A052468 Numerators in the Taylor series for arccosh(x) - log(2*x).

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

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

Eric W. Weisstein

Extensions

Updated by Frank Ellermann, May 22 2011

Cross-references edited by Johannes W. Meijer, Jul 05 2009

Status

approved