The OEIS is supported by the many generous donors to the OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A024199 a(n) = (2n-1)!! * Sum_{k=0..n-1}(-1)^k/(2k+1). 20
 0, 1, 2, 13, 76, 789, 7734, 110937, 1528920, 28018665, 497895210, 11110528485, 241792844580, 6361055257725, 163842638377950, 4964894559637425, 147721447995130800, 5066706567801827025, 171002070002301095250, 6548719685561840296125, 247199273204273879989500 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 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 REFERENCES A. E. Jolliffe, Continued Fractions, in Encyclopaedia Britannica, 11th ed., pp. 30-33; see p. 31. LINKS 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 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) EXAMPLE 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 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 *) PROG (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 CROSSREFS 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. Sequence in context: A192700 A007509 A077413 * A037523 A037732 A090187 Adjacent sequences: A024196 A024197 A024198 * A024200 A024201 A024202 KEYWORD nonn,easy AUTHOR Clark Kimberling EXTENSIONS Edited by N. J. A. Sloane, Jul 19 2002 New name from Cyril Damamme, Jul 19 2015 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified April 21 13:26 EDT 2024. Contains 371870 sequences. (Running on oeis4.)