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.

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]

Links

Crossrefs

Cf. A005258, A156554, A177316, A352654.
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

A352655 a(n) = (1/2)*(A005258(n) + A005258(n-1)).

Original entry on oeis.org

2, 11, 83, 699, 6252, 58106, 554633, 5399099, 53356322, 533627511, 5388927513, 54859837434, 562267554552, 5796123147756, 60047675871333, 624801952898619, 6526036790730942, 68395815476047901, 718992874207884953, 7578808590187108199
Offset: 1

Views

Author

Peter Bala, Apr 17 2022

Keywords

Comments

The Apéry numbers A005258 satisfy the supercongruences A005258(p) == 3 (mod p^3) and A005258(p-1) == 1 (mod p^3) for primes p >= 5. It easily follows that a(p) == 2 (mod p^3) for primes p >= 3. We conjecture that the stronger supercongruences a(p) == 2 (mod p^5) hold for primes p >= 5. See A212334 for the corresponding conjecture for the Apéry numbers A005259.
Conjecture: for r >= 2, and all primes p >= 5, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ). - Peter Bala, Oct 13 2022

Examples

			Examples of superconguences:
a(5) - 2 = 6252 - 2 = 2*(5^5) == 0 (mod 5^5).
a(7) - 2 = 554633 - 2 = 3*(7^5)*11 == 0 (mod 7^5).
a(11) - 2 = 5388927513 - 2 = (11^5)*33461 == 0 (mod 11^5).

Crossrefs

Cf. A005258, A103882, A108628, A208675, A212334, A352654.

Programs

  • Maple
    seq((1/2)*add((2*n^2 - k*n + k^2)/(n*(n + k)) * binomial(n, k)^2 * binomial(n + k, k), k = 0..n), n = 1..20);
  • PARI
    f(n) = sum(k=0, n, binomial(n, k)^2 * binomial(n+k, k)); \\ A005258
a(n) = (f(n) + f(n-1))/2; \\ Michel Marcus, Apr 20 2022

Formula

a(n) = (1/2)*Sum_{k = 0..n} (2*n^2 - k*n + k^2)/(n*(n + k)) * binomial(n,k)^2 * binomial(n + k,k).
a(n) = (1/2)*Sum_{k = 0..n-1} (4*n + k)*(n - k)/(n*(n + k)) * binomial(n,k)^2* binomial(n + k,k) for n >= 1.
a(n) = (1/2)*(A108628(n-1) + 3*A208675(n)) for n >= 1.
a(n) = (1/2)*(2*A103882(n) - A352654(n)).
a(n) ~ 5^(3/4)*(13 + 5*sqrt(5))/(20*sqrt(22 + 10*sqrt(5))*Pi*n) * ((11 + 5*sqrt(5))/2)^n.
(11*n^2 - 31*n + 22)*n^2*a(n) = (121*n^4 - 462*n^3 + 607*n^2 - 322*n + 64)*a(n-1) + (11*n^2 - 9*n + 2)*(n - 2)^2*a(n-2) with a(1) = 2 and a(2) = 11.
The g.f. A(x) = 2*x + 11*x^2 + 83*x^3 + ... satisfies the differential equation
(x^5 + 13*x^4 + 22*x^3 + 9*x^2 - x)*A''(x) + (x^4 + 4*x^3 + 26*x^2 + 22*x - 1)*A'(x) + (2*x^2 - 16*x + 4)*A(x) + x^2 - 8*x + 2 = 0, with A(0) = 2 and A'(0) = 11.

A361714 a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(n,k)*binomial(n+k-1,k)^2.

Original entry on oeis.org

0, 1, 7, 82, 1063, 14376, 199204, 2806770, 40053031, 577468684, 8397778882, 123029274666, 1814016998116, 26898142793068, 400836647993292, 5999796281063082, 90162110212198695, 1359731143731297396, 20571691450059355174, 312134224830052880826, 4748435338386591995938
Offset: 0

Views

Author

Peter Bala, Mar 21 2023

Keywords

Comments

Conjecture 1: the supercongruence a(p) == a(1) (mod p^5) holds for all primes p >= 7 (checked up to p = 199).
Conjecture 2: for r >= 2, the supercongruence a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) holds for all primes p >= 7.
Compare with the Apéry numbers A005258(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n,k) * binomial(n+k,k)^2, which satisfy the weaker supercongruences A005258(p^r) == A005258(p^(r-1)) (mod p^(3*r)) for positive integer r and all primes p >= 5.

Examples

			Examples of supercongruence:
a(11) - a(1) = 23029274666 - 1 = 5*(11^5)*152783 == 0 (mod 11^5).
a(13) - a(1) = 26898142793068 - 1 = (3^2)*7*(13^5)*1149913 == 0 (mod 13^5).
a(5^2) - a(5) = 3994642669575050040375014376 - 14376 = (2^6)*(3^6)*(5^9)*103* 425601520324429 == 0 (mod 5^9).

Links

Crossrefs

Cf. A005258, A352654, A361712, A361713, A361715, A361717.

Programs

  • Maple
    seq(add((-1)^(n+k+1)*binomial(n,k)*binomial(n+k-1,k)^2, k = 0..n-1), n = 0..20);
# Alternative:
A361714 := n -> binomial(2*n-1, n)^2 - (-1)^n*hypergeom([-n, n, n], [1, 1], 1):
seq(simplify(A361714(n)), n = 0..20); # Peter Luschny, Mar 27 2023
  • Mathematica
    A361714[n_] := Binomial[2*n-1, n]^2 - (-1)^n*HypergeometricPFQ[{-n, n, n}, {1, 1}, 1]; Array[A361714, 20, 0] (* Paolo Xausa, Jul 11 2024 *)

Formula

a(n) = binomial(2*n-1,n)^2 - (1/5)*(A005258(n) - 3*A005258(n-1)) for n >= 1.
P-recursive:
(395*n^10 - 6083*n^9 + 39816*n^8 - 144606*n^7 + 318639*n^6 - 436307*n^5 + 362870*n^4 - 167820*n^3 + 33096*n^2)*a(n) = (10665*n^10 - 174906*n^9 + 1243697*n^8 - 5033114*n^7 + 12789951*n^6 - 21235254*n^5 + 23221451*n^4 - 16437246*n^3 + 7182940*n^2 - 1753656*n + 185472)*a(n-1) - (69125*n^10 - 1202775*n^9 + 9159576*n^8 - 40005738*n^7 + 110271201*n^6 - 198723383*n^5 + 234346978*n^4 - 175661976*n^3 + 78402944*n^2 - 18529392*n + 1901088)*a(n-2) - 4*(n - 3)^2*(1580*n^8 - 19592*n^7 + 101515*n^6 - 284307*n^5 + 464411*n^4 - 444309*n^3 + 236490*n^2 - 62500*n + 7000)*a(n-3) with a(0) = 0, a(1) = 1 and a(2) = 7.
a(n) = binomial(2*n-1, n)^2 - (-1)^n*hypergeom([-n, n, n], [1, 1], 1). This is another way to write the first formula. - Peter Luschny, Mar 27 2023
Showing 1-3 of 3 results.

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