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 1-3 of 3 results.

A101600 Expansion of g.f. c(3x)^2, where c(x) is the g.f. of A000108.

Original entry on oeis.org

1, 6, 45, 378, 3402, 32076, 312741, 3127410, 31899582, 330595668, 3471254514, 36848701764, 394807518900, 4263921204120, 46370143094805, 507343918566690, 5580783104233590, 61682339573108100, 684673969261499910, 7629224228913856140, 85308598196036755020
Offset: 0

Views

Author

Paul Barry, Dec 08 2004

Keywords

Crossrefs

Programs

  • Maple
    Z[0]:=1: for k to 30 do Z[k]:=simplify(1/(1-3*z*Z[k-1])) od: g:=sum((Z[j]-Z[j-1]), j=1..30): gser:=series(g, z=0, 27): seq(coeff(gser, z, n)/3, n=1..19); # Zerinvary Lajos, May 21 2008
  • Mathematica
    a[n_] := 3^n * CatalanNumber[n + 1]; Array[a, 20, 0] (* Amiram Eldar, May 15 2022 *)

Formula

G.f.: 4/(1+sqrt(1-12*x))^2.
a(n) = 3^n * A000108(n+1).
(n+2)*a(n) -6*(2*n+1)*a(n-1)=0. - R. J. Mathar, Nov 15 2011
O.g.f. A(x) = 1/x*series reversion( x/(1 + 3*x)^2 ). 1 + x*A'(x)/A(x) = 1/sqrt(1 - 12*x) is the o.g.f. for A098658. - Peter Bala, Jul 17 2015
From Amiram Eldar, May 15 2022: (Start)
Sum_{n>=0} 1/a(n) = 87/121 + 648*arcsin(1/(2*sqrt(3)))/(121*sqrt(11)).
Sum_{n>=0} (-1)^n/a(n) = 93/169 + 648*arcsinh(1/(2*sqrt(3)))/(169*sqrt(13)). (End)
E.g.f.: BesselI(1,6*z)*exp(6*z)/(3*z) where BesselI is the modified Bessel function of type I. - Karol A. Penson, Feb 17 2025

A101601 G.f.: c(3x)^3, c(x) the g.f. of A000108.

Original entry on oeis.org

1, 9, 81, 756, 7290, 72171, 729729, 7505784, 78298974, 826489170, 8811646074, 94753804536, 1026499549140, 11192793160815, 122744496427425, 1352917116177840, 14979996753469110, 166542316847391870, 1858400773709785470, 20806975169765062200, 233671377667405024620
Offset: 0

Views

Author

Paul Barry, Dec 08 2004

Keywords

Crossrefs

Programs

  • Mathematica
    terms = 18;
    c[x_] = (1 - Sqrt[1 - 4x])/(2x) + O[x]^terms // Normal;
    CoefficientList[c[3x]^3, x][[1 ;; terms]] (* Jean-François Alcover, Dec 15 2017 *)

Formula

G.f.: 8/(1+sqrt(1-12*x))^3.
a(n) = (3*n+3)/(n+3) * 3^n * C(n+1).
Conjecture: (n+3)*a(n) -3*(5*n+7)*a(n-1) +18*(2*n-1)*a(n-2)=0. - R. J. Mathar, Nov 15 2011
From Amiram Eldar, May 15 2022: (Start)
Sum_{n>=0} 1/a(n) = 51/121 + 964*arcsin(1/(2*sqrt(3)))/(121*sqrt(11)).
Sum_{n>=0} (-1)^n/a(n) = 57/169 + 1204*arcsinh(1/(2*sqrt(3)))/(169*sqrt(13)). (End)

A101602 G.f.: c(3x)^4, c(x) the g.f. of A000108.

Original entry on oeis.org

1, 12, 126, 1296, 13365, 138996, 1459458, 15466464, 165297834, 1780130520, 19301700924, 210564010080, 2309623985565, 25458117777540, 281857732537050, 3133071216411840, 34953325758094590, 391242268149428520, 4392583646950402020, 49454259823789423200
Offset: 0

Views

Author

Paul Barry, Dec 08 2004

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := ((8*n + 12)/(3*n + 12)) * ((3*n + 3)/(n + 3))* 3^n* CatalanNumber[n + 1]; Array[a, 20, 0] (* Amiram Eldar, May 15 2022 *)

Formula

G.f.: 16/(1+sqrt(1-12x))^4.
a(n)=((8n+12)/(3n+12))((3n+3)/(n+3))3^n*C(n+1).
Conjecture: (n+4)*a(n) - 6*(3*n+7)*a(n-1) + 36*(2*n+1)*a(n-2) = 0. - R. J. Mathar, Nov 15 2011
From Amiram Eldar, May 15 2022: (Start)
Sum_{n>=0} 1/a(n) = 1479/484 - 2691*arcsin(1/(2*sqrt(3)))/(121*sqrt(11)).
Sum_{n>=0} (-1)^n/a(n) = 7569*arcsinh(1/(2*sqrt(3)))/(169*sqrt(13)) - 1767/676. (End)
Showing 1-3 of 3 results.