cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

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
Offset: 0

Views

Author

Wolfdieter Lang, Jul 29 2014

Keywords

Comments

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

Examples

			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
		

Crossrefs

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).

Programs

  • Mathematica
    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 *)
  • 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

Formula

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

Views

Author

Clark Kimberling, Mar 05 2009

Keywords

Comments

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)

Examples

			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
		

References

  • 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.

Crossrefs

Programs

  • Mathematica
    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 *)
  • PARI
    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

Formula

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)

Extensions

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

A228565 Triangle read by rows: coefficients of descending powers of the polynomial V(n,x) = cos((2n+1)(arccos(x)/2))/cos(arccos(x)/2), n >= 0.

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, 1024, -512, -2304, 1024, 1792, -672, -560, 160, 60, -10, -1, 2048, -1024, -5120, 2304, 4608, -1792, -1792, 560, 280, -60, -12, 1, 4096, -2048, -11264, 5120, 11520, -4608, -5376, 1792, 1120, -280, -84, 12, 1
Offset: 0

Views

Author

Jonny Griffiths, Aug 25 2013

Keywords

Comments

V(n,x) is related to the Dirichlet kernel and its associated polynomials. V(n,x) arises in studying recurrences connecting the Chebyshev polynomials of the first and second kinds. It differs from A180870 above only in the signs of terms.
Chebyshev polynomials V(n,x) of the third kind (see, for example, Mason and Handscomb, Chapter 1, Definition 1.3). See A180870 for Chebyshev polynomials of the fourth kind. Cf. A155751. - Peter Bala, Jan 17 2014

Examples

			V(0,x) = 1, V(1,x) = 2x-1, V(2,x) = 4x^2-2x-1, V(3,x) = 8x^3 -4x^2 - 4x + 1, V(4,x) = 16x^4 - 8x^3 - 12x^2 + 4x + 1, V(5,x) = 32x^5 - 16x^4 - 32x^3 + 12x^2 + 6x - 1, V(6,x) =64x^6 - 32x^5 - 80x^4 + 32x^3 + 24x^2 - 6x - 1, ...
Triangle begins:
     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;
  1024, -512, -2304, 1024, 1792, -672, -560, 160,  60, -10,  -1;
  ...
		

References

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

Crossrefs

Programs

  • Maple
    A228565 := proc(n,k)
        local t,Vn,x ;
        t := arccos(x) ;
        Vn := cos((n+1/2)*t)/cos(t/2) ;
        coeftayl(%,x=0,n-k) ;
    end proc:
    for n from 0 to 10 do
        for k from 0 to n do
            printf("%d,",A228565(n,k)) ;
        end do:
        printf("\n") ;
    end do: # R. J. Mathar, Mar 12 2014
  • Mathematica
    V[n_] := Cos[(2*n + 1)*(ArcCos[x]/2)]/Cos[ArcCos[x]/2];
    row[n_] := CoefficientList[V[n] + O[x]^(n + 1), x] // Reverse;
    Table[row[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Nov 20 2017 *)

Formula

V(n+1,x) = 2xV(n,x) - V(n-1,x) with V(0,x) = 1, V(1,x) = 2x-1.
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 + ....
In terms of the Chebyshev polynomials T(n,x) of the first kind and Chebyshev polynomials U(n,x) of the second kind we have
V(n,x) = U(n,x) - U(n-1,x);
V(n,x) + V(n-1,x) = 2*T(n,x);
V(n,x) = 1/u*T(2*n+1,u) with u = sqrt((1 + x)/2).
Also binomial(2*n,n)*V(n,x) = 2^(2*n)*Jacobi_P(n,-1/2,1/2,x). (End)

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

Views

Author

Jonny Griffiths, Aug 28 2013

Keywords

Comments

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),....

Examples

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

Crossrefs

Formula

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

Views

Author

Jonny Griffiths, Aug 28 2013

Keywords

Comments

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.

Examples

			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 ...
		

Crossrefs

Programs

  • Mathematica
    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 *)

Formula

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

A376498 Array read by ascending antidiagonals: A(n, k) = 2^k*Sum_{j=1..n} cos((2*j - 1)*Pi/(2*n + 1))^k.

Original entry on oeis.org

0, 1, 0, 2, 1, 0, 3, 1, 1, 0, 4, 1, 3, 1, 0, 5, 1, 5, 4, 1, 0, 6, 1, 7, 4, 7, 1, 0, 7, 1, 9, 4, 13, 11, 1, 0, 8, 1, 11, 4, 19, 16, 18, 1, 0, 9, 1, 13, 4, 25, 16, 38, 29, 1, 0, 10, 1, 15, 4, 31, 16, 58, 57, 47, 1, 0, 11, 1, 17, 4, 37, 16, 78, 64, 117, 76, 1, 0
Offset: 0

Views

Author

Cheng-Jun Li, Sep 25 2024

Keywords

Comments

It is only a conjecture that the A(n, k) are always integers.

Examples

			Array starts:
[0] 0, 0,  0, 0,  0,  0,   0,  0,   0,   0,    0,    0, ...  [A000004]
[1] 1, 1,  1, 1,  1,  1,   1,  1,   1,   1,    1,    1, ...  [A000012]
[2] 2, 1,  3, 4,  7, 11,  18, 29,  47,  76,  123,  199, ...  [A000032]
[3] 3, 1,  5, 4, 13, 16,  38, 57, 117, 193,  370,  639, ...  [A096975]
[4] 4, 1,  7, 4, 19, 16,  58, 64, 187, 247,  622,  925, ...  [A094649]
[5] 5, 1,  9, 4, 25, 16,  78, 64, 257, 256,  874, 1013, ...  [A189234]
[6] 6, 1, 11, 4, 31, 16,  98, 64, 327, 256, 1126, 1024, ...  [A216605]
[7] 7, 1, 13, 4, 37, 16, 118, 64, 397, 256, 1378, 1024, ...
[8] 8, 1, 15, 4, 43, 16, 138, 64, 467, 256, 1630, 1024, ...
[9] 9, 1, 17, 4, 49, 16, 158, 64, 537, 256, 1882, 1024, ...
		

Crossrefs

Rows: A000004 (n=0), A000012 (n=1), A000032 (n=2), A096975 (n=3), A094649 (n=4), A189234 (n=5), A216605 (n=6, with alternate signs).
Columns: A001477 (k=0), A057427 (k=1).
Cf. A180870.

Programs

  • PARI
    A(n, k) = 2^k*sum(j=1, n, (cos((2*j-1)*Pi/(2*n+1)))^k, x=0)

Formula

A(n + k, 2*k - 1) = A(k, 2*k-1) = 4^(k-1).
Let P_n(x) be the polynomial: Sum_{k=0..n} x^k*A180870(n, k). Let R_n(x) be the polynomial Product_{k=0..n} x-Roots(P_n, k)^m. A(n, k) = abs([x^1] R_n(x))/2^(m*(n-1)), for n > 0. - Thomas Scheuerle, Oct 07 2024
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