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 A052468 Numerators in the Taylor series for arccosh(x) - log(2*x). 4
 1, 3, 5, 35, 63, 77, 429, 6435, 12155, 46189, 88179, 676039, 1300075, 5014575, 646323, 300540195, 583401555, 756261275, 4418157975, 6892326441, 22427411435, 263012370465, 514589420475, 2687300306925, 15801325804719, 61989816618513, 121683714103007 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS A055786 is the preferred version of this sequence. LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..1000 Eric Weisstein's World of Mathematics, Inverse Hyperbolic Secant Eric Weisstein's World of Mathematics, Inverse Hyperbolic Cosecant Eric Weisstein's World of Mathematics, Inverse Hyperbolic Cosine Eric Weisstein's World of Mathematics, Inverse Hyperbolic Sine FORMULA 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 ). a(n) = numerator((2*n-1)!/(4^n * (n!)^2)). - Johannes W. Meijer, Jul 06 2009 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 a(n) = numerator(binomial(2n,n)/(n*2^(2n-1))). - Daniel Suteu, Oct 30 2017 EXAMPLE 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 + ... 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 MATHEMATICA a [n_]:=Numerator[(2 n - 1)! / (2^(2 n) n!^2)]; Array[a, 40] (* Vincenzo Librandi, Jul 10 2017 *) PROG (MAGMA) [Numerator(Factorial(2*n-1)/( 2^(2*n)* Factorial(n)^2)): n in [1..30]]; // Vincenzo Librandi, Jul 10 2017 (PARI) {a(n) = numerator((2*n-1)!/(4^n*(n!)^2))}; \\ G. C. Greubel, May 18 2019 (Sage) [numerator(factorial(2*n-1)/(4^n*(factorial(n))^2)) for n in (1..30)] # G. C. Greubel, May 18 2019 (GAP) List([1..30], n-> NumeratorRat( Factorial(2*n-1)/(4^n*(Factorial(n))^2) )) # G. C. Greubel, May 18 2019 CROSSREFS See A055786 for further information. a(n)/A052469(n) = (1/(2*n))*A001790(n)/A046161(n) for n=>1. Equals A162441(n+1)/(2n+1) for n=>1. - Johannes W. Meijer, Jul 06 2009 Sequence in context: A261659 A346715 A259853 * A055786 A001790 A173092 Adjacent sequences:  A052465 A052466 A052467 * A052469 A052470 A052471 KEYWORD nonn,easy,frac AUTHOR EXTENSIONS Updated by Frank Ellermann, May 22 2011 Cross-references edited by Johannes W. Meijer, Jul 05 2009 STATUS approved

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Last modified July 2 14:02 EDT 2022. Contains 355007 sequences. (Running on oeis4.)