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.

A320825 Expansion of -(10*x^2 - 6*x + 1)*sqrt(1 - 4*x)/(3*x - 1)^2.

Original entry on oeis.org

-1, 2, 1, 0, -3, -10, -17, 28, 435, 2710, 13489, 60392, 254211, 1028250, 4046977, 15621932, 59463209, 224047866, 838012755, 3118339056, 11563677321, 42790868982, 158180470803, 584617335420, 2161733579313, 8001589660746, 29660171058675, 110136151678696, 409773163539325
Offset: 0

Views

Author

Peter Luschny, Oct 22 2018

Keywords

Crossrefs

Programs

  • Magma
    m:=40; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!(-(10*x^2-6*x+1)*Sqrt(1-4*x)/(1-3*x)^2)); // G. C. Greubel, Oct 27 2018
  • Maple
    c := n -> catalan(n)*(n-3)*(n+1)/((2*n-3)*(2*n-1)):
    h := n -> hypergeom([1, -n], [5/2 - n], 3/4): A320825 := n -> c(n)*h(n):
    seq(simplify(A320825(n)), n=0..28);
  • Mathematica
    CoefficientList[Series[-(10x^2 - 6x + 1) Sqrt[1 - 4x]/(3x - 1)^2, {x, 0, 28}], x]
  • PARI
    x='x+O('x^40); Vec(-(10*x^2-6*x+1)*sqrt(1-4*x)/(1-3*x)^2) \\ G. C. Greubel, Oct 27 2018
    

Formula

a(n) = c(n)*h(n) where c(n) = Catalan(n)*((n-3)*(n+1))/((2*n-3)*(2*n-1)) and h(n) = hypergeom([1, -n], [5/2 - n], 3/4).
A320825(n) = A320826(n) + A320827(n).
D-finite with recurrence: n*a(n) +(-13*n+15)*a(n-1) +4*(16*n-37)*a(n-2) +2*(-71*n+245)*a(n-3) +60*(2*n-9)*a(n-4)=0. - R. J. Mathar, Jan 23 2020

A320826 Expansion of x*(1 - 4*x)^(3/2)/(3*x - 1)^2.

Original entry on oeis.org

0, 1, 0, -3, -14, -51, -168, -521, -1542, -4365, -11740, -29439, -65670, -112273, -28344, 1018689, 6961550, 34606929, 151831044, 623095683, 2453975622, 9402575805, 35339538912, 130994480547, 480676041954, 1750847208621, 6343667488692, 22899720430251, 82466180250590
Offset: 0

Views

Author

Peter Luschny, Oct 22 2018

Keywords

Crossrefs

Programs

  • Magma
    m:=30; R:=PowerSeriesRing(Rationals(), m); [0] cat Coefficients(R!(x*(1-4*x)^(3/2)/(1-3*x)^2)); // G. C. Greubel, Oct 27 2018
  • Maple
    c := n -> (-4)^(n-1)*binomial(3/2, n-1):
    h := n -> hypergeom([2, 1 - n], [7/2 - n], 3/4):
    A320826 := n -> c(n)*h(n): seq(simplify(A320826(n)), n=0..28);
  • Mathematica
    CoefficientList[Series[(x (1 -  4 x)^(3/2))/(3 x - 1)^2, {x, 0, 28}], x]
  • PARI
    x='x+O('x^30); concat([0], Vec(x*(1-4*x)^(3/2)/(1-3*x)^2)) \\ G. C. Greubel, Oct 27 2018
    

Formula

a(n) = c(n)*h(n) where c(n) = Catalan(n)*(3*n*(n + 1))/(2*(2*n-5)*(2*n-3)*(2*n-1)) = (-4)^(n-1)*binomial(3/2, n-1) and h(n) = hypergeom([2, 1 - n], [7/2 - n], 3/4).
A320826(n) = A320825(n) - A320827(n).

A291292 Necklace Catalan numbers.

Original entry on oeis.org

1, 1, 1, 3, 10, 34, 116, 396, 1353, 4631, 15895, 54757, 189465, 658835, 2303381, 8098783, 28642314, 101894922, 364614216, 1312191768, 4748561094, 17275277322, 63163858146, 232041604038, 856219298484, 3172442815476, 11799466553232, 44041859928944, 164924424558532, 619454123593948
Offset: 0

Views

Author

Sachin J. Valera, Oct 05 2018

Keywords

Comments

The n-th term is the number of ways to 'parenthesize' n beads arranged on a necklace. This can be proved.

Crossrefs

Programs

  • GAP
    Concatenation([1,1,1,3],List([4..30],n->3^(n-2)+(Sum([0..n-4],i->(3^i)*(2*(n-i-3))/((n-i-1)*(n-i))*Binomial(2*(n-i-2),n-i-2))))); # Muniru A Asiru, Oct 05 2018
  • Maple
    a:=n->`if`(n<=2,1,`if`(n=2,3,3^(n-2)+add((3^i)*(2*(n-i-3))/((n-i-1)*(n-i))*binomial(2*(n-i-2),n-i-2),i=0..n-4))); seq(a(n),n=0..30); # Muniru A Asiru, Oct 05 2018
    # Alternative:
    ogf := x -> 3/2 + (x - sqrt(1 - 4*x))*(2*x - 1)/(6*x - 2):
    ser := series(ogf(x),x,32):
    seq(coeff(ser, x, n), n=0..29); # Peter Luschny, Oct 25 2018
    # Derivation of the recurrence (requires Maple 2022):
    FormalPowerSeries:-FindRE(3/2 + (x - sqrt(1 - 4*x))*(2*x - 1)/(6*x - 2),x,a(n)); # Georg Fischer, Oct 21 2022
  • Mathematica
    Flatten[{1, 1, Table[3^(n - 2) + Sum[3^i*2*(n - i - 3)/((n - i - 1)*(n - i)) * Binomial[2*(n - i - 2), n - i - 2], {i, 0, n - 4}], {n, 2, 30}]}] (* Vaclav Kotesovec, Oct 22 2018 *)
  • PARI
    a(n) = if (n<=2, 1, if (n==2, 3, 3^(n-2) + sum(i=0, (n-4), (3^i)*(2*(n-i-3))/((n-i-1)*(n-i))*binomial(2*(n-i-2), n-i-2)))); \\ Michel Marcus, Oct 05 2018
    

Formula

a(n) = 3^(n-2) + (Sum_{i=0..n-4} 3^i*(2*(n-i-3))/((n-i-1)*(n-i))*binomial(2*(n-i-2), n-i-2)), for n >= 4. Initial terms are a(1)=a(2)=1, a(3)=3.
a(n) ~ 2^(2*n-1) / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Oct 22 2018
From Peter Luschny, Oct 25 2018: (Start)
a(n) = (3^(n-2) + (-4)^n*binomial(3/2, n)*((4/3)*n - 2 + hypergeom([1, -n], [5/2 - n], 3/4)))/2) for n >= 3.
a(n) = [x^n] (3/2) + (x - sqrt(1 - 4*x))*(2*x - 1)/(6*x - 2). (End)
Recurrence: 12*(2*n-7)*(n-4)*a(n-3) + (-158*n^2+744*n-862)*a(n-2) + 2*(n-1)*(79*n-143)*a(n-1) - 6*n*(11*n-9)*a(n) + (n+1)*(13*n+2)*a(n+1) - (n+1)*(n+2)*a(n+2) = 0. - Georg Fischer, Oct 21 2022

Extensions

More terms from Michel Marcus, Oct 05 2018
Showing 1-3 of 3 results.