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A074143
a(1) = 1; a(n) = n * Sum_{k=1..n-1} a(k).
13
1, 2, 9, 48, 300, 2160, 17640, 161280, 1632960, 18144000, 219542400, 2874009600, 40475635200, 610248038400, 9807557760000, 167382319104000, 3023343138816000, 57621363351552000, 1155628453883904000, 24329020081766400000, 536454892802949120000
OFFSET
1,2
COMMENTS
a(n) is also the number of elements of the alternating semigroup (A^c_n) for F(n, p) if p = n - 1 (cf. A001710). - Bakare Gatta Naimat, Jan 15 2016
LINKS
Stephen Lipscomb, Symmetric inverse semigroups, Mathematical surveys and monographs, Vol.46 Amer. Math. Soc. (1996).
Michael Penn, Australian Mathematical Olympiad 2018 Question 5, Youtube video, 2020.
FORMULA
a(n) = n^2*a(n-1)/(n-1) for n > 2.
a(n) = n*ceiling[n!/2] = n*A001710(n-1) = A001710(n+1)-A001710(n) = ceiling[A001563(n)/2] - Henry Bottomley, Nov 27 2002
a(n) = ((n+1)!-n!)/2 for n > 1. - Vladimir Joseph Stephan Orlovsky, Apr 03 2011
G.f.: (U(0) + x)/(2*x) where U(k)= 1 - 1/(k+1 - x*(k+1)^2*(k+2)/(x*(k+1)*(k+2) - 1/U(k+1))) ; (continued fraction, 3-step). - Sergei N. Gladkovskii, Sep 27 2012
G.f.: 1/2 + Q(0), where Q(k)= 1 - 1/(k+2 - x*(k+2)^2*(k+3)/(x*(k+2)*(k+3)-1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Apr 19 2013
a(n) = sum(j = 0..n, (-1)^(n-j)*binomial(n, j)*(j)^(n+1))/(n+1), n > 1, a(1) = 1. - Vladimir Kruchinin, Jun 01 2013
a(n) = numerator(n!/2*n). - Vincenzo Librandi, Apr 15 2014
a(n) is F(n;p) = n^2(n-1)!/2 if p = n-1 in A^c_n. For instance for n=4 and p=n-1: F(4; 4-1)= 4^2(4-1)!/2 = 16*6/2 = 48. - Bakare Gatta Naimat, Nov 18 2015
MAPLE
seq(sum(mul(j, j=3..n), k=1..n), n=1..19); # Zerinvary Lajos, Jun 01 2007
a := n -> `if`(n=1, 1, n!*n/2): seq(a(n), n=1..19); # Peter Luschny, Jan 22 2016
MATHEMATICA
A074143[1] = 1; A074143[n_] := A074143[n] = n * Sum[a[k], {k, n - 1}]; Array[A074143, 20] (* T. D. Noe, Apr 05 2011 *)
Table[Numerator[n!/2 n], {n, 40}] (* Vincenzo Librandi, Apr 15 2014)
PROG
(Magma) [Numerator(Factorial(n)/2*n): n in [1..30]]; // Vincenzo Librandi, Apr 15 2014
(SageMath)
def b(n): return 1/2 if (n==1) else n^2*b(n-1)/(n-1)
def A074143(n): return b(n) + int(n==1)/2
[A074143(n) for n in range(1, 41)] # G. C. Greubel, Nov 29 2022
CROSSREFS
A diagonal of A254040.
Sequence in context: A368961 A100427 A214404 * A198892 A357790 A205571
KEYWORD
nonn
AUTHOR
Amarnath Murthy, Aug 28 2002
EXTENSIONS
More terms from Henry Bottomley, Nov 27 2002
STATUS
approved