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.

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

A268138 a(n) = (Sum_{k=0..n-1} A001850(k)*A001003(k+1))/n.

Original entry on oeis.org

1, 5, 51, 747, 13245, 264329, 5721415, 131425079, 3159389817, 78729848397, 2019910325499, 53087981674275, 1423867359013749, 38855956977763857, 1076297858301372687, 30203970496501504239, 857377825323716359665, 24586286492003180067989, 711463902659879056604995, 20756358426519694831851227
Offset: 1

Views

Author

Zhi-Wei Sun, Jan 26 2016

Keywords

Comments

Conjecture: (i) All the terms are odd integers. Also, p | a(p) for any odd prime p.
(ii) Let D_n(x) = Sum_{k=0..n} binomial(n,k)*binomial(n+k,k)*x^k = Sum_{k=0..n} binomial(n,k)^2*x^k*(x+1)^(n-k) for n >= 0, and s_n(x) = Sum_{k=1..n} (binomial(n,k)*binomial(n,k-1)/n)*x^(k-1)*(x+1)^(n-k) = (Sum_{k=0..n} binomial(n,k)*binomial(n+k,k)*x^k/(k+1))/(x+1) for n > 0. Then, for any positive integer n, all the coefficients of the polynomial (1/n)*Sum_{k=0..n-1} D_k(x)*s_{k+1}(x) are integral and the polynomial is irreducible over the field of rational numbers.
The conjecture was essentially proved by the author in arXiv:1602.00574, except for the irreducibility of (Sum_{k=0..n-1} D_k(x)*s_{k+1}(x))/n. - Zhi-Wei Sun, Feb 01 2016

Examples

			a(3) = 51 since (A001850(0)*A001003(1) + A001850(1)*A001003(2) + A001850(2)*A001003(3))/3 = (1*1 + 3*3 + 13*11)/3 = 153/3 = 51.
		

Crossrefs

Programs

  • Maple
    A001850 := n -> LegendreP(n, 3); seq(((3*(2*n+1)*A001850(n)*A001850(n-1)-n*A001850(n-1)^2)/(n+1) - A001850(n)^2)/4, n=1..20); # Mark van Hoeij, Nov 12 2022
    # Alternative (which also gives an integer for n = 0):
    f := n -> hypergeom([-n, -n], [1], 2):          # A001850
    h := n -> hypergeom([-n,  n], [1], 2):          # A182626
    g := n -> hypergeom([-n,  n, 1/2], [1, 1], -8): # A358388
    a := n -> (f(n)*((3*n + 1)*f(n) - (-1)^n*(6*n + 3)*h(n)) - n*g(n))/(2*n + 2):
    seq(simplify(a(n)), n = 1..20); # Peter Luschny, Nov 13 2022
  • Mathematica
    d[n_]:=Sum[Binomial[n,k]Binomial[n+k,k],{k,0,n}]
    s[n_]:=Sum[Binomial[n,k]Binomial[n,k-1]/n*2^(k-1),{k,1,n}]
    a[n_]:=Sum[d[k]s[k+1],{k,0,n-1}]/n
    Table[a[n],{n,1,20}]

Formula

a(n) = ((3*(2*n+1)*A001850(n)*A001850(n-1) - n*A001850(n-1)^2)/(n+1) - A001850(n)^2)/4. - Mark van Hoeij, Nov 12 2022
G.f.: (1-(1+1/x)*Int((1-34*x+x^2)^(1/2) * hypergeom([-1/2,1/2],[1], -32*x/(1-34*x+x^2))/((1-x)*(1+x)^2),x))/4. - Mark van Hoeij, Nov 28 2024

A358387 a(n) = 3 * h(n - 1) * h(n) for n >= 1, where h(n) = hypergeom([-n, -n], [1], 2), and a(0) = 1.

Original entry on oeis.org

1, 9, 117, 2457, 60669, 1620729, 45385461, 1311647913, 38774378493, 1165936210281, 35529105456117, 1094291069720121, 34000718751227133, 1064200845293945433, 33516300131277352821, 1061218377653812515657, 33757038339556757274621, 1078167326486278065165513
Offset: 0

Views

Author

Peter Luschny, Nov 15 2022

Keywords

Crossrefs

Cf. A358388.

Programs

  • Maple
    h := n -> hypergeom([-n, -n], [1], 2):
    A358387 := n -> ifelse(n = 0, 1, 3*h(n-1)*h(n)):
    seq(simplify(A358387(n)), n = 0..17);
  • Python
    def A358387gen() -> Generator:
        b, a, n = 1, 3, 1
        yield b
        while True:
            yield  3 * a * b
            n += 1
            aa = a * (6 * n - 3)
            bb = b * (n - 1)
            b, a = a, (aa - bb) // n
    A358387 = A358387gen()
    print([next(A358387) for n in range(18)])

Formula

a(n) ~ 3*sqrt(2) * (1 + sqrt(2))^(4*n) / (8*Pi*n). - Vaclav Kotesovec, Jan 08 2024
Showing 1-3 of 3 results.