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.

Showing 1-2 of 2 results.

A099023 Diagonal of Euler-Seidel matrix with start sequence e.g.f. 1-tanh(x).

Original entry on oeis.org

1, -1, 4, -46, 1024, -36976, 1965664, -144361456, 13997185024, -1731678144256, 266182076161024, -49763143319190016, 11118629668610842624, -2925890822304510631936, 895658946905031792553984
Offset: 0

Views

Author

Ralf Stephan, Sep 23 2004

Keywords

Comments

T(2n,n), where T is A008280 (signed).

Crossrefs

Programs

  • Mathematica
    A099023List[n_] := Module[{e, dim, m, k}, dim = 2 n; e[0, 0] = 1; For[m = 1, m <= dim - 1, m++, If[EvenQ[m], e[m, 0] = 1; For[k = m - 1, k >= -1, k--, e[k, m - k] = e[k + 1, m - k - 1] - e[k, m - k - 1]], e[0, m] = 1; For[k = 1, k <= m + 1, k++, e[k, m - k] = e[k - 1, m - k + 1] + e[k - 1, m - k]]]]; Table[e[k, k], {k, 0, (dim + 1)/2 - 1}]];
    A099023List[15] (* Jean-François Alcover, Jun 11 2019, after Peter Luschny *)
  • Sage
    # Variant of an algorithm of L. Seidel (1877).
    def A099023_list(n) :
        dim = 2*n; E = matrix(ZZ, dim); E[0,0] = 1
        for m in (1..dim-1) :
            if m % 2 == 0 :
                E[m,0] = 1;
                for k in range(m-1,-1,-1) :
                    E[k,m-k] = E[k+1,m-k-1] - E[k,m-k-1]
            else :
                E[0,m] = 1;
                for k in range(1,m+1,1) :
                    E[k,m-k] = E[k-1,m-k+1] + E[k-1,m-k]
        return [E[k,k] for k in range((dim+1)//2)]
    # Peter Luschny, Jul 14 2012

Formula

|a(n)| = A000657(n) - Sean A. Irvine, Dec 22 2010
G.f.: 1/G(0) where G(k) = 1 + x*(k+1)*(4*k+1)/(1 + x*(k+1)*(4*k+3)/G(k+1) ) ; (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 05 2013
G.f.: G(0)/(1+x), where G(k) = 1 - x^2*(k+1)^2*(4*k+1)*(4*k+3)/( x^2*(k+1)^2*(4*k+1)*(4*k+3) - (1 + x*(8*k^2+4*k+1))*(1 + x*(8*k^2+20*k+13))/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Feb 01 2014

A009744 Expansion of e.g.f. tan(x)*sin(x) (even powers only).

Original entry on oeis.org

0, 2, 4, 62, 1384, 50522, 2702764, 199360982, 19391512144, 2404879675442, 370371188237524, 69348874393137902, 15514534163557086904, 4087072509293123892362, 1252259641403629865468284, 441543893249023104553682822, 177519391579539289436664789664
Offset: 0

Views

Author

Keywords

Crossrefs

Programs

  • Maple
    seq((2*i)!*coeff(series(tan(x)*sin(x),x,30),x,2*i),i=0..14); # Peter Luschny, Jul 14 2012
  • Mathematica
    nn = 30; t = Range[0, nn]! CoefficientList[Series[Tan[x]*Sin[x], {x, 0, nn}], x]; Take[t, {1, nn, 2}] (* T. D. Noe, Jul 15 2012 *)
  • PARI
    x='x+O('x^50); v=Vec(serlaplace(tan(x)*sin(x))); concat([0], vector(#v\2,n,v[2*n-1])) \\ G. C. Greubel, Mar 04 2018
  • Sage
    # Variant of an algorithm of L. Seidel (1877) with a(0) = 1.
    def A009744_list(n) :
        dim = 2*n; E = matrix(ZZ, dim); E[0, 0] = 1
        for m in (1..dim-1) :
            if m % 2 == 0 :
                E[m, 0] = 1;
                for k in range(m-1, -1, -1) :
                    E[k, m-k] = E[k+1, m-k-1] - E[k, m-k-1]
            else :
                E[0, m] = 1;
                for k in range(1, m+1, 1) :
                    E[k, m-k] = E[k-1, m-k+1] + E[k-1, m-k]
        return [(-1)^(k//2)*E[0,k] for k in range(dim) if is_even(k)]
    A009744_list(14)  # Peter Luschny, Jul 14 2012
    

Formula

G.f.: 1/G(0) - 1/(1+x) where G(k) = 1 - x*(2*k+1)^2/(1 - x*(2*k+2)^2/G(k+1) ); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 05 2013
G.f.: 1/G(0) - 1/(1+x) where G(k) = 1 - x*(k+1)^2/G(k+1); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 09 2013
a(n) ~ (2*n)! * 4^(n+1) / Pi^(2*n+1). - Vaclav Kotesovec, Jan 24 2015
Conjectural o.g.f.: Sum_{n >= 0} 4*x/2^n * Sum_{k = 0..n} (-1)^k*(k+1)*binomial(n,k)/( (1 + x*(2*k + 1)^2)*(1 + x*(2*k + 3)^2) ) = 2*x + 4*x^2 + 62*x^3 + 1384*x^4 + .... - Peter Bala, Mar 03 2015
From Peter Luschny, Jun 13 2021: (Start)
a(n) = (-1)^n*(Euler(2*n) - 1).
a(n) ~ 4^(2*n + 3/2)*exp(1/(24*n) - 2*n)*(n/Pi)^(2*n + 1/2). (End)

Extensions

Extended and signs tested by Olivier Gérard, Mar 15 1997
Showing 1-2 of 2 results.