%I #67 Jun 27 2023 12:01:42

%S 0,0,2,16,116,888,7416,67968,682272,7467840,88678080,1136712960,

%T 15655438080,230672171520,3621985113600,60392753971200,

%U 1065907048550400,19855856150323200,389354639411404800,8017578241634304000,172991656889856000000,3903132531903897600000

%N Number of key comparisons to sort all n! permutations of n elements by quicksort.

%C From _Petros Hadjicostas_, May 04 2020: (Start)

%C Depending on the assumptions used in the literature, the average number to sort n items in random order by quicksort appears as -C*n + 2*(1+n)*HarmonicNumber(n), where C = 2, 3, or 4. (To get the number of key comparisons to sort all n! permutations of n elements by quicksort, multiply this number by n!.)

%C For this sequence, we have C = 4. The corresponding number of key comparisons to sort all n! permutations of n elements by quicksort when C = 3 is A063090(n). Thus, a(n) = A063090(n) - n!*n.

%C For more details, see my comments and references for sequences A093418, A096620, and A115107. (End)

%H Daniel Krenn, <a href="/A288964/b288964.txt">Table of n, a(n) for n = 0..100</a>

%H Pamela E. Harris, Jan Kretschmann, and J. Carlos Martínez Mori, <a href="https://arxiv.org/abs/2306.13065">Lucky Cars and the Quicksort Algorithm</a>, arXiv:2306.13065 [math.CO], 2023.

%H <a href="/index/So#sorting">Index entries for sequences related to sorting</a>

%F a(n) = n!*(2*(n+1)*H(n) - 4*n).

%F c(n) = a(n) / n! satisfies c(n) = (n-1) + 2/n * Sum_{i < n} c(i).

%F a(n) = 2*A002538(n-1), n >= 2. - _Omar E. Pol_, Jun 20 2017

%F E.g.f.: -2*log(1-x)/(1-x)^2 - 2*x/(1-x)^2. - _Daniel Krenn_, Jun 20 2017

%F a(n) = ((2*n^2-3*n-1)*a(n-1) -(n-1)^2*n*a(n-2))/(n-2) for n >= 3, a(n) = n*(n-1) for n < 3. - _Alois P. Heinz_, Jun 21 2017

%F From _Petros Hadjicostas_, May 03 2020: (Start)

%F a(n) = A063090(n) - n!*n for n >= 1.

%F a(n) = n!*A115107(n)/A096620(n) = n!*(-n + A093418(n)/A096620(n)). (End)

%p a:= proc(n) option remember; `if`(n<3, n*(n-1),

%p ((2*n^2-3*n-1)*a(n-1)-(n-1)^2*n*a(n-2))/(n-2))

%p end:

%p seq(a(n), n=0..25); # _Alois P. Heinz_, Jun 21 2017

%t a[n_] := n! (2(n+1)HarmonicNumber[n] - 4n);

%t a /@ Range[0, 25] (* _Jean-François Alcover_, Oct 29 2020 *)

%o (SageMath) [n.factorial() * (2*(n+1)*sum(1/k for k in srange(1, n+1)) - 4*n) for n in srange(21)]

%o (SageMath) # alternative using the recurrence relation

%o @cached_function

%o def c(n):

%o if n <= 1:

%o return 0

%o return (n - 1) + 2/n*sum(c(i) for i in srange(n))

%o [n.factorial() * c(n) for n in srange(21)]

%Y Cf. A063090, A093418, A096620, A115107, A117627, A117628, A159324, A288965.

%K nonn

%O 0,3

%A _Daniel Krenn_, Jun 20 2017