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A032184
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a(n) = 2^n*(n-1)! for n > 1, a(1) = 1.
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12
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1, 4, 16, 96, 768, 7680, 92160, 1290240, 20643840, 371589120, 7431782400, 163499212800, 3923981107200, 102023508787200, 2856658246041600, 85699747381248000, 2742391916199936000, 93241325150797824000, 3356687705428721664000, 127554132806291423232000
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OFFSET
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1,2
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COMMENTS
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Previous name was: "CIJ" (necklace, indistinct, labeled) transform of 1, 3, 5, 7, ...
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LINKS
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FORMULA
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a(n) = 2^n*(n-1)! for n > 1, a(1) = 1.
E.g.f.: (1 + 2*x)/(1 - 2*x). - Paul Barry, May 26 2003 [This e.g.f. yields the sequence (a(n+1): n >= 0). - M. F. Hasler, Jan 15 2017]
a(n) + 2*(-n+1)*a(n-1) = 0. - R. J. Mathar, Nov 30 2012 [Valid for n >= 3; equivalently: a(n+1) = 2*n*a(n) for n > 1. - M. F. Hasler, Jan 15 2017]
G.f.: G(0) - 1, where G(k) = 1 + 1/(1 - 1/(1 + 1/((2*k + 2)*x*G(k+1)))); (continued fraction). - Sergei N. Gladkovskii, Jun 14 2013
Let s(n) = Sum_{k >= 1} 1/(2*k - 1)^n with n > 1, then s(n) = (-1)^n*PolyGamma(n-1, 1/2)/a(n). - Jean-François Alcover, Dec 18 2013
a(n) = round(-zeta(n)(1/2)) where zeta(n)(1/2) is the n-th derivative of the zeta function at 1/2. - Artur Jasinski, Feb 06 2021
Sum_{n>=1} 1/a(n) = (exp(1/2)+1)/2. - Amiram Eldar, Feb 02 2023
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MAPLE
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MATHEMATICA
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Join[{1}, Table[2^n (n-1)!, {n, 2, 20}]] (* Harvey P. Dale, Oct 08 2017 *)
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PROG
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CROSSREFS
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Apart from the initial term, same as A066318.
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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New name (using first formula) from Joerg Arndt, Mar 10 2022
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STATUS
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approved
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