

A096620


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


18



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
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OFFSET

0,4


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. 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 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_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 contating 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) = 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)


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
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 that yields numerators in A115107 and denominators in this sequence.]
Eric Weisstein's World of Mathematics, Quicksort. [He uses Havil's recurrence which yields numerators in sequence A093418 and denominators in this sequence.]
Eric Weisstein's World of Mathematics, Harmonic Number.
Wikipedia, Quicksort. [The article uses Cameron's recurrence which yields numerators in A115107 and denominators in this sequence.]


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 nth 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(1x))/(1x)^2 (due to Vladeta Jovovic, Jul 05 2004).
G.f. for fr_2(n) = E(Y_n): 2*(x + log(1x))/(1x)^2 (Cameron (1996), p. 68). (End)


EXAMPLE

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 recurrencesee 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 recurrencesee 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)


MATHEMATICA

Denominator[Table[2*(n + 1)*HarmonicNumber[n + 1]  1, {n, 0, 50}]] (* G. C. Greubel, Sep 01 2018 *)


PROG

(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
(MAGMA) [Denominator(2*(n+1)*HarmonicNumber(n+1) 1): n in [0..50]]; // G. C. Greubel, Sep 01 2018


CROSSREFS

Cf. A063090, A093418 (one set of numerators), A115107 (another set of numerators), A288964.
Essentially the same as A093419.
Sequence in context: A127780 A118413 A295320 * A093419 A160049 A331124
Adjacent sequences: A096617 A096618 A096619 * A096621 A096622 A096623


KEYWORD

nonn,frac


AUTHOR

Eric W. Weisstein, Jul 01 2004


EXTENSIONS

Offset corrected by Gary Detlefs, Sep 14 2011


STATUS

approved



