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%I #66 Sep 08 2022 08:45:14
%S 1,1,1,3,6,5,10,35,140,126,1260,1155,13860,12870,12012,45045,360360,
%T 340340,2042040,1939938,369512,117572,2586584,7436429,178474296,
%U 171609900,1487285800,1434168450,40156716600,38818159380,1164544781400,4512611027925,2187932619600
%N Denominator of -3*n + 2*(1+n)*HarmonicNumber(n).
%C Also, with initial term 0 (really this is A093419), denominator of q_n = -4*n + 2*(1+n)*HarmonicNumber(n) (Cameron). Cf. A115107.
%C Average time to quicksort n items in random order.
%C From _Petros Hadjicostas_, Oct 25 2019: (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. See, for example, A115107 and A288964. Other variations of the above formula are possible.
%C 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.
%C 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)).
%C 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).
%C 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)).
%C (End)
%D Peter J. Cameron, Combinatorics: Topics, Techniques and Algorithms, Cambridge Univ. Press, 1996; see pp. 66-68.
%D J. H. Conway and R. K. Guy, The Book of Numbers. New York: Springer-Verlag, 1996, pp. 143 and 258-259.
%D Julian Havil, Gamma: Exploring Euler's constant, Princeton University Press, 2003; see pp. 128-130.
%D D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 3, 1998; see pp. 427-431 and 454-455.
%H G. C. Greubel, <a href="/A096620/b096620.txt">Table of n, a(n) for n = 0..1000</a>
%H S. B. Ekhad and D. Zeilberger, <a href="https://arxiv.org/abs/1903.03708">A detailed analysis of quicksort running time</a>, arXiv:1903.03708 [math.PR], 2019. [They have the recurrence with n-1, rather than n, from which they get -4*n + 2*(1+n)*HarmonicNumber(n) for the average number of comparisons.]
%H M. Kauers and P. Paule, <a href="https://doi.org/10.1007/978-3-7091-0445-3">The Concrete Tetrahedron</a>, Springer, 2011; see p. 4. [They agree with Cameron's recurrence that yields numerators in A115107 and denominators in this sequence.]
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Quicksort.html">Quicksort</a>. [He uses Havil's recurrence which yields numerators in sequence A093418 and denominators in this sequence.]
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HarmonicNumber.html">Harmonic Number</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Quicksort">Quicksort</a>. [The article uses Cameron's recurrence which yields numerators in A115107 and denominators in this sequence.]
%F a(n) = Denominator(2*(n+1)*HarmonicNumber(n+1)-1). - _Gary Detlefs_, Sep 14 2011
%F 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
%F From _Petros Hadjicostas_, Oct 25 2019: (Start)
%F G.f. for fr_1(n) = E(X_n): -(x + 2*log(1-x))/(1-x)^2 (due to _Vladeta Jovovic_, Jul 05 2004).
%F G.f. for fr_2(n) = E(Y_n): -2*(x + log(1-x))/(1-x)^2 (Cameron (1996), p. 68). (End)
%e Extended by _Petros Hadjicostas_, Oct 25 2019: (Start)
%e 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).
%e 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).
%e 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.
%e (End)
%t Denominator[Table[2*(n + 1)*HarmonicNumber[n + 1] - 1, {n, 0, 50}]] (* _G. C. Greubel_, Sep 01 2018 *)
%o (PARI) {h(n) = sum(k=1,n,1/k)};
%o for(n=0,50, print1(denominator(2*(n+1)*h(n+1) -1), ", ")) \\ _G. C. Greubel_, Sep 01 2018
%o (Magma) [Denominator(2*(n+1)*HarmonicNumber(n+1) -1): n in [0..50]]; // _G. C. Greubel_, Sep 01 2018
%o (Python)
%o from fractions import Fraction
%o from itertools import count, islice
%o def agen(): # generator of terms
%o Hn = Fraction(0, 1)
%o for n in count(0):
%o yield (-3*n + 2*(1+n)*Hn).denominator
%o Hn += Fraction(1, n+1)
%o print(list(islice(agen(), 30))) # _Michael S. Branicky_, Apr 17 2022
%Y Cf. A063090, A093418 (one set of numerators), A115107 (another set of numerators), A288964.
%Y Essentially the same as A093419.
%K nonn,frac
%O 0,4
%A _Eric W. Weisstein_, Jul 01 2004
%E Offset corrected by _Gary Detlefs_, Sep 14 2011
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