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a(n) = n!*2^n*hypergeom([-n], [-n], 1/2). - Peter Luschny, Jul 28 2024

a[n_] := n! 2^n Hypergeometric1F1[-n, -n, 1/2];

Table[a[n], {n, 0, 19}] (* Peter Luschny, Jul 28 2024 *)

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a(n) = 2^n*KummerU(-n, -n, 1/2). - Peter Luschny, May 10 2022

seq(simplify(2^n*KummerU(-n, -n, 1/2)), n = 0..19); # Peter Luschny, May 10 2022

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nxt[{n_, a_}]:={n+1, 2*a(n+1)+1}; NestList[nxt, {0, 1}, 20][[All, 2]] (* Harvey P. Dale, Jan 06 2022 *)

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a(n) is a specific instance of sequences having the form b(0) = x, b(n) = a*n*b(n-1) + k for n >= 1. (Here x = 1, a = 2, and k = 1). Sequences of this form have a closed form of b(n) = n!*a^n*x + k*Sum_{j=1..n} n!*a^(n-j)/j!. - Gary Detlefs, Mar 26 2018

a(n) = 2*n*a(n-1) + 1 with a(0) = 1.

a(n) is a specific instance of sequences having the form ab(0) = x, ab(n) = a*n*ab(n-1) + k for n >= 1. (Here x = 1, a = 2, and k = 1). Sequences of this form have a closed form of n!*a^n*x + k*sum(Sum_{j=1..n} n!*a^(n-j)/j!, j = 1..n). - Gary Detlefs, Mar 26 2018

Guo-Niu Han, <a href="/A196265/a196265.pdf">Enumeration of Standard Puzzles</a> . [Cached copy]

Guo-Niu Han, <a href="https://arxiv.org/abs/2006.14070">Enumeration of Standard Puzzles</a>, arXiv:2006.14070 [math.CO], 2020.

a(n) = floor( n! * e^(1/2) * 2^n ) = n! * Sum( _{k=0..n} 2^(n-k) / k!, k=0..n) (i.e. binomial transform of (2n)!! = n!*2^n) = n! * (e^(1/2) * 2^n - Sum( _{k >= n+1} 2^(n-k) / k!, k=n+1...)). - Michael Somos, Mar 26 1999

a(n) = A056541(n) + A000165(n). - Henry Bottomley, Jun 20 2000

E.g.f.: exp(x)/(1 - 2*x). - Vladeta Jovovic, Aug 11 2002

a(n) = Sum[_{k=0..n} P(n, k)*2^k, {where P(n,k, 0, ) = n!/(n}]-k)!. - Ross La Haye, Aug 29 2005

G.f.: 1/(1 - x -2x 2*x/(1 -2x 2*x/(1 - x -4x 4*x/(1 -4x 4*x/(1 - x -6x 6*x/(1 -6x 6*x/(1 - x -8x 8*x/(1 -8x 8*x/(1 - x -10x 10*x/(1 - ... (continued fraction).

0 = a(n)*(2*a(n+1) - 5*a(n+2) + a(n+3)) + a(n+1)*(a(n+1) + a(n+2) - a(n+3)) + a(n+2)*a(n+2) if n > -2. - Michael Somos, Jan 02 2014

a(n) + (-2*n-1)*a(n-1) + 2*(n-1)*a(n-2) = 0. - R. J. Mathar, Jan 31 2014

a(n) = Integral_{x >= 0} (2*x + 1)^n*exp(-x) dx. (Cf. A000354.)

The e.g.f. y = exp(x)/(1 - 2*x) satisfies the differential equation (1 - 2*x)*y' = (3 - 2*x)*y. _R. J. Mathar_'s recurrence above follows easily from this.

The sequence b(n) := 2^n*n! also satisfies _R. J. Mathar_'s recurrence with b(0) = 1, and b(1) = 2. This leads to the continued fraction representation a(n) = 2^n*n!*( 1 + 1/(2 - 2/(5 - 4/(7 - ... - (2*n - 2)/(2*n + 1) )))) for n >= 2. Taking the limit gives the continued fraction representation exp(1/2) = 1 + 1/(2 - 2/(5 - 4/(7 - ... - (2*n - 2)/((2*n + 1) - ... )))). (End)

a(3) = 2*3*a(2) + 1 = 6*13 + 1 = 79.

Cf. A000522, A010845, A007566, A000165, A000354, A000522, A007566, A010845, A019774, A097817, A097819.

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