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.

A383949 Expansion of 1/sqrt((1-x)^3 * (1-5*x)).

Original entry on oeis.org

1, 4, 15, 60, 255, 1128, 5117, 23600, 110115, 518220, 2455101, 11693124, 55934385, 268535400, 1293178275, 6243968880, 30217425795, 146529719100, 711810105725, 3463284659300, 16874328961245, 82322471522280, 402079323279975, 1965900162652800, 9621179345962525, 47127880914834148
Offset: 0

Views

Author

Seiichi Manyama, Aug 19 2025

Keywords

Crossrefs

Programs

  • Magma
    R := PowerSeriesRing(Rationals(), 34); f := 1/Sqrt((1- x)^3 * (1-5*x)); coeffs := [ Coefficient(f, n) : n in [0..33] ]; coeffs; // Vincenzo Librandi, Aug 27 2025
  • Mathematica
    CoefficientList[Series[1/Sqrt[(1-x)^3*(1-5*x)],{x,0,33}],x] (* Vincenzo Librandi, Aug 27 2025 *)
  • PARI
    my(N=30, x='x+O('x^N)); Vec(1/sqrt((1-x)^3*(1-5*x)))
    

Formula

n*a(n) = (6*n-2)*a(n-1) - 5*n*a(n-2) for n > 1.
a(n) = (1/4)^n * Sum_{k=0..n} 5^(n-k) * (2*k+1) * binomial(2*k,k) * binomial(2*(n-k),n-k).
a(n) = Sum_{k=0..n} (-1)^k * 5^(n-k) * (2*k+1) * binomial(2*k,k) * binomial(n+1,n-k).
a(n) = Sum_{k=0..n} binomial(2*k,k) * binomial(n+1,n-k).

A097790 a(n)=5a(n-1)+C(n+3,3),n>0, a(0)=1.

Original entry on oeis.org

1, 9, 55, 295, 1510, 7606, 38114, 190690, 953615, 4768295, 23841761, 119209169, 596046300, 2980232060, 14901160980, 74505805716, 372529029549, 1862645148885, 9313225745755, 46566128730315, 232830643653346
Offset: 0

Views

Author

Paul Barry, Aug 24 2004

Keywords

Comments

Partial sums of A052244.

Programs

  • Mathematica
    nxt[{n_,a_}]:={n+1,5a+Binomial[n+4,3]}; NestList[nxt,{0,1},20][[All,2]] (* or *) LinearRecurrence[{9,-26,34,-21,5},{1,9,55,295,1510},30] (* Harvey P. Dale, Sep 20 2022 *)

Formula

G.f.: 1/((1-5*x)*(1-x)^4).
a(n) = 5^(n+4)/256-(32*n^3+312*n^2+1012*n+1107)/768.
a(n) = Sum_{k=0..n} binomial(n+4, k+4)*4^k.

A229611 Expansion of 1/((1-x)^3*(1-11x)).

Original entry on oeis.org

1, 14, 160, 1770, 19485, 214356, 2357944, 25937420, 285311665, 3138428370, 34522712136, 379749833574, 4177248169405, 45949729863560, 505447028499280, 5559917313492216, 61159090448414529, 672749994932559990, 7400249944258160080, 81402749386839761090
Offset: 0

Views

Author

Yahia Kahloune, Sep 26 2013

Keywords

Comments

This sequence was chosen to illustrate a method of matching generating functions and closed-form solutions: The general term associated with the generating function 1/((1-s*x)^3*(1-r*x)) with r>s>=1 is a(n) = [r^(n+3) - s^(n+1)*(s^2 + (r-s)*s*binomial(n+3,1) +(r-s)^2*binomial(n+3,2))] / (r-s)^3 .

Examples

			a(3) = (11^6 - (50*3^2+260*3 + 331))/1000 = 1770 .
		

Crossrefs

Programs

  • Magma
    [(11^(n+3) - (50*n^2 + 260*n + 331))/1000: n in [0..25]]; // Vincenzo Librandi, Sep 27 2013
  • Mathematica
    CoefficientList[Series[1/((1 - x)^3 (1 - 11 x)), {x, 0, 20}], x] (* Vincenzo Librandi, Sep 27 2013 *)
    LinearRecurrence[{14,-36,34,-11},{1,14,160,1770},30] (* Harvey P. Dale, Apr 09 2016 *)

Formula

a(n) = (11^(n+3) - (1 + 10*C(n+3,1) + 100*C(n+3,2)))/1000 = (11^(n+3) - (50*n^2 + 260*n + 331))/1000.
a(n) = 14*a(n-1) -36*a(n-2) +34*a(n-3) -11*a(n-4). - Vincenzo Librandi, Sep 27 2013
Showing 1-3 of 3 results.