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Exponential generating function = (1+2*x)/(1-2*x)^3.
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%I #31 Mar 29 2024 11:44:13

%S 1,8,72,768,9600,138240,2257920,41287680,836075520,18579456000,

%T 449622835200,11771943321600,331576403558400,9998303861145600,

%U 321374052679680000,10969567664799744000,396275631890890752000

%N Exponential generating function = (1+2*x)/(1-2*x)^3.

%H Robert Israel, <a href="/A014479/b014479.txt">Table of n, a(n) for n = 0..395</a>

%H Ben Adenbaum, Jennifer Elder, Pamela E. Harris, and J. Carlos Martínez Mori, <a href="https://arxiv.org/abs/2403.07989">Boolean intervals in the weak Bruhat order of a finite Coxeter group</a>, arXiv:2403.07989 [math.CO], 2024. See p. 8.

%F a(n) = A014477(n) * n!. - _Franklin T. Adams-Watters_, Nov 02 2006

%F G.f.: Sum_{n>=0} (2*n+1)^(n+1) * x^n / (1 + (2*n+1)*x)^(n+1). - _Paul D. Hanna_, Jan 02 2013

%F From _Vladimir Reshetnikov_, Oct 28 2015: (Start)

%F a(n) = 2^n*(n+1)^2*n!.

%F Recurrence: a(0) = 1, n*a(n) = 2*(n+1)^2*a(n-1). (End)

%F From _Amiram Eldar_, Dec 10 2022: (Start)

%F Sum_{n>=0} 1/a(n) = 2*(Ei(1/2) - gamma + log(2)), where Ei(x) is the exponential integral and gamma is Euler's constant (A001620).

%F Sum_{n>=0} (-1)^n/a(n) = 2*(gamma - Ei(-1/2) - gamma - log(2)). (End)

%p seq(add(count(Composition(k))*count(Permutation(k)),k=1..n),n=1..17); # _Zerinvary Lajos_, Oct 17 2006

%p seq(2^n*(n+1)^2*n!, n=0..30); # _Robert Israel_, Oct 28 2015

%t Table[2^n (n+1)^2 n!, {n, 0, 20}] (* _Vladimir Reshetnikov_, Oct 28 2015 *)

%o (PARI) {a(n)=polcoeff( sum(m=0,n,(2*m+1)^(m+1)*x^m / (1 + (2*m+1)*x +x*O(x^n))^(m+1)),n)} \\ _Paul D. Hanna_, Jan 02 2013

%o for(n=0,20,print1(a(n),", "))

%o (PARI) vector(30, n, n--; n!*(n+1)^2*2^n) \\ _Altug Alkan_, Oct 28 2015

%Y Cf. A001620, A014477, A187735.

%K nonn

%O 0,2

%A _N. J. A. Sloane_.