%I A001879 M4251 N1775
%S A001879 1,6,45,420,4725,62370,945945,16216200,310134825,6547290750,151242416325,
%T A001879 3794809718700,102776096548125,2988412653476250,92854250304440625,3070380543400170000,
%U A001879 107655217802968460625,3989575718580595893750,155815096120119939628125
%N A001879 (2n+2)!/(n!2^(n+1)).
%C A001879 Contribution from Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), 
               Oct 06 2008: (Start)
%C A001879 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.
%C A001879 The e.g.f. g(x)=(1+4)/(1-4*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). W. Lang, Oct 06 2008. (End)
%C A001879 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. [From Emeric Deutsch 
               (deutsch(AT)duke.poly.edu), Jun 05 2009]
%D A001879 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, 
               Academic Press, 1995 (includes this sequence).
%D A001879 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 
               (includes this sequence).
%D A001879 J. Riordan, Combinatorial Identities, Wiley, 1968, p. 77 (Problem 10, 
               values of Bessel polynomials).
%F A001879 E.g.f.: (1+x)/(1-2x)^(5/2).
%F A001879 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
%F A001879 Contribution from Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), 
               Oct 06 2008: (Start)
%F A001879 a(n)=(n+1)*(2*n+1)!! with the double factorials (2*n+1)!!=A001147(n+1).
%F A001879 Three term recurrence a(n) = 6*a(n-1) + ((2*n-1)^2)*a(n-2), a(-1):=0, 
               a(0)=1. W. Lang, Oct 06 2008. (End)
%p A001879 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);# [From Zerinvary 
               Lajos (zerinvarylajos(AT)yahoo.com), Apr 04 2009]
%o A001879 (PARI) a(n)=if(n<0,0,(2*n+2)!/n!/2^(n+1))
%Y A001879 Cf. A002544, A001814, A001876-A001878.
%Y A001879 Second column of triangle A001497. Equals [A001147(n+1)-A001147(n)]/2.
%Y A001879 Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), Oct 16 
               2009: (Start)
%Y A001879 Equals row sums of A163938.
%Y A001879 (End)
%Y A001879 Sequence in context: A101600 A135148 A137974 this_sequence A019577 A097814 
               A084064
%Y A001879 Adjacent sequences: A001876 A001877 A001878 this_sequence A001880 A001881 
               A001882
%K A001879 nonn,easy
%O A001879 0,2
%A A001879 N. J. A. Sloane (njas(AT)research.att.com).
%E A001879 Entry revised Aug 31 2004 (thanks to Ralf Stephan and Michael Somos).