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.

A119468 Triangle read by rows: T(n,k) = Sum_{j=0..n-k} binomial(n,2j)*binomial(n-2j,k).

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 4, 6, 3, 1, 8, 16, 12, 4, 1, 16, 40, 40, 20, 5, 1, 32, 96, 120, 80, 30, 6, 1, 64, 224, 336, 280, 140, 42, 7, 1, 128, 512, 896, 896, 560, 224, 56, 8, 1, 256, 1152, 2304, 2688, 2016, 1008, 336, 72, 9, 1, 512, 2560, 5760, 7680, 6720, 4032, 1680, 480, 90, 10, 1
Offset: 0

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Author

Paul Barry, May 21 2006

Keywords

Comments

Product of Pascal's triangle A007318 and A119467. Row sums are A007051. Diagonal sums are A113225.
Variant of A080928, A115068 and A082137. - R. J. Mathar, Feb 09 2010
Matrix inverse of the Euler tangent triangle A081733. - Peter Luschny, Jul 18 2012
Central column: T(2*n,n) = A069723(n). - Peter Luschny, Jul 22 2012
Subtriangle of the triangle in A198792. - Philippe Deléham, Nov 10 2013

Examples

			Triangle begins
    1;
    1,    1;
    2,    2,    1;
    4,    6,    3,    1;
    8,   16,   12,    4,    1;
   16,   40,   40,   20,    5,    1;
   32,   96,  120,   80,   30,    6,    1;
   64,  224,  336,  280,  140,   42,    7,   1;
  128,  512,  896,  896,  560,  224,   56,   8,  1;
  256, 1152, 2304, 2688, 2016, 1008,  336,  72,  9,  1;
  512, 2560, 5760, 7680, 6720, 4032, 1680, 480, 90, 10, 1;
		

Crossrefs

A082137 read as triangle with rows reversed.

Programs

  • Maple
    A119468_row := proc(n) local s,t,k;
      s := series(exp(z*x)/(1-tanh(x)),x,n+2);
      t := factorial(n)*coeff(s,x,n); seq(coeff(t,z,k), k=(0..n)) end:
    for n from 0 to 7 do A119468_row(n) od; # Peter Luschny, Aug 01 2012
    # Alternatively:
    T := (n, k) -> 2^(n-k-1+0^(n-k))*binomial(n,k):
    for n from 0 to 9 do seq(T(n,k), k=0..n) od; # Peter Luschny, Nov 10 2017
  • Mathematica
    A[k_] := Table[If[m < n, 1, -1], {m, k}, {n, k}]; a = Join[{{1}}, Table[(-1)^n*CoefficientList[CharacteristicPolynomial[A[n], x], x], {n, 1, 10}]]; Flatten[a] (* Roger L. Bagula and Gary W. Adamson, Jan 25 2009 *)
    Table[Sum[Binomial[n,2j]Binomial[n-2j,k],{j,0,n-k}],{n,0,10},{k,0,n}]//Flatten (* Harvey P. Dale, Dec 14 2022 *)
  • Sage
    R = PolynomialRing(QQ, 'x')
    def p(n,x) :
      return 1 if n==0 else add((-1)^n*binomial(n,k)*(x^(n-k)-1) for k in range(n))
    def A119468_row(n):
        x = R.gen()
        return [abs(cf) for cf in list((p(n,x-1)-p(n,x+1))/2+x^n)]
    for n in (0..8) : print(A119468_row(n)) # Peter Luschny, Jul 22 2012

Formula

G.f.: (1 - x - xy)/(1 - 2x - 2x*y + 2x^2*y + x^2*y^2).
Number triangle T(n,k) = Sum_{j=0..n} binomial(n,j)*binomial(j,k)*(1+(-1)^(j-k))/2.
Define matrix: A(n,m,k) = If[m < n, 1, -1];
p(x,k) = CharacteristicPolynomial[A[n,m,k],x]; then t(n,m) = coefficients(p(x,n)). - Roger L. Bagula and Gary W. Adamson, Jan 25 2009
E.g.f.: exp(x*z)/(1-tanh(x)). - Peter Luschny, Aug 01 2012
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - 2*T(n-2,k-1) - T(n-2,k-2) for n >= 2, T(0,0) = T(1,0) = T(1,1) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Nov 10 2013
E.g.f.: [(e^(2t)+1)/2] e^(tx) = e^(P.(x)t), so this is an Appell sequence with lowering operator D = d/dx and raising operator R = x + 2/(e^(-2D)+1), i.e., D P_n(x) = n P_{n-1}(x) and R p_n(x) = P_{n+1}(x) where P_n(x) = [(x+2)^n + x^n]/2. Also, (P.(x)+y)^n = P_n(x+y), umbrally. R = x + 1 + D - 2 D^3/3! + ... contains the e.g.f.(D) mod signs of A009006 and A155585 and signed, aerated A000182, the zag numbers, so the unsigned differential component 2/[e^(2D)+1] = 2 Sum_{n >= 0} Eta(-n) (-2D)^n/n!, where Eta(s) is the Dirichlet eta function, and 2 *(-2)^n Eta(-n) = (-1)^n (2^(n+1)-4^(n+1)) Zeta(-n) = (2^(n+1)-4^(n+1)) B(n+1)/(n+1) with Zeta(s), the Riemann zeta function, and B(n), the Bernoulli numbers. The polynomials PI_n(x) of A081733 are the umbral compositional inverses of this sequence, i.e., P_n(PI.(x)) = x^n = PI_n(P.(x)) under umbral composition. Aside from the signs and the main diagonals, multiplying this triangle by 2 gives the face-vectors of the hypercubes A038207. - Tom Copeland, Sep 27 2015
T(n,k) = 2^(n-k-1+0^(n-k))*binomial(n, k). - Peter Luschny, Nov 10 2017