A288964 Number of key comparisons to sort all n! permutations of n elements by quicksort.

0, 0, 2, 16, 116, 888, 7416, 67968, 682272, 7467840, 88678080, 1136712960, 15655438080, 230672171520, 3621985113600, 60392753971200, 1065907048550400, 19855856150323200, 389354639411404800, 8017578241634304000, 172991656889856000000, 3903132531903897600000

Offset: 0

Author

Daniel Krenn, Jun 20 2017

Keywords

• nonn

Comments

From Petros Hadjicostas, May 04 2020: (Start)

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!.)

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.

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

Links

• Daniel Krenn, Table of n, a(n) for n = 0..100
• Pamela E. Harris, Jan Kretschmann, and J. Carlos Martínez Mori, Lucky Cars and the Quicksort Algorithm, arXiv:2306.13065 [math.CO], 2023.
• Index entries for sequences related to sorting

Crossrefs

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

Programs

Maple

a:= proc(n) option remember; `if`(n<3, n*(n-1),
      ((2*n^2-3*n-1)*a(n-1)-(n-1)^2*n*a(n-2))/(n-2))
    end:
seq(a(n), n=0..25);  # Alois P. Heinz, Jun 21 2017

Mathematica

a[n_] := n! (2(n+1)HarmonicNumber[n] - 4n);
a /@ Range[0, 25] (* Jean-François Alcover, Oct 29 2020 *)

SageMath

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

SageMath

# alternative using the recurrence relation
@cached_function
def c(n):
    if n <= 1:
        return 0
    return (n - 1) + 2/n*sum(c(i) for i in srange(n))
[n.factorial() * c(n) for n in srange(21)]

Formula

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

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

a(n) = 2*A002538(n-1), n >= 2. - Omar E. Pol, Jun 20 2017

E.g.f.: -2*log(1-x)/(1-x)^2 - 2*x/(1-x)^2. - Daniel Krenn, Jun 20 2017

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

From Petros Hadjicostas, May 03 2020: (Start)

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

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