 31 2011

Edited (name reformulated, Wikipedia link added) by Wolfdieter Lang, Jul 26 2014

A228637 The number triangle associated with the polynomials V_n(x).

Original entry on oeis.org

1, -1, 1, -1, 1, 1, 1, 1, 3, 1, 1, 1, 11, 5, 1, -1, 1, 41, 29, 7, 1, -1, 1, 153, 169, 55, 9, 1, 1, 1, 571, 985, 433, 89, 11, 1, 1, 1, 2131, 5741, 3409, 881, 131, 13, 1, -1, 1, 7953, 33461, 26839, 8721, 1561, 181, 15, 1

Offset: 0

Jonny Griffiths, Aug 28 2013

V(n) is the polynomial with integer coefficients in x given by cos((2n+1)(arccos(x)/2))/(arccos(x)/2). The triangle here is given by V_0(0), V_1(0), V_0(1), V_2(0), V_1(1), V_0(2), V_3(0), V_2(1), V_1(2), V_0(3), V_4(0),....

			V_0(x)=1, V_1(x)=2x-1, V_2(x)=4x^2-2x-1,  ...

Cf. A028297, A101124, A133156, A228161, A228565, A180870.

The terms are given by the recurrence relation V_{n+1}(x) = 2xV_n(x)-V_{n-1}(x), V_0(x) = 1, V_1(x)=2x-1.

A228356 The triangle associated with the family of polynomials W_n(x).

Original entry on oeis.org

1, 1, 1, -1, 3, 1, -1, 5, 5, 1, 1, 7, 19, 7, 1, 1, 9, 71, 41, 9, 1, -1, 11, 265, 239, 71, 11, 1, -1, 13, 989, 1393, 559, 109, 13, 1, 1, 15, 3691, 8119, 4401, 1079, 155, 15, 1, 1, 17, 13775, 47321, 34649, 10681, 1847, 209, 17, 1

Offset: 0

Jonny Griffiths, Aug 28 2013

W_n(x) is the family of polynomials in x with integer coefficients given by W_n(x) = sin((2n+1)arccos(x)/2)/(sin(arccos(x)/2)).

These polynomials are intimately linked with the Chebyshev polynomials of the first and second kinds, and represent the polynomials associated with the Dirichlet kernel.

			The triangle is given here as W_0(0)=1, W_1(0)=1, W_0(1)=1, W_2(0)=-1, W_1(1)=3, W_0(2)=1, W_3(0)=-1, W_2(1)=5 ...

Cf. A028297, A101124, A133156, A228161, A228565, A228637, A180870.

W[0, ] = 1; W[1, x] := 2 x + 1; W[n_, x_] := W[n, x] = 2 x W[n - 1, x] - W[n - 2, x]; Table[W[n - x, x] , {n, 0, 9}, {x, 0, n}] // Flatten (* Jean-François Alcover, Jun 11 2017 *)

W_{n+1} = 2xW_n(x) - W_{n-1}, W_0(x)=1, W_1(x)=2x+1.

cp's OEIS Frontend

A244419 Coefficient triangle of polynomials related to the Dirichlet kernel. Rising powers. Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)).

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A157751 Triangle of coefficients of polynomials F(n,x) in descending powers of x generated by F(n,x)=(x+1)*F(n-1,x)+F(n-1,-x), with initial F(0,x)=1.

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A180870 D(n, x) is the Dirichlet kernel sin((n+1/2)x)/sin(x/2). The triangle gives in row n the coefficients of descending powers of x of the polynomial D(n, arccos(x)).

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Maple

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A228637 The number triangle associated with the polynomials V_n(x).

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A228356 The triangle associated with the family of polynomials W_n(x).

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