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.

A103882 a(n) = Sum_{i=0..n} C(n+1,i)*C(n-1,i-1)*C(2n-i,n).

Original entry on oeis.org

1, 2, 12, 92, 780, 7002, 65226, 623576, 6077196, 60110030, 601585512, 6078578508, 61908797418, 634756203018, 6545498596110, 67830161708592, 705951252118284, 7375213677918294, 77310179609631564, 812839595630249540, 8569327862277434280, 90562666977432643862
Offset: 0

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Author

Ralf Stephan, Feb 20 2005

Keywords

Comments

Number of permutations of n copies of 1..3 with all adjacent differences <= 1 in absolute value. - R. H. Hardin, May 06 2010 [Cf. A177316. - Peter Bala, Jan 14 2020]

Crossrefs

Equals A103881(n, n).
Row n=3 of A331562.

Programs

  • Magma
    [1] cat [&+[Binomial(n+1, i)*Binomial(n-1, i-1) * Binomial(2*n-i, n): i in [0..n]]:n in  [1..21]]; // Marius A. Burtea, Jan 19 2020
    
  • Magma
    [&+[Binomial(n, k)^2*Binomial(n+k-1, k): k in [0..n]]:n in  [0..21]]; // Marius A. Burtea, Jan 19 2020
    
  • Maple
    a:= proc(n) option remember; `if`(n<2, n+1,
          ((n-1)*(55*n^3-143*n^2+102*n-24)*a(n-1)+
          n*(5*n-3)*(n-2)^2*a(n-2))/((n-1)*(5*n-8)*n^2))
        end:
    seq(a(n), n=0..30);  # Alois P. Heinz, Jun 29 2015
    # Alternative:
    a := n -> hypergeom([-n, -n, n], [1, 1], 1):
    seq(simplify(a(n)), n=0..21); # Peter Luschny, Jan 19 2020
  • Mathematica
    Drop[Table[Sum[Sum[Multinomial[r, g, n + 1 - r - g] Binomial[n - 1,n - r] Binomial[n - 1, n - g], {g, 1, n}], {r, 1, n}], {n, 0, 18}], 1] (* Geoffrey Critzer, Jun 29 2015 *)
    Table[Sum[Binomial[n+1,k]Binomial[n-1,k-1]Binomial[2n-k,n],{k,0,n}],{n,0,30}] (* Harvey P. Dale, Jun 19 2021 *)
  • PARI
    a(n) = polcoef(pollegendre(n, (1 + x)/(1 - x)) + O(x^(n+1)), n); \\ Michel Marcus, Dec 20 2020
    
  • Python
    def A103882(n):
        if n == 0: return 1
        m, g = 1, 0
        for k in range(n+1):
            g += m*n//(n+k)
            m *= (n+k+1)*(n-k)**2
            m //= (k+1)**3
        return g # Chai Wah Wu, Oct 04 2022
    
  • SageMath
    def A103882(n): return hypergeometric([-n,-n,n], [1,1], 1).simplify()
    [A103882(n) for n in range(31)] # G. C. Greubel, May 24 2023

Formula

a(n) = (A005258(n-1) + 3*A005258(n))/5 (Apéry numbers). - Mark van Hoeij, Jul 13 2010
n^2*(n-1)*(5*n-8)*a(n) = (n-1)*(55*n^3-143*n^2+102*n-24)*a(n-1) + n*(n-2)^2*(5*n-3)*a(n-2). - Alois P. Heinz, Jun 29 2015
a(n) ~ phi^(5*n + 3/2) / (2*Pi*5^(1/4)*n), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jul 21 2019
From Peter Bala, Jan 14 2020: (Start)
a(n) = Sum_{k = 0..n} C(n,k)^2*C(n+k-1,k). Cf. A005258.
For any prime p >= 5, a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for all positive integers n and k (follows from known supercongruences satisfied by the Apéry numbers A005258 - see Straub, Example 3.4). (End)
a(n) = hypergeometric([-n, -n, n], [1, 1], 1). - Peter Luschny, Jan 19 2020
From Peter Bala, Dec 19 2020: (Start)
a(n) = Sum_{k = 1..n} C(n,k)*C(n+k,k)*C(n-1,k-1) for n >= 1.
a(n) = [x^n] P(n, (1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. Cf. A156554. (End)
a(n) = Sum_{k = 0..n} binomial(2*n-k-1,n-k)*binomial(n,k)^2. Cf. A108628. - Peter Bala, Mar 24 2022
From Peter Bala, Apr 15 2022: (Start)
a(-n) = (-1)^n*A352654(n).
a(n) = [x^n*y^n*z^(n-1)] 1/(1 - x - y - z + x*z + y*z - x*y*z) for n >= 1.
a(n) = B(n,n,n-1) in the notation of Straub, see equation 24.
a(n) = [x^n*y^n*z^(n-1)] (x + y + z)^n*(x + y)^n*(y + z)^(n-1) for n >= 1. (End)
D-finite with recurrence 9*n^2*a(n) -3*(31*n^2-27*n+6)*a(n-1) -2*(37*n^2-138*n+108)*a(n-2) -(n-3)*(17*n-56)*a(n-3) -(n-4)^2*a(n-4) = 0. - R. J. Mathar, Aug 01 2022
a(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n-1, n-k)*binomial(n+k, k)*binomial(n+k-1, k). - Peter Bala, Aug 13 2023
a(n) = Sum_{k = 0..n} (-1)^k * binomial(n+1, k)*binomial(2*n-k, n-k)^2. - Peter Bala, Oct 05 2024

Extensions

a(0)=1 prepended by Alois P. Heinz, Jun 29 2015