cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 14 results. Next

A093418 Numerator of -3*n + 2*(1+n)*HarmonicNumber(n).

Original entry on oeis.org

0, 1, 3, 17, 53, 62, 163, 717, 3489, 3727, 43391, 45596, 619313, 644063, 667229, 2756003, 24124223, 24784883, 160941559, 164719333, 33664415, 11451017, 268428987, 819836496, 20845424563, 21181779967, 193553388003, 196368607553, 5773568883787, 5849866645327
Offset: 0

Views

Author

Amarnath Murthy, Mar 30 2004

Keywords

Comments

Numerator of the average time to quicksort n items in random order.
From Petros Hadjicostas, Oct 20 2019: (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. See, for example, A115107 and A288964. Other variations of the above formula are possible.
Let X_n be the random number of comparisons needed to sort a list of numbers in the symmetric group S_n and let R_n be the rank of the pivot. According to Havil (2003, pp. 128-130), we have X_n = n + X_{R_n-1} + X_{n-R_n} because it takes 1 unit of comparison time to pick the pivot and n-1 comparisons to divide the data into two lists of numbers (less than the pivot and greater than the pivot). No matter how we pick the pivot, we have to assume that R_n is jointly independent of (X_1, ..., X_n). We let X_0 = 0.
Denoting expectation by E(.) and conditional expectation by E(.|.), we have E(X_n) = Sum_{r = 1..n} E(n + X_{R_n-1} + X_{n-R_n} | R_n=r) * P(R_n=r) = n + (1/n) * (E(X_{r-1}) + E(X_{n-r})) The last step follows from the assumed independence of R_n from (X_1, ..., X_n). This simplifies to E(X_n) = n + (2/n) * Sum_{r = 0..n-1} E(X_r). As in Havil (2003), solving the recurrence we get E(X_n) = fr(n) = -3*n + 2*(1+n)*HarmonicNumber(n). Here a(n) = numerator(E(X_n)) and A096620(n) = denominator(E(X_n)).
Note that E(X_n)*n! = (-3*n + 2*(1+n)*HarmonicNumber(n)) * n! = A063090(n), and according to the documentation of that sequence, A063090(n)/(n*n!) is the "average number of comparisons needed to find a node in a binary search tree containing n nodes inserted in a random order". See Knuth (1998, Vol. 3, pp. 430-431 and Exercise 6 on pp. 454-455).
Other authors (e.g., Cameron (1996)) do not count the choice of the pivot as a comparison time. In such a case, if we let Y_n be the modified number of comparisons used by quicksort to sort a random list of length n, we get the modified recurrence Y_n = n - 1 + Y_{R_n-1} + Y_{n-R_n}, from which we get E(Y_n) = n - 1 + (2/n) * Sum_{r = 0..n-1} E(Y_r). Solving this modified recurrence, we get E(Y_n) = -4*n + + 2*(1+n)*HarmonicNumber(n). In such a case, A115107(n) = numerator(E(Y_n)) = numerator(-4*n + 2*(1+n)*HarmonicNumber(n)) and A288964(n) = n! * E(Y_n) = n! * (-4*n + 2*(1+n)*HarmonicNumber(n)).
(End)

Examples

			fr(n) = 0, 1, 3, 17/3, 53/6, 62/5, 163/10, 717/35, 3489/140, ... = A093418/A096620.
		

References

  • Peter J. Cameron, Combinatorics: Topics, Techniques and Algorithms, Cambridge Univ. Press, 1996; see pp. 66-68.
  • J. H. Conway and R. K. Guy, The Book of Numbers. New York: Springer-Verlag, 1996, pp. 143 and 258-259.
  • Julian Havil, Gamma: Exploring Euler's constant, Princeton University Press, 2003; see pp. 128-130.
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 3, 1998; see pp. 427-431 and 454-455.

Crossrefs

The denominators in A096620 are essentially the same as the numbers in A093419, which are the denominators of A093412.

Programs

  • Magma
    [Numerator(2*(n+1)*HarmonicNumber(n) -3*n): n in [0..50]]; // G. C. Greubel, Sep 01 2018
    
  • Maple
    a := n -> numer(2*(n + 1)*(Psi(n + 1) + gamma) - 3*n);
    seq(a(n), n=0..26); # Peter Luschny, Aug 26 2019
  • Mathematica
    Numerator[Table[2*Sum[Sum[1/i, {i, 1, k}], {k, 1, n}]-n, {n, 0, 20}]] (* or *) Numerator[Table[2*Sum[HarmonicNumber[k], {k, 1, n}]-n, {n, 0, 20}]]
    Numerator[Table[2*(n+1)*HarmonicNumber[n] - 3*n, {n, 0, 50}]] (* G. C. Greubel, Sep 01 2018 *)
  • PARI
    {h(n) = sum(k=1,n,1/k)};
    for(n=0,50, print1(numerator(2*(n+1)*h(n) -3*n), ", ")) \\ G. C. Greubel, Sep 01 2018
    
  • Python
    from fractions import Fraction
    from itertools import count, islice
    def agen(): # generator of terms
        Hn = Fraction(0, 1)
        for n in count(0):
            yield (-3*n + 2*(1+n)*Hn).numerator
            Hn += Fraction(1, n+1)
    print(list(islice(agen(), 27))) # Michael S. Branicky, Apr 17 2022

Formula

G.f. for fractions fr(n): -(x+2*log(1-x))/(1-x)^2. - Vladeta Jovovic, Jul 05 2004
a(n) = 1 + (1/n!) * Sum_{k=0..n} (-1)^(n-k) * Stirling1(n, k) * (k-1) * 2^k. - Vladeta Jovovic, Jul 05 2004
a(n) = numerator(-n + 2 * H^{2}(n)), where H^{2}(n) = Sum_{k=1..n} HarmonicNumber(k) is second-order harmonic number. - Alexander Adamchuk, Nov 01 2004

Extensions

Edited by Eric W. Weisstein, Jul 01 2004
Offset changed to 0 by Petros Hadjicostas, Aug 26 2019

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

Original entry on oeis.org

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

Views

Author

Daniel Krenn, Jun 20 2017

Keywords

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)

Crossrefs

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)

A096620 Denominator of -3*n + 2*(1+n)*HarmonicNumber(n).

Original entry on oeis.org

1, 1, 1, 3, 6, 5, 10, 35, 140, 126, 1260, 1155, 13860, 12870, 12012, 45045, 360360, 340340, 2042040, 1939938, 369512, 117572, 2586584, 7436429, 178474296, 171609900, 1487285800, 1434168450, 40156716600, 38818159380, 1164544781400, 4512611027925, 2187932619600
Offset: 0

Views

Author

Eric W. Weisstein, Jul 01 2004

Keywords

Comments

Also, with initial term 0 (really this is A093419), denominator of q_n = -4*n + 2*(1+n)*HarmonicNumber(n) (Cameron). Cf. A115107.
Average time to quicksort n items in random order.
From Petros Hadjicostas, Oct 25 2019: (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. See, for example, A115107 and A288964. Other variations of the above formula are possible.
Let X_n be the random number of comparisons needed to sort a list of numbers in the symmetric group S_n and let R_n be the rank of the pivot. According to Havil (2003, pp. 128-130), we have X_n = n + X_{R_n-1} + X_{n-R_n} because it takes 1 unit of comparison time to pick the pivot and n-1 comparisons to divide the data into two lists of numbers (less than the pivot and greater than the pivot). No matter how we pick the pivot, we have to assume R_n is jointly independent of (X_1, ..., X_n). We let X_0 = 0.
Denoting expectation by E(.) and conditional expectation by E(.|.), we have E(X_n) = Sum_{r = 1..n} E(n + X_{R_n-1} + X_{n-R_n} | R_n=r) * P(R_n=r) = n + (1/n) * (E(X_{r-1}) + E(X_{n-r})) The last step follows from the assumed independence of R_n from (X_1, ..., X_n). This simplifies to E(X_n) = n + (2/n) * Sum_{r = 0..n-1} E(X_r). As in Havil (2003), solving the recurrence we get E(X_n) = fr_1(n) = -3*n + 2*(1+n)*HarmonicNumber(n). Here A093418(n) = numerator(E(X_n)) = numerator(fr_1(n)) and a(n) = denominator(E(X_n)) = denominator(fr_1(n)).
Note that E(X_n)*n! = (-3*n + 2*(1+n)*HarmonicNumber(n)) * n! = A063090(n), and according to the documentation of that sequence, A063090(n)/(n*n!) is the "average number of comparisons needed to find a node in a binary search tree containing n nodes inserted in a random order". See Knuth (1998, Vol. 3, pp. 430-431 and Exercise 6 on pp. 454-455).
Other authors (e.g., Cameron (1996)) do not count the choice of the pivot as a comparison time. In such a case, if we let Y_n be the modified number of comparisons used by quicksort to sort a random list of length n, we get the modified recurrence Y_n = n - 1 + Y_{R_n-1} + Y_{n-R_n}, from which we get E(Y_n) = n - 1 + (2/n) * Sum_{r = 0..n-1} E(Y_r). Solving this modified recurrence, we get E(Y_n) = fr_2(n) = -4*n + + 2*(1+n)*HarmonicNumber(n). In such a case, A115107(n) = numerator(E(Y_n)) = numerator(-4*n + 2*(1+n)*HarmonicNumber(n)) and A288964(n) = n! * E(Y_n) = n! * (-4*n + 2*(1+n)*HarmonicNumber(n)). In addition, a(n) = denominator(E(Y_n)) = denominator(fr_2(n)).
(End)

Examples

			Extended by _Petros Hadjicostas_, Oct 25 2019: (Start)
fr_1(n) = 0, 1, 3, 17/3, 53/6, 62/5, 163/10, 717/35, 3489/140, ... = -3*n + 2*(1+n)*HarmonicNumber(n) = A093418(n)/a(n) = A288964(n)/n! + n (Havil's recurrence, which is related to Knuth's recurrence--see comments above).
fr_2(n) = 0, 0, 1, 8/3, 29/6, 37/5, 103/10, 472/35, 2369/140, ... = -4*n + 2*(1+n)*HarmonicNumber(n) = A115107(n)/a(n) = A288964/n! (Cameron's recurrence, which is the same as Kauers and Paule's recurrence--see comments above).
Both fr_1(n) and fr_2(n) are equal to the average time to quicksort n items in random order but under slightly different assumptions.
(End)
		

References

  • Peter J. Cameron, Combinatorics: Topics, Techniques and Algorithms, Cambridge Univ. Press, 1996; see pp. 66-68.
  • J. H. Conway and R. K. Guy, The Book of Numbers. New York: Springer-Verlag, 1996, pp. 143 and 258-259.
  • Julian Havil, Gamma: Exploring Euler's constant, Princeton University Press, 2003; see pp. 128-130.
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 3, 1998; see pp. 427-431 and 454-455.

Crossrefs

Cf. A063090, A093418 (one set of numerators), A115107 (another set of numerators), A288964.
Essentially the same as A093419.

Programs

  • Magma
    [Denominator(2*(n+1)*HarmonicNumber(n+1) -1): n in [0..50]]; // G. C. Greubel, Sep 01 2018
    
  • Mathematica
    Denominator[Table[2*(n + 1)*HarmonicNumber[n + 1] - 1, {n, 0, 50}]] (* G. C. Greubel, Sep 01 2018 *)
  • PARI
    {h(n) = sum(k=1,n,1/k)};
    for(n=0,50, print1(denominator(2*(n+1)*h(n+1) -1), ", ")) \\ G. C. Greubel, Sep 01 2018
    
  • Python
    from fractions import Fraction
    from itertools import count, islice
    def agen(): # generator of terms
        Hn = Fraction(0, 1)
        for n in count(0):
            yield (-3*n + 2*(1+n)*Hn).denominator
            Hn += Fraction(1, n+1)
    print(list(islice(agen(), 30))) # Michael S. Branicky, Apr 17 2022

Formula

a(n) = Denominator(2*(n+1)*HarmonicNumber(n+1)-1). - Gary Detlefs, Sep 14 2011
a(n) = Denominator((H(n+1) + H(n))/(H(n+1) - H(n))), where H(n) is the n-th harmonic number. - Gary Detlefs, Oct 03 2011
From Petros Hadjicostas, Oct 25 2019: (Start)
G.f. for fr_1(n) = E(X_n): -(x + 2*log(1-x))/(1-x)^2 (due to Vladeta Jovovic, Jul 05 2004).
G.f. for fr_2(n) = E(Y_n): -2*(x + log(1-x))/(1-x)^2 (Cameron (1996), p. 68). (End)

Extensions

Offset corrected by Gary Detlefs, Sep 14 2011

A063090 a(n)/(n*n!) is the average number of comparisons needed to find a node in a binary search tree containing n nodes inserted in a random order.

Original entry on oeis.org

1, 6, 34, 212, 1488, 11736, 103248, 1004832, 10733760, 124966080, 1575797760, 21403457280, 311623441920, 4842481190400, 80007869491200, 1400671686758400, 25902542427955200, 504597366114508800, 10328835149402112000
Offset: 1

Views

Author

Rob Arthan, Aug 06 2001

Keywords

Comments

a(n) is the sum over all permutations, p, of {1, ..., n} of the number of comparisons required to find all the entries in the tree formed when the order of insertion is p(1), p(2), ... p(n). To derive the formula given, first group the trees according to the value of k = p(1). For a given k, p determines a permutation of {1, ..., k-1} that gives the structure of the left subtree. By symmetry, the contribution of the right subtrees will be the same as the left subtrees. Now count and simplify.
a(n) mod n is n-2 or 0 depending on whether n is prime or not. - Gary Detlefs, May 28 2012

References

  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 3, p. 427, C(n).

Crossrefs

Programs

  • Magma
    [Factorial(n)*((2*n+2)*HarmonicNumber(n) - 3*n): n in [1..30]]; // G. C. Greubel, Sep 01 2018
  • Maple
    A[1]:= 1:
    for n from 2 to 30 do A[n]:= (2*n-1)*(n-1)!+(n+1)*A[n-1] od:
    seq(A[n],n=1..30); # Robert Israel, Sep 21 2014
  • Mathematica
    a[n_] := n!*((2*n+2)*HarmonicNumber[n] - 3*n); Table[a[n], {n, 1, 20}] (* Jean-François Alcover, Sep 20 2012, after Gary Detlefs *)
  • PARI
    {h(n) = sum(k=1,n, 1/k)};
    for(n=1,30, print1(n!*(2*(n+1)*h(n) - 3*n), ", ")) \\ G. C. Greubel, Sep 01 2018
    

Formula

a(1) = 1, a(n) = n*n! + 2 * Sum_{k=1}^{n-1} (n-1)!/k! * a(k).
a(n) = (2*n - 1)*(n - 1)! + (n + 1)*a(n-1).
E.g.f.: -(x+2*log(1-x))/(1-x)^2. - Vladeta Jovovic, Sep 15 2003
a(n) = Sum_{k=0..n} |Stirling1(n, k)|*A000337(k). - Vladeta Jovovic, Jul 06 2004
a(n) = 2*(n+1)*abs(Stirling1(n+1, 2))-3*n*n!. - Vladeta Jovovic, Jul 06 2004
a(n) = n!*((2*n+2)*h(n) - 3*n), where h(n) is the n-th harmonic number. - Gary Detlefs, May 28 2012
a(n) = A288964(n) + n!*n (because this sequence and A288964 have the same definition related to quicksort but under slightly different assumptions). - Petros Hadjicostas, May 03 2020

Extensions

More terms from Vladeta Jovovic, Aug 08 2001
Missing brackets in the formula in the name inserted by Rob Arthan, Sep 21 2014

A067699 Number of comparisons made in a version of the sorting algorithm QuickSort for an array of size n with n identical elements.

Original entry on oeis.org

0, 4, 8, 14, 18, 24, 30, 38, 42, 48, 54, 62, 68, 76, 84, 94, 98, 104, 110, 118, 124, 132, 140, 150, 156, 164, 172, 182, 190, 200, 210, 222, 226, 232, 238, 246, 252, 260, 268, 278, 284, 292, 300, 310, 318, 328, 338, 350, 356, 364, 372, 382
Offset: 1

Views

Author

Karla J. Oty (oty(AT)uscolo.edu), Feb 05 2002

Keywords

References

  • Thomas H. Cormen, Charles E. Leiserson and Ronald L. Rivest. Introduction to Algorithms. McGraw-Hill Book Company, 2000. (Gives description of this version of QuickSort.)

Crossrefs

Programs

  • Python
    from functools import cache
    @cache
    def b(n): return 0 if n == 0 else b(n//2) + b((n-1)//2) + n + 2 + (n&1)
    def a(n): return b(n-1)
    print([a(n) for n in range(1, 53)]) # Michael S. Branicky, Aug 08 2022

Formula

a(n) = 2*ceiling((n+1)/2) + a(ceiling(n/2)) + a(floor(n/2)) with a(1) = 0, a(2) = 4 and a(3) = 8.
From Ralf Stephan, Oct 24 2003: (Start)
a(n) = A076178(n-1) + 4*(n-1) for n >= 1.
a(n) = b(n-1) for n >= 1, where b(0) = 0, b(2*n) = b(n) + b(n-1) + 2*n + 2, b(2*n+1) = 2*b(n) + 2*n + 4.
(End)

A330852 Numerator of the rational number A(n) that appears in the formula for the n-th cumulant k(n) = (-1)^n*2^n*(A(n) - (n - 1)!*zeta(n)) of the limiting distribution of the number of comparisons in quicksort, for n >= 2, with A(0) = 1 and A(1) = 0.

Original entry on oeis.org

1, 0, 7, 19, 937, 85981, 21096517, 7527245453, 19281922400989, 7183745930973701, 3616944955616896387, 273304346447259998403709, 76372354431694636659849988531, 25401366514997931592208126670898607, 110490677504100075544188675746430710672527, 37160853195949529205295416197788818165029489819
Offset: 0

Views

Author

Petros Hadjicostas, Apr 28 2020

Keywords

Comments

Hennequin conjectured his cumulant formula in his 1989 paper and proved it in his 1991 thesis.
First he calculates the numbers (B(n): n >= 0), with B(0) = 1 and B(0) = 0, given for p >= 0 by the recurrence
Sum_{r=0..p} Stirling1(p+2, r+1)*B(p-r)/(p-r)! + Sum_{r=0..p} F(r)*F(p-r) = 0, where F(r) = Sum_{i=0..r} Stirling1(r+1,i+1)*G(r-i) and G(k) = Sum_{a=0..k} (-1)^a*B(k-a)/(a!*(k-a)!*2^a).
Then A(n) = L_n(B(1),...,B(n)), where L_n(x_1,...,x_n) are the logarithmic polynomials of Bell.
Hoffman and Kuba (2019, 2020) gave an alternative proof of Hennequin's cumulant formula and gave an alternative calculation for the constants (-2)^n*A(n), which they denote by a_n. See also Finch (2020).
The Maple program below is based on Tan and Hadjicostas (1993), where the numbers (A(n): n >= 0) are also tabulated.
The rest of the references give the theory of the limiting distribution of the number of comparisons in quicksort (and for that reason we omit any discussion of that topic).

Examples

			The first few fractions A(n) are
1, 0, 7/4, 19/8, 937/144, 85981/3456, 21096517/172800, 7527245453/10368000, 19281922400989/3810240000, 7183745930973701/177811200000, ...
The first few fractions (-2)^n*A(n) (= a_n in Hoffman and Kuba and in Finch) are
1, 0, 7, -19, 937/9, -85981/108, 21096517/2700, -7527245453/81000, 19281922400989/14883750, -7183745930973701/347287500, ...
		

References

  • Pascal Hennequin, Analyse en moyenne d'algorithmes, tri rapide et arbres de recherche, Ph.D. Thesis, L'École Polytechnique Palaiseau (1991), p. 83.

Crossrefs

Programs

  • Maple
    # Produces the sequence (B(n): n >= 0)
    B := proc(m) option remember: local v, g, f, b:
    if m = 0 then v := 1: end if: if m = 1 then v := 0: end if:
    if 2 <= m then
    g := proc(k) add((-1)^a*B(k - a)/(a!*(k - a)!*2^a), a = 0 .. k): end proc:
    f := proc(r) add(Stirling1(r + 1, i + 1)*g(r - i), i = 0 .. r): end proc:
    b := proc(p) (-1)^p*(add(Stirling1(p + 2, r + 1)*B(p - r)/(p - r)!, r = 1 .. p) + add(f(rr)*f(p - rr), rr = 1 .. p - 1) + 2*(-1)^p*p!*add((-1)^a*B(p - a)/(a!*(p - a)!*2^a), a = 1 .. p) + 2*add(Stirling1(p + 1, i + 1)*g(p - i), i = 1 .. p))/(p - 1): end proc:
    v := simplify(b(m)): end if: v: end proc:
    # Produces the sequence (A(n): n >= 0)
    A := proc(m) option remember: local v:
    if m = 0 then v := 1: end if: if m = 1 then v := 0: end if:
    if 2 <= m then v := -(m - 1)!*add(A(k + 1)*B(m - 1 - k)/(k!*(m - 1 - k)!), k = 0 .. m - 2) + B(m): end if: v: end proc:
    # Produces the sequence of numerators of the A(n)'s
    seq(numer(A(n)), n = 0 .. 15);
  • Mathematica
    B[m_] := B[m] = Module[{v, g, f, b}, If[m == 0, v = 1]; If[m == 1, v = 0]; If[2 <= m, g[k_] := Sum[(-1)^a*B[k - a]/(a!*(k - a)!*2^a), {a, 0, k}]; f[r_] := Sum[StirlingS1[r + 1, i + 1]*g[r - i], {i, 0, r}]; b[p_] := (-1)^p*(Sum[StirlingS1[p + 2, r + 1]*B[p - r]/(p - r)!, {r, 1, p}] + Sum[f[rr]*f[p - rr], {rr, 1, p - 1}] + 2*(-1)^p*p!*Sum[(-1)^a*B[p - a]/(a!*(p - a)!*2^a), {a, 1, p}] + 2*Sum[StirlingS1[p + 1, i + 1]*g[p - i], {i, 1, p}])/(p - 1); v = Simplify[b[m]]]; v];
    A[m_] := A[m] = Module[{v}, If[ m == 0, v = 1]; If[m == 1, v = 0]; If[2 <= m , v = -(m - 1)!*Sum[A[k + 1]*B[m - 1 - k]/(k!*(m - 1 - k)!), {k , 0 , m - 2}] + B[m]]; v];
    Table[Numerator[A[n]], {n, 0, 15}] (* Jean-François Alcover, Aug 17 2020, translated from Maple *)

A330860 Denominator of the rational number A(n) that appears in the formula for the n-th cumulant k(n) = (-1)^n*2^n*(A(n) - (n - 1)!*zeta(n)) of the limiting distribution of the number of comparisons in quicksort, for n >= 2, with A(0) = 1 and A(1) = 0.

Original entry on oeis.org

1, 1, 4, 8, 144, 3456, 172800, 10368000, 3810240000, 177811200000, 9957427200000, 75278149632000000, 1912817782149120000000, 53023308921173606400000000, 17742659631203112173568000000000, 426249654980023566857797632000000000, 9600207854287580784554747166720000000000
Offset: 0

Views

Author

Petros Hadjicostas, Apr 28 2020

Keywords

Comments

Hennequin conjectured his cumulant formula in his 1989 paper and proved it in his 1991 thesis.
First he calculates the numbers (B(n): n >= 0), with B(0) = 1 and B(0) = 0, given for p >= 0 by the recurrence
Sum_{r=0..p} Stirling1(p+2, r+1)*B(p-r)/(p-r)! + Sum_{r=0..p} F(r)*F(p-r) = 0, where F(r) = Sum_{i=0..r} Stirling1(r+1,i+1)*G(r-i) and G(k) = Sum_{a=0..k} (-1)^a*B(k-a)/(a!*(k-a)!*2^a).
Then A(n) = L_n(B(1),...,B(n)), where L_n(x_1,...,x_n) are the logarithmic polynomials of Bell.
Hoffman and Kuba (2019, 2020) gave an alternative proof of Hennequin's cumulant formula and gave an alternative calculation for the constants (-2)^n*A(n), which they denote by a_n. See also Finch (2020).
The Maple program given in A330852 is based on Tan and Hadjicostas (1993), where the numbers (A(n): n >= 0) are also tabulated.
For a list of references about the theory of the limiting distribution of the number of comparisons in quicksort, which we do not discuss here, see the ones for sequence A330852.

Examples

			The first few fractions A(n) are
1, 0, 7/4, 19/8, 937/144, 85981/3456, 21096517/172800, 7527245453/10368000, 19281922400989/3810240000, 7183745930973701/177811200000, ...
The first few fractions (-2)^n*A(n) (= a_n in Hoffman and Kuba and in Finch) are
1, 0, 7, -19, 937/9, -85981/108, 21096517/2700, -7527245453/81000, 19281922400989/14883750, -7183745930973701/347287500, ...
		

References

  • Pascal Hennequin, Analyse en moyenne d'algorithmes, tri rapide et arbres de recherche, Ph.D. Thesis, L'École Polytechnique Palaiseau (1991), p. 83.

Crossrefs

Programs

  • Maple
    # The function A is defined in A330852.
    # Produces the sequence of denominators of the A(n)'s.
    seq(denom(A(n)), n = 0 .. 40);

A329001 Numerator of the rational constant term c(n) of the n-th moment mu(n) of the limiting distribution of the number of comparisons in quicksort (for n >= 0).

Original entry on oeis.org

1, 0, 7, -19, 2260, -229621, 74250517, -30532750703, 90558126238639, -37973078754146051, 21284764359226368337, -1770024989560214080011109, 539780360793818428471498394131, -194520883210026428577888559667954807, 911287963487139630688627952818633149408727
Offset: 0

Views

Author

Petros Hadjicostas, Apr 30 2020

Keywords

Comments

Let X_n be the number of comparisons needed to sort a list of n distinct numbers by quicksort. Then E(X_n) = A115107(n)/A096620(n) = -4*n + + 2*(1+n)*HarmonicNumber(n) or E(X_n) = A093418(n)/A096620(n) = -3*n + + 2*(1+n)*HarmonicNumber(n) depending on the assumptions.
Régnier (1989) and Rösler (1991) proved that (X_n - E(X_n))/n converges a.s. (and in other modes of convergence) to a non-degenerate r.v. X. Rösler proved the existence of all moments for X and Tan and Hadjicostas (1995) proved that it has a density w.r.t. to the Lebesgue measure. Fill and Janson (2000, 2002) proved many other properties of the distribution of X.
Hennequin (1989, 1991) calculated moments and cumulants of X. He proved that mu(n) = E(X^n) is a linear combination of a finite number of zeta values at positive integer arguments with rational coefficients plus a rational constant c(n). It is the numerators of this constant term c(n) of mu(n) that we tabulate here, while the denominators are in A330876.
Hennequin (1991) proved that the n-th cumulant of X for n >= 2 is k(n) = (-2)^n*(A(n) - (n-1)!*zeta(n)), where A(n) = A330852(n)/A330860(n) with A(0) = 1 and A(1) = 0. Also, (-2)^n*A(n) = A330875(n)/A330876(n); see Hoffman and Kuba (2019-2020) and Finch (2020).
Actually, Hoffman and Kuba (2019-2020, Proposition 17) express the constants c(n) in terms of "tiered binomial coefficients". Finch (2020) uses their results to write a Mathematica program with which he calculates at least c(2) - c(8). The values c(2) - c(8) have also been calculated indirectly by Ekhad and Zeilberger (2019).
Because moments and cumulants are connected via the relationship mu(n) = Sum_{s=0..n-1} binomial(n-1,s)*k(s+1)*mu(n-1-s) (e.g., see Tan and Hadjicostas (1993)), we easily deduce that c(n) = Sum_{s=0..n-1} binomial(n-1,s)*(-2)^(s+1)*A(s+1)*c(n-1-s) for n >= 1 because c(1) = A(1) = mu(1) = k(1) = 0 and k(n) = (-2)^n*A(n) - (-2)^n*(n-1)!*zeta(n) for n >= 2 and c(n) is the constant term of mu(n).

Examples

			The first few c(n) are {1, 0, 7, -19, 2260/9, -229621/108, 74250517/2700, -30532750703/81000, 90558126238639/14883750, -37973078754146051/347287500, ...}.
		

References

  • Pascal Hennequin, Analyse en moyenne d'algorithmes, tri rapide et arbres de recherche, Ph.D. Thesis, L'École Polytechnique Palaiseau (1991).

Crossrefs

Programs

  • Maple
    # The function A is defined in A330852.
    # Produces the constants c(n):
    c := proc(p) local v, a: option remember:
    if p = 0 then v := 1; end if: if p = 1 then v := 0: end if:
    if 2 <= p then v := (p - 1)!*add((-2)^(a + 1)*A(a + 1)*c(p - 1 - a)/(a!*(p - 1 - a)!), a = 0 .. p - 1): end if:
    v: end proc:
    # Produces the numerators of c(n):
    seq(numer(c(n)), n=0..15);
  • Mathematica
    (* The function A is defined in A330852. *)
    c[p_] := c[p] = Module[{v, a}, If[p == 0, v = 1;]; If[p == 1, v = 0]; If[2 <= p, v = (p - 1)!*Sum[(-2)^(a + 1)*A[a + 1]*c[p - 1 - a]/(a!*(p - 1 - a)!), {a, 0, p - 1}]]; v];
    Table[Numerator[c[n]], {n, 0, 15}] (* Jean-François Alcover, Mar 28 2021, after Maple code *)

Formula

Recurrence for the fractions: c(n) = Sum_{s=0..n-1} binomial(n-1,s)*(-2)^(s+1)*A(s+1)*c(n-1-s) for n >= 1, with c(0) = 1 = A(0) and c(1) = 0 = A(1), where A(n) = A330852(n)/A330860(n) and (-2)^n*A(n) = A330875(n)/A330876(n).

A330876 Denominator of the fraction fr(n) that appears in the n-th cumulant k(n) = fr(n) - (-2)^n*(n-1)!*zeta(n) of the limiting distribution of the number of comparisons in quicksort, for n >= 2, starting with fr(0) = 1 and fr(1) = 0.

Original entry on oeis.org

1, 1, 1, 1, 9, 108, 2700, 81000, 14883750, 347287500, 9724050000, 36756909000000, 466996528845000000, 6472571889791700000000, 1082926002881049327000000000, 13008107146607164515924000000000
Offset: 0

Views

Author

Petros Hadjicostas, Apr 29 2020

Keywords

Comments

Hennequin conjectured his cumulant formula in his 1989 paper and proved it in his 1991 thesis.
First he calculates the numbers (B(n): n >= 0), with B(0) = 1 and B(0) = 0, given for p >= 0 by the recurrence
Sum_{r=0..p} Stirling1(p+2, r+1)*B(p-r)/(p-r)! + Sum_{r=0..p} F(r)*F(p-r) = 0, where F(r) = Sum_{i=0..r} Stirling1(r+1,i+1)*G(r-i) and G(k) = Sum_{a=0..k} (-1)^a*B(k-a)/(a!*(k-a)!*2^a).
Then fr(n) = (-2)^n*L_n(B(1),...,B(n)), where L_n(x_1,...,x_n) are the logarithmic polynomials of Bell.
Hoffman and Kuba (2019, 2020) gave an alternative proof of Hennequin's cumulant formula and gave an alternative calculation for the constants fr(n), which they denote by a_n. See also Finch (2020).
Tan and Hadjicostas (1993) used a Maple program (an update of which can be found in A330852) to tabulate the numbers (fr(n)/(-2)^n: n >= 0).
Sequence A330852 contains additional references that give the theory of the limiting distribution of the number of comparisons in quicksort (and for that reason we omit any discussion of that topic).

Examples

			The first few fractions fr(n) are: 1, 0, 7, -19, 937/9, -85981/108, 21096517/2700, -7527245453/81000, 19281922400989/14883750, -7183745930973701/347287500, ...
		

References

  • Pascal Hennequin, Analyse en moyenne d'algorithmes, tri rapide et arbres de recherche, Ph.D. Thesis, L'École Polytechnique Palaiseau (1991), p. 83.

Crossrefs

Formula

a(n) = denominator((-2)^n*A330852(n)/A330860(n)).

A330875 Numerator of the fraction fr(n) that appears in the n-th cumulant k(p) = fr(n) - (-2)^n*(n-1)!*zeta(n) of the limiting distribution of the number of comparisons in quicksort, for n >= 2, starting with fr(0) = 1 and fr(1) = 0.

Original entry on oeis.org

1, 0, 7, -19, 937, -85981, 21096517, -7527245453, 19281922400989, -7183745930973701, 3616944955616896387, -273304346447259998403709, 76372354431694636659849988531, -25401366514997931592208126670898607, 110490677504100075544188675746430710672527
Offset: 0

Views

Author

Petros Hadjicostas, Apr 29 2020

Keywords

Comments

Hennequin conjectured his cumulant formula in his 1989 paper and proved it in his 1991 thesis.
First he calculates the numbers (B(n): n >= 0), with B(0) = 1 and B(0) = 0, given for p >= 0 by the recurrence
Sum_{r=0..p} Stirling1(p+2, r+1)*B(p-r)/(p-r)! + Sum_{r=0..p} F(r)*F(p-r) = 0, where F(r) = Sum_{i=0..r} Stirling1(r+1,i+1)*G(r-i) and G(k) = Sum_{a=0..k} (-1)^a*B(k-a)/(a!*(k-a)!*2^a).
Then fr(n) = (-2)^n*L_n(B(1),...,B(n)), where L_n(x_1,...,x_n) are the logarithmic polynomials of Bell.
Hoffman and Kuba (2019, 2020) gave an alternative proof of Hennequin's cumulant formula and gave an alternative calculation for the constants fr(n), which they denote by a_n. See also Finch (2020).
Tan and Hadjicostas (1993) used a Maple program (an update of which can be found in A330852) to tabulate the numbers (fr(n)/(-2)^n: n >= 0).
Sequence A330852 contains additional references that give the theory of the limiting distribution of the number of comparisons in quicksort (and for that reason we omit any discussion of that topic).

Examples

			The first few fractions fr(n) are 1, 0, 7, -19, 937/9, -85981/108, 21096517/2700, -7527245453/81000, 19281922400989/14883750, -7183745930973701/347287500, ...
		

References

  • Pascal Hennequin, Analyse en moyenne d'algorithmes, tri rapide et arbres de recherche, Ph.D. Thesis, L'École Polytechnique Palaiseau (1991), p. 83.

Crossrefs

Formula

a(n) = numerator((-2)^n*A330852(n)/A330860(n)).
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