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The product of (n!)^2/8 and the variance of the random number of comparisons needed to sort a list of n distinct items using quicksort.
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%I #8 May 17 2020 05:22:28

%S 0,0,0,1,58,3312,219528,17445312,1665090432,189515635200,

%T 25472408256000,4002577182720000,728259627506688000,

%U 152076300005855232000,36154290403284480000000,9714114050265240698880000,2930311008702414783774720000,986466808456816565267988480000,368586443487759607372452986880000

%N The product of (n!)^2/8 and the variance of the random number of comparisons needed to sort a list of n distinct items using quicksort.

%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; see Theorem 2.

%H V. Iliopoulos, <a href="https://arxiv.org/abs/1503.02504">The quicksort algorithm and related topics</a>, arXiv:1503.02504 [cs.DS], 2015.

%H V. Iliopoulos and D. Penman, <a href="https://arxiv.org/abs/1006.4063">Variance of the number of comparisons of randomized quicksort</a>, arXiv:1006.4063 [math.PR], 2010.

%F a(n) = ((n!)^2/8)*A330895(n)/A330907(n).

%F a(n) = ((n!)^2/8)*(n*(7*n + 13) - 2*(n + 1)*Sum_{k=1..n} 1/k - 4*(n + 1)^2*Sum_{k=1..n} 1/k^2).

%o (PARI) lista(nn) = {my(va = vector(nn)); for(n=1, nn, va[n] = ((n!)^2/8)*(n*(7*n+13) - 2*(n+1)*sum(k=1, n, 1/k) - 4*(n+1)^2*sum(k=1, n, 1/k^2))); concat(0, va); }

%Y Cf. A330895, A330907.

%K nonn

%O 0,5

%A _Petros Hadjicostas_, May 16 2020