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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
OFFSET
1,2
COMMENTS
A055786 is the preferred version of this sequence.
LINKS
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
KEYWORD
nonn,easy,frac
EXTENSIONS
Updated by Frank Ellermann, May 22 2011
Cross-references edited by Johannes W. Meijer, Jul 05 2009
STATUS
approved