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.

A330799 Evaluation of the Motzkin polynomials at 1/2 and normalized with 2^n.

Original entry on oeis.org

1, 3, 13, 59, 285, 1419, 7245, 37659, 198589, 1059371, 5705517, 30976571, 169338781, 931239243, 5147825421, 28587660123, 159406327677, 892113040491, 5009160335085, 28210229053563, 159304938535773, 901845743050635, 5117144607546573, 29096321095698843, 165765778648482621
Offset: 0

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Author

Peter Luschny, Jan 01 2020

Keywords

Crossrefs

Programs

  • Magma
    m:=30;
    R:=PowerSeriesRing(Rationals(), m+2);
    A330799:= func< n | Coefficient(R!( 2/(1-4*x+Sqrt((1-6*x)*(1+2*x))) ), n) >;
    [A330799(n): n in [0..m]]; // G. C. Greubel, Sep 14 2023
  • Maple
    a := proc(n) option remember; if n < 3 then return [1, 3, 13][n+1] fi;
    (-84*(n - 2)*a(n-3) - 2*(8*n + 5)*a(n-2) + (11*n + 5)*a(n-1))/(n + 1) end:
    seq(a(n), n=0..24);
    # Alternative:
    gf := 2/(1 - 4*x + sqrt((1 - 6*x)*(2*x + 1))):
    ser := series(gf, x, 30): seq(coeff(ser,x,n), n=0..24);
    # Or:
    series((x^2+x)/(7*x^2+4*x+1), x, 30): gfun:-seriestoseries(%, 'revogf'):
    convert(%, polynom) / x: seq(coeff(%, x, n), n=0..24);
  • Mathematica
    With[{C = Binomial}, A064189[n_, k_] := Sum[C[n, j]* (C[n-j, j+k] - C[n-j, j+k+2]), {j, 0, n}]];
    a[n_] := 2^n*Sum[A064189[n, k]/2^k, {k, 0, n}];
    Table[a[n], {n, 0, 24}] (* Jean-François Alcover, Sep 25 2022 *)
    (* Second program *)
    A330799[n_]:= Coefficient[Series[2/(1-4*x+Sqrt[(1-6*x)*(1+2*x)]), {x, 0,50}], x, n]; Table[A330799[n], {n,0,30}] (* G. C. Greubel, Sep 14 2023 *)
  • SageMath
    R. = PowerSeriesRing(QQ)
    f = (x^2 + x)/(7*x^2 + 4*x+1)
    f.reverse().shift(-1).list()
    

Formula

a(n) = Sum_{k=0..n} A201641(n,k).
a(n) = 2^n*Sum_{k=0..n} A064189(n,k)/2^k.
a(n) = (-84*(n - 2)*a(n-3) - 2*(8*n + 5)*a(n-2) + (11*n + 5)*a(n-1))/(n + 1).
a(n) = [x^n] 2/(1 - 4*x + sqrt((1 - 6*x)*(2*x + 1))).
a(n) = [x^n] reverse((x^2 + x)/(7*x^2 + 4*x+1))/x.