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 1-4 of 4 results.

A087547 a(n) = n!*2^(n+1) * (Integral_{x = 0..1} 1/(1+x^2)^(n+1) dx) - Pi*(2*n)!/(2^(n+1)*n!).

Original entry on oeis.org

0, 1, 4, 22, 160, 1464, 16224, 211632, 3179520, 54092160, 1028113920, 21594021120, 496702402560, 12418039065600, 335293281792000, 9723592350259200, 301432670532403200, 9947299050359193600, 348155822449999872000, 12881771833023700992000, 502389223133024747520000
Offset: 0

Views

Author

Al Hakanson (hawkuu(AT)excite.com), Oct 24 2003

Keywords

Comments

a(n)/A001147 gives an approximation for Pi/2 with (n-1)/3 + 1 digits of accuracy. - Aaron Kastel, Nov 13 2012
a(n) is the number of linear chord diagrams on 2n vertices with one marked chord such that none of the remaining n-1 chords are excluded by (i.e., are outside and do not contain) the marked chord, see [Young]. - Donovan Young, Aug 11 2020

Examples

			a(3) = 22.
		

Crossrefs

Programs

  • Magma
    [0] cat [n eq 1 select 1 else (2*n-1)*Self(n-1)+Factorial(n-1): n in [1..25]]; // Vincenzo Librandi, Nov 07 2014
    
  • Magma
    I:=[1,4]; [0] cat [n le 2 select I[n]  else (3*n-2)*Self(n-1)-(n-1)*(2*n-3)*Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Feb 19 2015
  • Maple
    f := proc(n) 4*n!*2^(n-1) * (int (1/(1+x^2)^(n+1),x=0..1)) - Pi*(2*n)!/(2^(n+1)*n!); end; # N. J. A. Sloane, Oct 30 2003
  • Mathematica
    f[n_] := Simplify[n!*2^(n + 1)*(Integrate[ 1/(1 + x^2)^(n + 1), {x, 0, 1}]) - Pi(2n)!/(2^(n + 1)*n!)]; Table[ f[n], {n, 0, 20}] (* Robert G. Wilson v, Oct 31 2003 *)
    CoefficientList[Normal[Series[1/Sqrt[1-2*x]*ArcTan[x/Sqrt[1-2*x]],{x,0,10}]]/.{x^n_.->x^n*n!},x] (* Donovan Young, Aug 11 2020 *)

Formula

a(n) = (2n-1)*a(n-1) + (n-1)!. - Aaron Kastel, Nov 13 2012
From Peter Bala, Jun 21 2013: (Start)
a(n) = (2*n)!/(n!*2^n)*(Sum_{k = 0..n-1} 2^k*k!^2/(2*k+1)!). Thus a(n)/ ((2*n)!/(n!*2^n)) -> Pi/2 as n -> infinity since Sum_{k >= 0} 2^k*k!^2/(2*k+1)! = Pi/2.
It appears that a(n) = Sum_{k = 1..n} 2^(k-1)*(k-1)!*(n+k-1)!/(2*k-1)!. Cf. A167571.
a(n) = (2*n)!/(n!*2^n)*(Pi/2) - 2^(n+1)*n!*(Integral_{x = 0..1} x^(2*n)/(1 + x^2)^(n+1) dx). Cf. A068102. (End)
From Peter Bala, Feb 18 2015: (Start)
Recurrence equation: a(n) = (3*n - 2)*a(n-1) - (n - 1)*(2*n - 3)*a(n-2) with a(1) = 1 and a(2) = 4.
The sequence b(n) = A001147(n), beginning [1, 3, 15, 105, 945, ... ], satisfies the same second-order recurrence equation. This leads to the generalized continued fraction expansion lim_{n -> infinity} a(n)/b(n) = Pi/2 = 1 + 1/(3 - 6/(7 - 15/(10 - ... - n*(2*n - 1)/((3*n + 1) - ... )))). (End)
E.g.f.: arctan(x/sqrt(1 - 2*x))/sqrt(1 - 2*x). - Donovan Young, Aug 11 2020
From Sela Fried, Apr 13 2023: (Start)
a(n) = (n - 1)!*Sum_{k=0..n - 1} binomial(2*n - 1, 2*k)/binomial(n - 1, k).
a(n)/(n-1)! = 1 + (2*n - 1)/2*Integral_{t = 0..Pi/2} (1 + sin(2*t))^(n - 1) - (1 - sin(2*t))^(n - 1)dt.
Bala's conjecture is true. (End)
a(n) ~ Pi * 2^(n - 1/2) * n^n / exp(n). - Vaclav Kotesovec, Apr 13 2023
a(n) = (n - 1)!/2 * Sum_{k=0..n - 1} binomial(2*n, 2*k + 1)/binomial(n - 1, k). - Sela Fried, May 18 2023

Extensions

More terms from N. J. A. Sloane, Oct 30 2003

A336599 Triangle read by rows: T(n,k) is the number of linear chord diagrams on 2n vertices with one marked chord such that exactly k of the remaining n-1 chords are contained within the marked chord.

Original entry on oeis.org

1, 5, 1, 33, 9, 3, 279, 87, 39, 15, 2895, 975, 495, 255, 105, 35685, 12645, 6885, 4005, 2205, 945, 509985, 187425, 106785, 66465, 41265, 23625, 10395, 8294895, 3133935, 1843695, 1198575, 795375, 513135, 301455, 135135, 151335135, 58437855, 35213535, 23601375, 16343775, 11263455, 7453215, 4459455, 2027025
Offset: 1

Views

Author

Donovan Young, Jul 29 2020

Keywords

Examples

			Triangle begins:
     1;
     5,    1;
    33,    9,    3;
   279,   87,   39,  15;
  2895,  975,  495, 255, 105;
...
For n = 2 and k = 1, let the four vertices be {1,2,3,4}. The marked chord can only be (1,4) and it contains one other chord, namely (2,3), hence T(2,1) = 1.
		

Crossrefs

Row sums are n*A001147(n) for n > 0.
Leading diagonal is A001147(n-1) for n > 0.
The first column is A129890(n-1) for n > 0.
The second column is A035101(n+1) for n > 0.

Programs

  • Mathematica
    CoefficientList[Normal[Series[(Sqrt[1-2*y*x]-Sqrt[1-2*x])/(1-2*x)/(1-y),{x,0,10}]]/.{x^n_.->x^n*n!},{x,y}]

Formula

E.g.f.: (sqrt(1 - 2*y*x) - sqrt(1 - 2*x))/(1 - 2*x)/(1 - y).

A336598 Triangle read by rows: T(n,k) is the number of linear chord diagrams on 2n vertices with one marked chord such that exactly k of the remaining n-1 chords cross the marked chord.

Original entry on oeis.org

1, 4, 2, 21, 18, 6, 144, 156, 96, 24, 1245, 1500, 1260, 600, 120, 13140, 16470, 16560, 11160, 4320, 720, 164745, 207270, 231210, 194040, 108360, 35280, 5040, 2399040, 2976120, 3507840, 3402000, 2419200, 1149120, 322560, 40320
Offset: 1

Views

Author

Donovan Young, Jul 29 2020

Keywords

Examples

			Triangle begins:
     1;
     4,    2;
    21,   18,    6;
   144,  156,   96,  24;
  1245, 1500, 1260, 600, 120;
...
For n = 2 and k = 1, let the four vertices be {1,2,3,4}. The marked chord can be either (1,3), and so crossed once by (2,4), or (2,4), and so crossed once by (1,3). Hence T(2,1) = 2.
		

Crossrefs

Row sums are n*A001147(n) for n > 0.
First column is A233481(n) for n > 0.
Leading diagonal is A000142(n) for n > 0.
Sub-leading diagonal is n*A000142(n) for n > 1.

Programs

  • Mathematica
    CoefficientList[Normal[Series[x/Sqrt[1-2*x]/(1-x(1+y)),{x,0,10}]]/.{x^n_.->x^n*n!},{x,y}]
  • PARI
    T(n)={[Vecrev(p) | p<-Vec(serlaplace(x/sqrt(1 - 2*x + O(x^n))/(1 - x*(1 + y))))]}
    { my(A=T(8)); for(n=1, #A, print(A[n])) } \\ Andrew Howroyd, Jul 29 2020

Formula

T(n,k) = n*T(n-1,k) + n*T(n-1,k-1), with T(n,0) = A233481(n) for n > 0.
E.g.f.: x/sqrt(1 - 2*x)/(1 - x*(1 + y)).

A336600 Triangle read by rows: T(n,k) is the number of linear chord diagrams on 2n vertices with one marked chord such that exactly k of the remaining n-1 chords contain the marked chord.

Original entry on oeis.org

1, 5, 1, 32, 11, 2, 260, 116, 38, 6, 2589, 1344, 594, 174, 24, 30669, 17529, 9294, 3774, 984, 120, 422232, 257487, 153852, 76782, 28272, 6600, 720, 6633360, 4234320, 2746260, 1576980, 726480, 242640, 51120, 5040, 117193185, 77358600, 53170380, 33718500, 18171360, 7693200, 2340720, 448560, 40320
Offset: 1

Views

Author

Donovan Young, Jul 30 2020

Keywords

Examples

			Triangle begins:
     1;
     5,    1;
    32,   11,    2;
   260,  116,   38,   6;
  2589, 1344,  594, 174, 24;
...
For n = 2 and k = 1, let the four vertices be {1,2,3,4}. The marked chord can only be (2,3) and it is contained by one other chord, namely (1,4), hence T(2,1) = 1.
		

Crossrefs

Row sums are n*A001147(n) for n > 0.
Leading diagonal is A000142(n-1) for n > 0.
Sub-leading diagonal is A001344(n-2) for n > 1.

Programs

  • Mathematica
    CoefficientList[Normal[Series[Log[(1-x*(1+y))/(1-2*x)]/(1-y)/Sqrt[1-2*x],{x,0,10}]]/.{x^n_.->x^n*n!},{x,y}]

Formula

E.g.f.: log((1 - x*(1 + y))/(1 - 2*x))/(1 - y)/sqrt(1 - 2*x).
Showing 1-4 of 4 results.