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a(n) = n!*2^(n+1) * (Integral_{x = 0..1} 1/(1+x^2)^(n+1) dx) - Pi*(2*n)!/(2^(n+1)*n!).
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%I #72 Jan 01 2024 13:08:32

%S 0,1,4,22,160,1464,16224,211632,3179520,54092160,1028113920,

%T 21594021120,496702402560,12418039065600,335293281792000,

%U 9723592350259200,301432670532403200,9947299050359193600,348155822449999872000,12881771833023700992000,502389223133024747520000

%N a(n) = n!*2^(n+1) * (Integral_{x = 0..1} 1/(1+x^2)^(n+1) dx) - Pi*(2*n)!/(2^(n+1)*n!).

%C a(n)/A001147 gives an approximation for Pi/2 with (n-1)/3 + 1 digits of accuracy. - _Aaron Kastel_, Nov 13 2012

%C a(n) is the number of linear chord diagrams on 2n vertices with one marked chord such that none of the remaining n-1 chords are excluded by (i.e., are outside and do not contain) the marked chord, see [Young]. - _Donovan Young_, Aug 11 2020

%H Seiichi Manyama, <a href="/A087547/b087547.txt">Table of n, a(n) for n = 0..404</a>

%H Richard P. Brent, Hideyuki Ohtsuka, Judy-anne H. Osborn, and Helmut Prodinger, <a href="http://arxiv.org/abs/1411.1477">Some binomial sums involving absolute values</a>, arXiv:1411.1477 [math.CO], 2014 (see page 15).

%H Sela Fried and Toufik Mansour, <a href="https://arxiv.org/abs/2304.05728">Graph labelings obtainable by random walks</a>, arXiv:2304.05728 [math.CO], 2023.

%H Sela Fried and Toufik Mansour, <a href="https://arxiv.org/abs/2305.09971">Further results on random walk labelings</a>, arXiv:2305.09971 [math.CO], 2023.

%H Donovan Young, <a href="https://arxiv.org/abs/2007.13868">A critical quartet for queuing couples</a>, arXiv:2007.13868 [math.CO], 2020.

%F a(n) = (2n-1)*a(n-1) + (n-1)!. - _Aaron Kastel_, Nov 13 2012

%F From _Peter Bala_, Jun 21 2013: (Start)

%F a(n) = (2*n)!/(n!*2^n)*(Sum_{k = 0..n-1} 2^k*k!^2/(2*k+1)!). Thus a(n)/ ((2*n)!/(n!*2^n)) -> Pi/2 as n -> infinity since Sum_{k >= 0} 2^k*k!^2/(2*k+1)! = Pi/2.

%F It appears that a(n) = Sum_{k = 1..n} 2^(k-1)*(k-1)!*(n+k-1)!/(2*k-1)!. Cf. A167571.

%F a(n) = (2*n)!/(n!*2^n)*(Pi/2) - 2^(n+1)*n!*(Integral_{x = 0..1} x^(2*n)/(1 + x^2)^(n+1) dx). Cf. A068102. (End)

%F From _Peter Bala_, Feb 18 2015: (Start)

%F Recurrence equation: a(n) = (3*n - 2)*a(n-1) - (n - 1)*(2*n - 3)*a(n-2) with a(1) = 1 and a(2) = 4.

%F The sequence b(n) = A001147(n), beginning [1, 3, 15, 105, 945, ... ], satisfies the same second-order recurrence equation. This leads to the generalized continued fraction expansion lim_{n -> infinity} a(n)/b(n) = Pi/2 = 1 + 1/(3 - 6/(7 - 15/(10 - ... - n*(2*n - 1)/((3*n + 1) - ... )))). (End)

%F E.g.f.: arctan(x/sqrt(1 - 2*x))/sqrt(1 - 2*x). - _Donovan Young_, Aug 11 2020

%F From _Sela Fried_, Apr 13 2023: (Start)

%F a(n) = (n - 1)!*Sum_{k=0..n - 1} binomial(2*n - 1, 2*k)/binomial(n - 1, k).

%F a(n)/(n-1)! = 1 + (2*n - 1)/2*Integral_{t = 0..Pi/2} (1 + sin(2*t))^(n - 1) - (1 - sin(2*t))^(n - 1)dt.

%F Bala's conjecture is true. (End)

%F a(n) ~ Pi * 2^(n - 1/2) * n^n / exp(n). - _Vaclav Kotesovec_, Apr 13 2023

%F a(n) = (n - 1)!/2 * Sum_{k=0..n - 1} binomial(2*n, 2*k + 1)/binomial(n - 1, k). - _Sela Fried_, May 18 2023

%e a(3) = 22.

%p f := proc(n) 4*n!*2^(n-1) * (int (1/(1+x^2)^(n+1),x=0..1)) - Pi*(2*n)!/(2^(n+1)*n!); end; # _N. J. A. Sloane_, Oct 30 2003

%t f[n_] := Simplify[n!*2^(n + 1)*(Integrate[ 1/(1 + x^2)^(n + 1), {x, 0, 1}]) - Pi(2n)!/(2^(n + 1)*n!)]; Table[ f[n], {n, 0, 20}] (* _Robert G. Wilson v_, Oct 31 2003 *)

%t CoefficientList[Normal[Series[1/Sqrt[1-2*x]*ArcTan[x/Sqrt[1-2*x]],{x,0,10}]]/.{x^n_.->x^n*n!},x] (* _Donovan Young_, Aug 11 2020 *)

%o (Magma) [0] cat [n eq 1 select 1 else (2*n-1)*Self(n-1)+Factorial(n-1): n in [1..25]]; // _Vincenzo Librandi_, Nov 07 2014

%o (Magma) I:=[1,4]; [0] cat [n le 2 select I[n] else (3*n-2)*Self(n-1)-(n-1)*(2*n-3)*Self(n-2): n in [1..25] ]; // _Vincenzo Librandi_, Feb 19 2015

%Y Cf. A068102, A167571, A001147, A336601.

%K nonn,easy

%O 0,3

%A Al Hakanson (hawkuu(AT)excite.com), Oct 24 2003

%E More terms from _N. J. A. Sloane_, Oct 30 2003