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A052468
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Numerators in the Taylor series for arccosh(x) - log(2*x).
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4
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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
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OFFSET
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1,2
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COMMENTS
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A055786 is the preferred version of this sequence.
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LINKS
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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
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FORMULA
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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
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EXAMPLE
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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
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MATHEMATICA
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a [n_]:=Numerator[(2 n - 1)! / (2^(2 n) n!^2)]; Array[a, 40] (* Vincenzo Librandi, Jul 10 2017 *)
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PROG
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(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
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CROSSREFS
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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
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KEYWORD
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nonn,easy,frac
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AUTHOR
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Eric W. Weisstein
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EXTENSIONS
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Updated by Frank Ellermann, May 22 2011
Cross-references edited by Johannes W. Meijer, Jul 05 2009
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STATUS
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approved
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