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.

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A178808 a(n) = (1/n^2) * Sum_{k = 0..n-1} (2*k+1)*(D_k)^2, where D_0, D_1, ... are central Delannoy numbers given by A001850.

Original entry on oeis.org

1, 7, 97, 1791, 38241, 892039, 22092673, 571387903, 15271248769, 418796912007, 11725812711009, 333962374092543, 9648543623050593, 282164539499639559, 8338391167566634497, 248661515283002490879, 7474768663941435203073
Offset: 1

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Author

Zhi-Wei Sun, Jun 16 2010

Keywords

Comments

On Jun 14 2010, Zhi-Wei Sun conjectured that a(n) = (1/n^2) * Sum_{k = 0..n-1} (2*k+1)*(D_k)^2 is always an integer and that p^2*a(p) = p^2 - 4*p^3*q_p(2) - 2*p^4*q_p(2)^2 (mod p^5) for any prime p > 3, where q_p(2) denotes the Fermat quotient (2^(p-1) - 1)/p (see Sun, Remark 4.3, p. 26, 2014). He also conjectured that Sum_{k = 0..n-1} (2*k+1)*(-1)^k*(D_k)^2 == 0 (mod n*D_n/(3,D_n)) for all n = 1,2,3,....
The fact that a(n) is an integer follows directly from the formulas for a(n) in the formula section below. - Mark van Hoeij, Nov 13 2022

Examples

			For n = 3 we have a(3) = (D_0^2 + 3*D_1^2 + 5*D_2^2)/3^2 = (1 + 3*3^2 + 5*13^2)/3^2 = 97.
		

Crossrefs

Programs

  • Maple
    A001850 := n -> LegendreP(n, 3); seq((6*A001850(n)*A001850(n-1)-A001850(n)^2-A001850(n-1)^2)/8, n=1..20); # Mark van Hoeij, Nov 12 2022
    # Alternative:
    g := n -> hypergeom([n, -n, 1/2], [1, 1], -8): # A358388
    f := n -> hypergeom([-n, -n], [1], 2):         # A001850
    a := n -> (3*f(n)*f(n-1) - g(n)) / 4:
    seq(simplify(a(n)), n = 1..17); # Peter Luschny, Nov 13 2022
  • Mathematica
    DD[n_]:=Sum[Binomial[n+k,2k]Binomial[2k,k],{k,0,n}]; SS[n_]:= Sum[(2k+1)*DD[k]^2,{k,0,n-1}]/n^2; Table[SS[n],{n,1,25}]
    Table[Sum[(2k+1)*JacobiP[k,0,0,3]^2, {k, 0, n-1}]/n^2, {n, 1, 30}] (* G. C. Greubel, Jan 23 2019 *)
  • Python
    # prepends a(0) = 0
    def A178808List(size: int) -> list[int]:
        A358387 = A358387gen()
        A358388 = A358388gen()
        return [(next(A358387) - next(A358388)) // 4 for n in range(size)]
    print(A178808List(18)) # Peter Luschny, Nov 15 2022

Formula

a(n) ~ (1 + sqrt(2))^(4*n) / (16*Pi*n^2). - Vaclav Kotesovec, Jan 24 2019
G.f.: Integral(hypergeom([1/2, 1/2], [2], -32*x/(1 - 34*x + x^2))/((1 - x)*(1 - 34*x + x^2)^(1/2))). - Mark van Hoeij, Nov 10 2022
a(n) = (6*A001850(n)*A001850(n-1) - A001850(n)^2 - A001850(n-1)^2)/8. - Mark van Hoeij, Nov 12 2022
a(n) = (3*f(n)*f(n-1) - g(n))/4, where g(n) = hypergeom([n, -n, 1/2], [1, 1], -8) and f(n) = hypergeom([-n, -n], [1], 2). This formula also gives an integer value for n = 0. - Peter Luschny, Nov 13 2022

A358388 a(n) = hypergeom([n, -n, 1/2], [1, 1], -8).

Original entry on oeis.org

1, 5, 89, 2069, 53505, 1467765, 41817305, 1223277221, 36488826881, 1104851215205, 33853917808089, 1047387818876085, 32664869254856961, 1025606670801743061, 32387641973278794585, 1027864812983545977669, 32762392278424747311105, 1048268251830512324353221
Offset: 0

Views

Author

Peter Luschny, Nov 13 2022

Keywords

Crossrefs

Programs

  • Maple
    a := n -> hypergeom([n, -n, 1/2], [1, 1], -8):
    seq(simplify(a(n)), n = 0..17);
    # Alternative:
    A358388 := proc(n) local a;
    a := proc(n) option remember; if n < 3 then return [1, 1, 9][n + 1] fi;
    ((n - 3)^2*(2*n - 3)*a(n - 3) - (35*(n - 4)*n + 131)*((2*n - 5)*a(n - 2)
    + (3 - 2*n)*a(n - 1))) / ((n - 1)^2*(2*n - 5)) end:
    (a(n + 1) + a(n)) / 2 end: seq(A358388(n), n = 0..17);
  • Mathematica
    a[n_] := HypergeometricPFQ[{n, -n, 1/2}, {1, 1}, -8];
    Table[a[n], {n, 0, 17}] (* Jean-François Alcover, Nov 27 2023 *)
  • Python
    def A358388gen() -> Generator:
        c, b, a, n = 1, 1, 9, 2
        while True:
            yield (c + b) // 2
            n += 1
            f = 35 * (n - 4) * n + 131
            aa = a * f * (2 * n - 3)
            bb = b * f * (2 * n - 5)
            cc = c * (n - 3) ** 2 * (2 * n - 3)
            d = (aa - bb + cc) // ((n - 1) ** 2 * (2 * n - 5))
            c, b, a = b, a, d
    A358388 = A358388gen()
    print([next(A358388) for n in range(18)])

Formula

a(n) = Sum_{k=0..n} 2^(k - 1) * binomial(2*k, k)^2 * (binomial(n + k, 2*k) + binomial(n + k - 1, 2*k)).
a(n) = (i/Pi)*Integral_{t=0..1} hypergeom([n, -n],[1], -8*t)/sqrt(t*(t-1)).
a(n) ~ 3*sqrt(2) * (1 + sqrt(2))^(4*n) / (8*Pi*n). - Vaclav Kotesovec, Jan 08 2024
Showing 1-2 of 2 results.