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%I #45 Sep 08 2022 08:44:59
%S 1,3,5,35,63,77,429,6435,12155,46189,88179,676039,1300075,5014575,
%T 646323,300540195,583401555,756261275,4418157975,6892326441,
%U 22427411435,263012370465,514589420475,2687300306925,15801325804719,61989816618513,121683714103007
%N Numerators in the Taylor series for arccosh(x) - log(2*x).
%C A055786 is the preferred version of this sequence.
%H Vincenzo Librandi, <a href="/A052468/b052468.txt">Table of n, a(n) for n = 1..1000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/InverseHyperbolicSecant.html">Inverse Hyperbolic Secant</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/InverseHyperbolicCosecant.html">Inverse Hyperbolic Cosecant</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/InverseHyperbolicCosine.html">Inverse Hyperbolic Cosine</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/InverseHyperbolicSine.html">Inverse Hyperbolic Sine</a>
%F a(n)/A052469(n) = A001147(n)/(A000165(n)*2*n). E.g., a(6) = 77 = 1*3*5*7*9*11 / gcd( 1*3*5*7*9*11, 2*4*6*8*10*12*12 ).
%F a(n) = numerator((2*n-1)!/(4^n * (n!)^2)). - _Johannes W. Meijer_, Jul 06 2009
%F Let z(n) = 2*(2*n+1)!*4^(-n-1)/((n+1)!)^2, then a(n) = numerator(z(n)), A162442(n) = denominator(z(n)), and z(n) = 1/(n+1) - Sum_{k=0..n}(-1)^k*binomial(n,k)*z(k). - _Groux Roland_, Jan 04 2011
%F a(n) = numerator(binomial(2n,n)/(n*2^(2n-1))). - _Daniel Suteu_, Oct 30 2017
%e i*Pi/2 - arccosh(x) = i*x + (1/6)*i*x^3 + (3/40)*i*x^5 + (5/112)*i*x^7 + (35/1152)*i*x^9 + (63/2816)*i*x^11 + (231/13312)*i*x^13 + (143/10240)*i*x^15 + (6435/557056)*i*x^17 + ...
%e 0, 1, 0, 1/6, 0, 3/40, 0, 5/112, 0, 35/1152, 0, 63/2816, 0, 231/13312, 0, 143/10240, 0, 6435/557056, 0, 12155/1245184, 0, 46189/5505024, 0, ... = A052468/A052469
%t a [n_]:=Numerator[(2 n - 1)! / (2^(2 n) n!^2)]; Array[a, 40] (* _Vincenzo Librandi_, Jul 10 2017 *)
%o (Magma) [Numerator(Factorial(2*n-1)/( 2^(2*n)* Factorial(n)^2)): n in [1..30]]; // _Vincenzo Librandi_, Jul 10 2017
%o (PARI) {a(n) = numerator((2*n-1)!/(4^n*(n!)^2))}; \\ _G. C. Greubel_, May 18 2019
%o (Sage) [numerator(factorial(2*n-1)/(4^n*(factorial(n))^2)) for n in (1..30)] # _G. C. Greubel_, May 18 2019
%o (GAP) List([1..30], n-> NumeratorRat( Factorial(2*n-1)/(4^n*(Factorial(n))^2) )) # _G. C. Greubel_, May 18 2019
%Y See A055786 for further information.
%Y a(n)/A052469(n) = (1/(2*n))*A001790(n)/A046161(n) for n=>1.
%Y Equals A162441(n+1)/(2n+1) for n=>1. - _Johannes W. Meijer_, Jul 06 2009
%K nonn,easy,frac
%O 1,2
%A _Eric W. Weisstein_
%E Updated by _Frank Ellermann_, May 22 2011
%E Cross-references edited by _Johannes W. Meijer_, Jul 05 2009
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