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Revision History for A032184

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A032184

a(n) = 2^n*(n-1)! for n > 1, a(1) = 1.
(history; published version)

#71 by Joerg Arndt at Thu Feb 02 02:30:11 EST 2023

STATUS

reviewed

approved

#70 by Michel Marcus at Thu Feb 02 01:50:18 EST 2023

STATUS

proposed

reviewed

#69 by Amiram Eldar at Thu Feb 02 01:06:12 EST 2023

STATUS

editing

proposed

#68 by Amiram Eldar at Thu Feb 02 00:50:00 EST 2023

LINKS

<a href="/index/Ne#necklaces">Index entries for sequences related to necklaces</a>.

#67 by Amiram Eldar at Thu Feb 02 00:49:36 EST 2023

FORMULA

Sum_{n>=1} 1/a(n) = (exp(1/2)+1)/2. - Amiram Eldar, Feb 02 2023

STATUS

approved

editing

#66 by Alois P. Heinz at Sat Mar 12 16:18:52 EST 2022

STATUS

proposed

approved

#65 by Alois P. Heinz at Thu Mar 10 09:11:00 EST 2022

STATUS

editing

proposed

#64 by Alois P. Heinz at Thu Mar 10 09:10:08 EST 2022

FORMULA

E.g.f.: -x-log(1-2*x). - Alois P. Heinz, Mar 10 2022

STATUS

proposed

editing

#63 by Joerg Arndt at Thu Mar 10 08:37:26 EST 2022

STATUS

editing

proposed

#62 by Joerg Arndt at Thu Mar 10 08:36:38 EST 2022

FORMULA

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