A074143 a(1) = 1; a(n) = n * Sum_{k=1..n-1} a(k).
1, 2, 9, 48, 300, 2160, 17640, 161280, 1632960, 18144000, 219542400, 2874009600, 40475635200, 610248038400, 9807557760000, 167382319104000, 3023343138816000, 57621363351552000, 1155628453883904000, 24329020081766400000, 536454892802949120000
Offset: 1
Keywords
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..200
- Milan Janjić, Enumerative Formulas for Some Functions on Finite Sets
- 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.
Crossrefs
Programs
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Magma
[Numerator(Factorial(n)/2*n): n in [1..30]]; // Vincenzo Librandi, Apr 15 2014
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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
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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, 21}] (* Vincenzo Librandi, Apr 15 2014 *)
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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
Formula
a(n) = n^2 * a(n-1)/(n-1) for n > 2.
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). - 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
From Seiichi Manyama, Apr 27 2025: (Start)
E.g.f.: x/2 * (1 + 1/(1-x)^2).
a(n) = (n+2) * a(n-1) - (n-1) * a(n-2) for n > 3. (End)
From Amiram Eldar, May 04 2025: (Start)
Extensions
More terms from Henry Bottomley, Nov 27 2002
Comments