A135457 -id:A135457 - cp's OEIS frontend

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

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

Offset: 0

Clark Kimberling

(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

			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

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

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Peter Luschny, Is the Gamma function misdefined? Hadamard versus Euler - Who found the better Gamma function?
Wikipedia, William Brouncker, 2nd Viscount Brouncker

Cf. A004041, A000407.
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.

[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

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

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 *)

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)

Edited by N. J. A. Sloane, Jul 19 2002

New name from Cyril Damamme, Jul 19 2015

A167576 The first column of the ED3 array A167572.

Original entry on oeis.org

1, 5, 23, 167, 1473, 16413, 211479, 3192975, 54010305, 1030249845, 21566327895, 497334999735, 12405876372225, 335591130336525, 9716331072597975, 301633179343890975, 9941514351641143425, 348336799875365041125

Offset: 1

Johannes W. Meijer, Nov 10 2009

Basically a(n) measures the difference between the Euler factorial n! and the Luschny factorial L(n) at half-integer values. For the Luschny factorial see the link. The formula given in the Maple section is a variant of a formula given by Cyril Damamme in A135457. - Peter Luschny, Jul 18 2015

			G.f. = x + 5*x^2 + 23*x^3 + 167*x^4 + 1473*x^5 + 16413*x^6 + ...

G. C. Greubel, Table of n, a(n) for n = 1..400
Peter Luschny, Is the Gamma function misdefined? Hadamard versus Euler - Who found the better Gamma function?

Equals the first column of the ED3 array A167572.
Equals the first right hand column of A167583.
Other columns are A167577 and A167578.
Cf. A097801 (the 2*(-1)^n*(2*n-5)!! factor).
Cf. A007509 and A025547 (the sum((-1)^(k+n)/(2*k+1), k=0..n-1) factor).
Cf. A024199 and A135457.

L := x -> (1+x*(Psi(1-x/2)-Psi(1/2-x/2)))/(-x)!:
a := x -> (L(x-1/2)-(x-1/2)!)*2^(x-1)*sqrt(Pi):
seq(simplify(a(n)),n=1..18); # Peter Luschny, Jul 18 2015
a := proc(n) option remember: if n=1 then 1 else (2*n-1)*a(n-1)+2*(-1)^n*doublefactorial(2*n-5) fi: end: seq(a(n),n=1..18); # Johannes W. Meijer, Jul 20 2015

a[ n_] := If[ n < 1, 0, (2 n - 3)!! ((-1)^n - I (4 n - 2) Sum[ I^k / k, {k, 1, 2 n - 1, 2}])]; (* Michael Somos, Jul 20 2015 *)
a[ n_] := If[ n < 1, 0, (2 n - 3)!! ((-1)^n + (4 n - 2) Sum[ KroneckerSymbol[ -4, k]/ k, {k, 2 n - 1}])]; (* Michael Somos, Jan 31 2019 *)

{a(n) = if( n<1, 0, prod(k=1, n-1, 2*k - 1) * ((-1)^n - (4*n - 2) * sum(k=1, n, (-1)^k / (2*k - 1))))}; /* Michael Somos, Jul 20 2015 */

a(n) = (-1)^n*(2*n-3)!!*(1 + (4*n-2)*Sum_{k=0..n-1} (-1)^(k+n)/(2*k+1)).
a(n) = (2*n-1)*a(n-1) + 2*(-1)^n*(2*n-5)!! with a(1) = 1.
a(n) = 4*a(n-1) + (4*n^2 - 16*n + 15)*a(n-2) with a(1) = 1 and a(2) = 5 [Superseeker].
0 = a(n)*a(n+1)*(-440*a(n+2) - 220*a(n+3) + 55*a(n+4)) + a(n)*a(n+2)*(536*a(n+2) - 118*a(n+3) - 4*a(n+4)) + a(n)*a(n+3)*(-4*a(n+3) + a(n+4)) + a(n+1)^2*(-220*a(n+2) - 32*a(n+3) + 8*a(n+4)) + a(n+1)*a(n+2)*(+71*a(n+2) + 4*a(n+3) - 2*a(n+4)) + a(n+2)^2*(-4*a(n+2) + a(n+3)) if n>0. - Michael Somos, Jul 19 2015
a(n) = (-1 + (n-1/2)*LerchPhi(-1,1,n+1/2) + (-n+1/2)*LerchPhi(-1,1,-n+1/2))/(1-2*n)!!. - Johannes W. Meijer, Jul 20 2015
a(n) = A024199(n) + A135457(n). - Cyril Damamme, Jul 22 2015
a(n) = ((-1)^n/(2*n - 1) + Pi/2 - (-1)^n LerchPhi(-1, 1, n + 1/2)) (2*n - 1)!!. - Michael Somos, Jan 31 2019

cp's OEIS Frontend

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

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