

A093418


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


20



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
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OFFSET

0,3


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. 128130), we have X_n = n + X_{R_n1} + X_{nR_n} because it takes 1 unit of comparison time to pick the pivot and n1 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_n1} + X_{nR_n}  R_n=r) * P(R_n=r) = n + (1/n) * (E(X_{r1}) + E(X_{nr})) 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..n1} 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. 430431 and Exercise 6 on pp. 454455).
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_n1} + Y_{nR_n}, from which we get E(Y_n) = n  1 + (2/n) * Sum_{r = 0..n1} 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)


REFERENCES

Peter J. Cameron, Combinatorics: Topics, Techniques and Algorithms, Cambridge Univ. Press, 1996; see pp. 6668.
J. H. Conway and R. K. Guy, The Book of Numbers. New York: SpringerVerlag, 1996, pp. 143 and 258259.
Julian Havil, Gamma: Exploring Euler's constant, Princeton University Press, 2003; see pp. 128130.
D. E. Knuth, The Art of Computer Programming. AddisonWesley, Reading, MA, Vol. 3, 1998; see pp. 427431 and 454455.


LINKS

G. C. Greubel, Table of n, a(n) for n = 0..1000 [recomputed for offset 0 by Georg Fischer, Oct 13 2019]
S. B. Ekhad and D. Zeilberger, A detailed analysis of quicksort running time, arXiv:1903.03708 [math.PR], 2019. [They have the recurrence with n1, rather than n, from which they get 4*n + 2*(1+n)*HarmonicNumber(n) for the average number of comparisons.]
M. Kauers and P. Paule, The Concrete Tetrahedron, Springer, 2011; see p. 4. [They agree with Cameron's recurrence and the numerators are in A115107.]
Eric Weisstein's World of Mathematics, Quicksort. [He uses Havil's recurrence.]
Eric Weisstein's World of Mathematics, Harmonic Number.
Wikipedia, Quicksort. [Uses Cameron's recurrence and the numerators are in A115107.]


FORMULA

G.f. for fractions fr(n): (x+2*log(1x))/(1x)^2.  Vladeta Jovovic, Jul 05 2004
a(n) = 1 + (1/n!) * Sum_{k=0..n} (1)^(nk) * Stirling1(n, k) * (k1) * 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 secondorder harmonic number.  Alexander Adamchuk, Nov 01 2004


EXAMPLE

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


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 *)


PROG

(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
(MAGMA) [Numerator(2*(n+1)*HarmonicNumber(n) 3*n): n in [0..50]]; // 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


CROSSREFS

Cf. A001008, A002805, A063090, A096620, A115107, A288964.
The denominators in A096620 are essentially the same as the numbers in A093419, which are the denominators of A093412.
Sequence in context: A332869 A225727 A163943 * A173733 A294134 A258032
Adjacent sequences: A093415 A093416 A093417 * A093419 A093420 A093421


KEYWORD

nonn,frac


AUTHOR

Amarnath Murthy, Mar 30 2004


EXTENSIONS

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


STATUS

approved



