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.

A001879 a(n) = (2n+2)!/(n!*2^(n+1)).

Original entry on oeis.org

1, 6, 45, 420, 4725, 62370, 945945, 16216200, 310134825, 6547290750, 151242416325, 3794809718700, 102776096548125, 2988412653476250, 92854250304440625, 3070380543400170000, 107655217802968460625, 3989575718580595893750, 155815096120119939628125
Offset: 0

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Comments

From Wolfdieter Lang, Oct 06 2008: (Start)
a(n) is the denominator of the n-th approximant to the continued fraction 1^2/(6+3^2/(6+5^2/(6+... for Pi-3. W. Lang, Oct 06 2008, after an e-mail from R. Rosenthal. Cf. A142970 for the corresponding numerators.
The e.g.f. g(x)=(1+x)/(1-2*x)^(5/2) satisfies (1-4*x^2)*g''(x) - 2*(8*x+3)*g'(x) -9*g(x) = 0 (from the three term recurrence given below). Also g(x)=hypergeom([2,3/2],[1],2*x). (End)
Number of descents in all fixed-point-free involutions of {1,2,...,2(n+1)}. A descent of a permutation p is a position i such that p(i) > p(i+1). Example: a(1)=6 because the fixed-point-free involutions 2143, 3412, and 4321 have 2, 1, and 3 descents, respectively. - Emeric Deutsch, Jun 05 2009
First differences of A193651. - Vladimir Reshetnikov, Apr 25 2016
a(n-2) is the number of maximal elements in the absolute order of the Coxeter group of type D_n. - Jose Bastidas, Nov 01 2021

References

  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 77 (Problem 10, values of Bessel polynomials).
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Second column of triangle A001497. Equals (A001147(n+1)-A001147(n))/2.
Equals row sums of A163938.

Programs

  • Magma
    [Factorial(2*n+2)/(Factorial(n)*2^(n+1)): n in [0..20]]; // Vincenzo Librandi, Nov 22 2011
  • Maple
    restart: G(x):=(1-x)/(1-2*x)^(1/2): f[0]:=G(x): for n from 1 to 29 do f[n]:=diff(f[n-1],x) od:x:=0:seq(f[n],n=2..20); # Zerinvary Lajos, Apr 04 2009
  • Mathematica
    Table[(2n+2)!/(n!2^(n+1)),{n,0,20}] (* Vincenzo Librandi, Nov 22 2011 *)
  • PARI
    a(n)=if(n<0,0,(2*n+2)!/n!/2^(n+1))
    

Formula

E.g.f.: (1+x)/(1-2*x)^(5/2).
a(n)*n = a(n-1)*(2n+1)*(n+1); a(n) = a(n-1)*(2n+4)-a(n-2)*(2n-1), if n>0. - Michael Somos, Feb 25 2004
From Wolfdieter Lang, Oct 06 2008: (Start)
a(n) = (n+1)*(2*n+1)!! with the double factorials (2*n+1)!!=A001147(n+1).
D-finite with recurrence a(n) = 6*a(n-1) + ((2*n-1)^2)*a(n-2), a(-1)=0, a(0)=1. (End)
With interpolated 0's, e.g.f.: B(A(x)) where B(x)= x exp(x) and A(x)=x^2/2.
E.g.f.: -G(0)/2 where G(k) = 1 - (2*k+3)/(1 - x/(x - (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 06 2012
G.f.: (1-x)/(2*x^2*Q(0)) - 1/(2*x^2), where Q(k) = 1 - x*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 20 2013
From Karol A. Penson, Jul 12 2013: (Start)
Integral representation as n-th moment of a signed function w(x) of bounded variation on (0,infinity),
w(x) = -(1/4)*sqrt(2)*sqrt(x)*(1-x)*exp(-x/2)/sqrt(Pi):
a(n) = Integral_{x>=0} x^n*w(x), n>=0.
For x>1, w(x)>0. w(0)=w(1)=limit(w(x),x=infinity)=0. For x<1, w(x)<0.
Asymptotics: a(n)->(1/576)*2^(1/2+n)*(1152*n^2+1680*n+505)*exp(-n)*(n)^(n), for n->infinity. (End)
G.f.: 2F0(3/2,2;;2x). - R. J. Mathar, Aug 08 2015

Extensions

Entry revised Aug 31 2004 (thanks to Ralf Stephan and Michael Somos)
E.g.f. in comment line corrected by Wolfdieter Lang, Nov 21 2011

A024199 a(n) = (2n-1)!! * Sum_{k=0..n-1}(-1)^k/(2k+1).

Original entry on oeis.org

0, 1, 2, 13, 76, 789, 7734, 110937, 1528920, 28018665, 497895210, 11110528485, 241792844580, 6361055257725, 163842638377950, 4964894559637425, 147721447995130800, 5066706567801827025, 171002070002301095250, 6548719685561840296125, 247199273204273879989500
Offset: 0

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Comments

(2*n + 1)!!/a(n+1), n>=0, is the n-th approximant for William Brouncker's continued fraction for 4/Pi = 1 + 1^2/(2 + 3^2/(2 + 5^2/(2 + ... ))) See the C. Brezinski and J.-P. Delahaye references given under A142969 and A142970, respectively. The double factorials (2*n + 1)!! = A001147(n+1) enter. - Wolfdieter Lang, Oct 06 2008

Examples

			a(3) = (2*3 - 1)!! * Sum_{k=0..2} (-1)^k/(2k + 1) = 5!! * (1/(2*0 + 1) - 1/(2*1 + 1) + 1/(2*2 + 1)) = 5*3*1*(1/1 - 1/3 + 1/5) = 15 - 5 + 3 = 13. Notice that the first factor always cancels the common denominator of the sum. - _Michael B. Porter_, Jul 22 2016
		

References

  • A. E. Jolliffe, Continued Fractions, in Encyclopaedia Britannica, 11th ed., pp. 30-33; see p. 31.

Crossrefs

From Johannes W. Meijer, Nov 12 2009: (Start)
Cf. A007509 and A025547.
Equals first column of A167584.
Equals row sums of A167591.
Equals first right hand column of A167594.
(End)
Cf. A167576 and A135457.

Programs

  • Magma
    [0] cat [ n le 2 select (n) else 2*Self(n-1)+(2*n-3)^2*Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Feb 17 2015
  • Maple
    a := proc(n) option remember; if n=0 then 0 elif n=1 then 1 else 2*a(n-1)+(2*n-3)^2* a(n-2) fi end: seq(a(n), n=0..20); # Peter Luschny, Nov 16 2016 after N. J. A. Sloane
  • Mathematica
    f[k_] := (2 k - 1) (-1)^(k + 1)
    t[n_] := Table[f[k], {k, 1, n}]
    a[n_] := SymmetricPolynomial[n - 1, t[n]]
    Table[a[n], {n, 1, 22}]    (* A024199 signed *)
    (* Clark Kimberling, Dec 30 2011 *)
    RecurrenceTable[{a[n+1] == 2*a[n] + (2*n-1)^2*a[n-1],a[0] == 0, a[1] == 1},a,{n,0,20}] (* Vaclav Kotesovec, Mar 18 2014 *)
    CoefficientList[Series[Pi/4/Sqrt[1-2*x] - 1/2*Log[2*x+Sqrt[4*x^2-1]]/Sqrt[2*x-1], {x, 0, 20}], x] * Range[0, 20]! (* Vaclav Kotesovec, Mar 18 2014 *)

Formula

a(n) = s(1)s(2)...s(n)(1/s(1) - 1/s(2) + ... + c/s(n)) where c=(-1)^(n+1) and s(k) = 2k-1 for k = 1, 2, 3, ... (was previous definition). - Clark Kimberling
D-finite with recurrence a(0) = 0, a(1) = 1, a(n+1) = 2*a(n) + (2*n-1)^2*a(n-1). - N. J. A. Sloane, Jul 19 2002
a(n) + A024200(n) = A001147(n) = (2n-1)!!. - Max Alekseyev, Sep 23 2007
a(n)/A024200(n) -> Pi/(4-Pi) as n -> oo. - Max Alekseyev, Sep 23 2007
From Wolfdieter Lang, Oct 06 2008: (Start)
E.g.f. for a(n+1), n>=0: (sqrt(1-2*x)+arcsin(2*x)*sqrt(1+2*x)/2)/((1-4*x^2)^(1/2)*(1-2*x)). From the recurrence, solving (1-4*x^2)y''(x)-2*(8*x+1)*y'(x)-9*y=0 with inputs y(0)=1, y'(0)=2.
a(n+1) = A003148(n) + A143165(n), n>=0 (from the two terms of the e.g.f.). (End)
From Johannes W. Meijer, Nov 12 2009: (Start)
a(n) = (-1)^(n-1)*(2*n-3)!! + (2*n-1)*a(n-1) with a(0) = 0.
a(n) = (2*n-1)!!*sum((-1)^(k)/(2*k+1), k=0..n-1)
(End)
E.g.f.: Pi/4/sqrt(1-2*x) - 1/2*log(2*x+sqrt(4*x^2-1))/sqrt(2*x-1). - Vaclav Kotesovec, Mar 18 2014
a(n) ~ Pi * 2^(n-3/2) * n^n / exp(n). - Vaclav Kotesovec, Mar 18 2014
a(n) = (2*H(n+1/2)-Gamma(n+1/2))*2^(n-2)*sqrt(Pi) with H(x) the Hadamard factorial (see the link section). - Cyril Damamme, Jul 19 2015
a(n) = A135457(n) - (-1)^n A001147(n-1). - Cyril Damamme, Jul 19 2015
a(n) = (Pi + (-1)^n*(Psi(n/2 + 1/4) - Psi(n/2 + 3/4)))*Gamma(n+1/2)*2^(n-2)/sqrt(Pi). - Robert Israel, Jul 20 2015
a(n) = A167576(n) - A135457(n). - Cyril Damamme, Jul 22 2015
a(n)/A001147(n) -> Pi/4 as n -> oo. - Daniel Suteu, Jul 21 2016
From Peter Bala, Nov 15 2016: (Start)
Conjecture: a(n) = 1/2*Sum_{k = 0..2*n-1} (-1)^(n-k+1)*k!*(2*n - 2*k - 3)!!, where the double factorial of an odd integer (positive or negative) may be defined in terms of the gamma function as (2*N - 1)!! = 2^((N+1)/2)*Gamma(N/2 + 1)/sqrt(Pi).
E.g.f. 1/2*arcsin(2*x)/sqrt(1 - 2*x) = x + 2*x^2/2! + 13*x^3/3! + 76*x^4/4! + .... (End)

Extensions

Edited by N. J. A. Sloane, Jul 19 2002
New name from Cyril Damamme, Jul 19 2015

A167588 The second column of the ED4 array A167584.

Original entry on oeis.org

1, 6, 41, 372, 4077, 53106, 795645, 13536360, 257055705, 5400196830, 124170067665, 3104906420700, 83818724048325, 2431059231544650, 75354930324303525, 2486926158748693200, 87036225272850632625, 3220532233879435917750, 125594424461427237941625
Offset: 1

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Author

Johannes W. Meijer, Nov 10 2009

Keywords

Crossrefs

Equals the second column of the ED4 array A167584.
Other columns are A024199 and A167589.
Cf. A007509 and A025547 (the sum((-1)^(k+n)/(2*k+1), k=0..n-1) factor), A001147, A142970.

Programs

  • Mathematica
    Table[(1/2)*(-1)^(n)*(2*n - 3)!!*((n) + (4*n^2 - 1)*Sum[(-1)^(k + n)/(2*k + 1), {k, 0, n - 1}]), {n, 1, 50}] (* G. C. Greubel, Jun 17 2016 *)

Formula

a(n) = (1/2)*(-1)^(n)*(2*n-3)!!*(n+(4*n^2-1)*Sum_{k=0..n-1} ((-1)^(k+n)/(2*k+1))).
From Peter Bala, Nov 01 2016: (Start)
a(n) = (2*n + 1)!! * Sum_{k = 0..n-1} (-1)^(k-1)/((2*k - 1)*(2*k + 1)*(2*k + 3)).
a(n) ~ Pi * 2^(n-3/2) * ((n+1)/e)^(n+1).
E.g.f.: (4*x*sqrt(1 - 4*x^2) + 2*arcsin(2*x))/(8*(1 - 2*x)^(3/2)).
a(n) = 6*a(n-1) + (2*n - 5)*(2*n - 1)*a(n-2) with a(0) = 0, a(1) = 1.
The sequence b(n) := (2*n + 1)!! = (2*n + 2)!/((n + 1)!*2^(n+1)) satisfies the same recurrence with b(0) = 1 and b(1) = 3. This leads to the continued fraction representation a(n) = b(n)*[ 1/(3 - 3/(6 + 5/(6 + 21/(6 + ... + (2*n - 5)*(2*n - 1)/(6))))) ] for n >= 2.
As n -> infinity, a(n)/(A001147(n+1)) -> 1/2!*Pi/4 = 1/(3 - 3/(6 + 5/(6 + 21/(6 + ... + (2*n - 5)*(2*n - 1)/(6 + ...))))). Compare with the generalized continued fraction representation Pi = 3 + 1^2/(6 + 3^2/(6 + 5^2/(6 + ...))). See A142970. (End)

A254795 Numerators of the convergents of the generalized continued fraction 2 + 1^2/(4 + 3^2/(4 + 5^2/(4 + ... ))).

Original entry on oeis.org

2, 9, 54, 441, 4410, 53361, 747054, 12006225, 216112050, 4334247225, 95353438950, 2292816782025, 59613236332650, 1671463434096225, 50143903022886750, 1606276360166472225, 54613396245660055650, 1967688541203928475625, 74772164565749282073750
Offset: 0

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Author

Peter Bala, Feb 23 2015

Keywords

Comments

Brouncker gave the generalized continued fraction expansion 4/Pi = 1 + 1^2/(2 + 3^2/(2 + 5^2/(2 + ... ))). The sequence of convergents begins [1/1, 3/2, 15/13, 105/76, ... ]. The numerators of the convergents are in A001147, the denominators in A024199.
In extending Brouckner's result, Osler showed that 2 + 1^2/(4 + 3^2/(4 + 5^2/(4 + ... ))) = L^2/Pi, where L is the lemniscate constant A062539. The sequence of convergents to Osler's continued fraction begins [2/1, 9/4, 54/25, 441/200, 4410/2025, ...]. Here we list the (unreduced) numerators of these convergents. See A254796 for the sequence of denominators. See A254794 for the decimal expansion of L^2/Pi.

Crossrefs

Programs

  • Maple
    a[0] := 2: a[1] := 9:
    for n from 2 to 18 do a[n] := 4*a[n-1] + (2*n-1)^2*a[n-2] end do:
    seq(a[n], n = 0 .. 18);

Formula

a(2*n-1) = ( A008545(n) )^2 = ( Product {k = 0..n-1} 4*k + 3 )^2.
a(2*n) = (4*n + 2)*( A008545(n) )^2 = (4*n + 2)*( Product {k = 0..n-1} 4*k + 3 )^2.
a(n) = 4*a(n-1) + (2*n - 1)^2*a(n-2) with a(0) = 2, a(1) = 9.
a(2*n) = (4*n + 2)*a(2*n-1); a(2*n+1) = (4*n + 4)*a(2*n) + a(2*n-1).
Showing 1-4 of 4 results.