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 1-5 of 5 results.

A245551 Expansion of 1/(1 - 2*x - 3*x^2)^(5/2).

Original entry on oeis.org

1, 5, 25, 105, 420, 1596, 5880, 21120, 74415, 258115, 883883, 2994355, 10051860, 33479460, 110750580, 364177332, 1191186855, 3877914915, 12571302975, 40598200335, 130657125984, 419173385400, 1340928798300, 4278305877300, 13617034683525, 43243221276801, 137040737988105
Offset: 0

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Author

N. J. A. Sloane, Jul 30 2014

Keywords

Comments

From Petros Hadjicostas, Jun 03 2020: (Start)
For n >= 4, 2*a(n-4) counts 3-sets of leaves in "0,1,2" Motzkin rooted trees with n edges. "0,1,2" trees are rooted trees where each vertex has out-degree zero, one, or two. They are counted by the Motzkin numbers A001006.
For "0,1,2" trees, Salaam (2008) proved that the g.f. of the number of r-sets of leaves is A000108(r-1) * z^(2*r-2) * T(z)^(2*r-1), where T(z) = 1/sqrt(1 - 2*z - 3*z^2) is the g.f. of the central trinomial numbers A002426.
For r = 2, we get a shifted version of A102839. For r = 3, we get twice of a shifted version of the current sequence. (End)

Examples

			From _Petros Hadjicostas_, Jun 03 2020: (Start)
Out of the A001006(4) = 9 Motzkin trees with n = 4 edges, only the following 2*a(4-4) = 2 have 3-sets of leaves:
            A                    A
           / \                  / \
          /   \                /   \
         B     C              B     C
        / \                        / \
       /   \                      /   \
      D     E                    D     E
      {C, D, E}                {B, D, E}
(End)
		

Crossrefs

Programs

  • Maple
    A[0]:= 1: A[1]:= 5:
    for n from 2 to 100 do
    A[n]:= (2+3/n)*A[n-1] + (3+9/n)*A[n-2]
    od:
    seq(A[n],n=0..100); # Robert Israel, Aug 01 2014
  • Mathematica
    CoefficientList[Series[1/(1 - 2 x - 3 x^2)^(5/2), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 01 2014 *)
  • PARI
    x='x+O('x^50); Vec(1/(1-2*x-3*x^2)^(5/2)) \\ G. C. Greubel, Apr 06 2017

Formula

a(n) ~ 3^(n+3/2) * n^(3/2) / (8*sqrt(Pi)). - Vaclav Kotesovec, Jul 31 2014
a(n) = (2+3/n)*a(n-1) + (3+9/n)*a(n-2) for n >= 2. - Robert Israel, Aug 01 2014
a(n) = (binomial(n+4,2)/6) * Sum_{k=0..floor(n/2)} binomial(n+2,n-2*k) * binomial(2*k+2,k). - Seiichi Manyama, Jul 10 2024
a(n) = Sum_{k=0..n} (-2)^k * (3/2)^(n-k) * binomial(-5/2,k) * binomial(k,n-k). - Seiichi Manyama, Aug 23 2025

A375253 Expansion of (1 - 2*x + 2*x^2)/(1 - 2*x - 3*x^2)^(7/2).

Original entry on oeis.org

1, 5, 30, 140, 630, 2646, 10710, 41910, 159885, 597025, 2190188, 7914270, 28230020, 99567300, 347720040, 1203777072, 4135047615, 14105322315, 47813634330, 161154659820, 540353553894, 1803226621350, 5991410183850, 19827295283250, 65371101643575
Offset: 0

Views

Author

Seiichi Manyama, Aug 07 2024

Keywords

Crossrefs

Column k=4 of A091869 (with a different offset).

Programs

  • Mathematica
    a[n_]:=(1+n)(2+n)(3+n)(4+n)Hypergeometric2F1[(1-n)/2,-n/2,2,4]/24; Array[a,25,0] (* Stefano Spezia, Aug 07 2024 *)
  • PARI
    my(N=30, x='x+O('x^N)); Vec((1-2*x+2*x^2)/(1-2*x-3*x^2)^(7/2))

Formula

a(n) = (binomial(n+4,3)/4) * Sum_{k=0..floor(n/2)} binomial(n+1,n-2*k) * binomial(2*k+1,k).
a(n) = (binomial(n+4,3)/4) * A005717(n+1).
a(n) = ((n+4)/(n*(n+2))) * ((2*n+1)*a(n-1) + 3*(n+3)*a(n-2)).
a(n) = (1 + n)*(2 + n)*(3 + n)*(4 + n)*hypergeom([(1-n)/2, -n/2], [2], 4)/24. - Stefano Spezia, Aug 07 2024

A375260 Expansion of (1 - 3*x + 9*x^2 - 7*x^3)/(1 - 2*x - 3*x^2)^(7/2).

Original entry on oeis.org

1, 4, 30, 140, 665, 2856, 11844, 47160, 182655, 690580, 2560558, 9337692, 33573995, 119246960, 419034360, 1458687312, 5035531563, 17253821340, 58723235970, 198655153620, 668338862499, 2237229875496, 7454611712100, 24734393119800, 81748883914425
Offset: 0

Views

Author

Seiichi Manyama, Aug 08 2024

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec((1-3*x+9*x^2-7*x^3)/(1-2*x-3*x^2)^(7/2))

Formula

a(n) = binomial(n+3,3) * Sum_{k=0..floor(n/2)} binomial(n,n-2*k) * binomial(2*k,k).
a(n) = binomial(n+3,3) * A002426(n).
a(n) = A132885(n+6,3).
a(n) = ((n+3)/n^2) * ((2*n-1)*a(n-1) + 3*(n+2)*a(n-2)).

A374509 Expansion of 1/(1 - 2*x + 5*x^2)^(7/2).

Original entry on oeis.org

1, 7, 14, -42, -294, -462, 1386, 7722, 9009, -37037, -160160, -123760, 835380, 2848860, 1046520, -16550520, -45140865, 3533145, 296447690, 648593330, -393463070, -4895709390, -8489647530, 10975099590, 75528298755, 100311659721, -230350834728, -1097798696456
Offset: 0

Views

Author

Seiichi Manyama, Jul 09 2024

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_]:= Pochhammer[n+1, 6]*Hypergeometric2F1[(1-n)/2, -n/2, 4, -4]/6!; Array[a,28,0] (* Stefano Spezia, Jul 10 2024 *)
  • PARI
    a(n) = binomial(n+6, 3)/20*sum(k=0, n\2, (-1)^k*binomial(n+3, n-2*k)*binomial(2*k+3, k));

Formula

a(0) = 1, a(1) = 7; a(n) = ((2*n+5)*a(n-1) - 5*(n+5)*a(n-2))/n.
a(n) = (binomial(n+6,3)/20) * Sum_{k=0..floor(n/2)} (-1)^k * binomial(n+3,n-2*k) * binomial(2*k+3,k).
a(n) = Pochhammer(n+1, 6)*hypergeom([(1-n)/2, -n/2], [4], -4)/6!. - Stefano Spezia, Jul 10 2024
a(n) = (-1)^n * Sum_{k=0..n} 2^k * (5/2)^(n-k) * binomial(-7/2,k) * binomial(k,n-k). - Seiichi Manyama, Aug 23 2025

A375248 Expansion of (1 - x)/(1 - 2*x - 3*x^2)^(7/2).

Original entry on oeis.org

1, 6, 35, 168, 756, 3192, 12936, 50688, 193479, 722722, 2651649, 9581936, 34176324, 120526056, 420852204, 1456709328, 5002984791, 17062825626, 57827993685, 194871361608, 653285629920, 2179701604080, 7241015510820, 23958512912880, 78978801164445
Offset: 0

Views

Author

Seiichi Manyama, Aug 07 2024

Keywords

Crossrefs

First differences of A374506.

Programs

  • Mathematica
    a[n_]:=(1+n)(2+n)(3+n)(4+n)(5+n)Hypergeometric2F1[(1-n)/2,-n/2,3,4]/120; Array[a,25,0] (* Stefano Spezia, Aug 07 2024 *)
  • PARI
    my(N=30, x='x+O('x^N)); Vec((1-x)/(1-2*x-3*x^2)^(7/2))

Formula

a(n) = (binomial(n+5,3)/10) * Sum_{k=0..floor(n/2)} binomial(n+2,n-2*k) * binomial(2*k+2,k).
a(n) = (binomial(n+5,3)/10) * A014531(n+1).
a(n) = ((n+5)/(n*(n+4))) * ((2*n+3)*a(n-1) + 3*(n+4)*a(n-2)).
a(n) = (1 + n)*(2 + n)*(3 + n)*(4 + n)*(5 + n)*hypergeom([(1-n)/2, -n/2], [3], 4)/120. - Stefano Spezia, Aug 07 2024
Showing 1-5 of 5 results.