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.

A097189 Row sums of triangle A097186, in which the n-th row polynomial R_n(y) is formed from the initial (n+1) terms of g.f. A057083(y)^(n+1), where R_n(1/3) = 3^n for all n >= 0.

Original entry on oeis.org

1, 7, 55, 451, 3781, 32131, 275563, 2378971, 20640907, 179791327, 1571002291, 13762897435, 120832716655, 1062818450155, 9363143224315, 82600459304203, 729572125425661, 6450872644562491, 57092964352312951, 505729048454449651
Offset: 0

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Author

Paul D. Hanna, Aug 03 2004

Keywords

Crossrefs

Programs

  • GAP
    List([0..30], n-> 1 + Sum([0..n-1], k-> Sum([0..n-k], j-> (-1)^(n-k-j)*3^j*Binomial(j, n-k-j)*Binomial(n+j, n) )) ); # G. C. Greubel, Sep 17 2019
  • Magma
    R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( 3/((1-9*x) + 2*(1-9*x)^(2/3)) )); // G. C. Greubel, Sep 17 2019
    
  • Maple
    seq(coeff(series(3/((1-9*x) + 2*(1-9*x)^(2/3)), x, n+1), x, n), n = 0 .. 40); # G. C. Greubel, Sep 17 2019
  • Mathematica
    CoefficientList[Series[3/((1-9*x) + 2*(1-9*x)^(2/3)), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 04 2014 *)
  • Maxima
    a(n):=sum(sum(binomial(k,n-m-k)*3^k*(-1)^(n-m-k)*binomial(n+k,n),k,0,n-m),m,0,n-1)+1; /* Vladimir Kruchinin, Sep 09 2019 */
    
  • PARI
    a(n)=polcoeff(3/((1-9*x)+2*(1-9*x+x*O(x^n))^(2/3)),n,x)
    
  • Sage
    def A097189_list(prec):
        P. = PowerSeriesRing(QQ, prec)
        return P( 3/((1-9*x) + 2*(1-9*x)^(2/3)) ).list()
    A097189_list(30) # G. C. Greubel, Sep 17 2019
    

Formula

G.f.: A(x) = 3/((1-9*x) + 2*(1-9*x)^(2/3)).
G.f.: A(x) = A004988(x)/(1 - x*A097188(x)).
a(n) = 1 + Sum_{m=0..n-1} Sum_{k=0..n-m} C(k,n-m-k)*3^k*(-1)^(n-m-k)*C(n+k,n). - Vladimir Kruchinin, Sep 17 2019
Conjecture: n*(n-1)*a(n) - (19*n-18)*(n-1)*a(n-1) + 9*(11*n^2-31*n+22)*a(n-2) - 9*(3*n-4)*(3*n-5)*a(n-3) = 0. - R. J. Mathar, Nov 16 2012
a(n) ~ 3^(2*n+1)/(2*Gamma(2/3) * n^(1/3))*(1 - sqrt(3)*Gamma(2/3)^2 / (4*Pi*n^(1/3))). - Vaclav Kotesovec, Feb 04 2014