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 1-4 of 4 results.

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)).

A330895 Numerator of the variance of the random number of comparisons in quicksort applied to lists of length n.

Original entry on oeis.org

0, 0, 0, 2, 29, 46, 3049, 60574, 160599, 182789, 4913659, 1072364, 975570703, 1039388677, 5491155875, 92211937094, 6954047816459, 7225392149719, 2699717387790739, 2785123121790325, 573031978700759, 84009502802237, 45510401082365873, 46518885869845328
Offset: 0

Views

Author

Petros Hadjicostas, May 01 2020

Keywords

Examples

			The variances are: 0, 0, 0, 2/9, 29/36, 46/25, 3049/900, 60574/11025, 160599/19600, 182789/15876, 4913659/317520, 1072364/53361, ... = A330895/A330907.
		

References

  • D. E. Knuth, The Art of Computer Programming, Volume 3: Sorting and Searching, Addison-Wesley, 1973; see answer to Ex. 8(a) of Section 6.2.2.

Crossrefs

Programs

  • PARI
    lista(nn) = {my(va = vector(nn)); for(n=1, nn, va[n] = numerator(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);}

Formula

a(n) = numerator(fr(n)), where fr(n) = n*(7*n + 13) - 2*(n + 1)*Sum_{k=1..n} 1/k - 4*(n + 1)^2*Sum_{k=1..n} 1/k^2.

A334935 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.

Original entry on oeis.org

0, 0, 0, 1, 58, 3312, 219528, 17445312, 1665090432, 189515635200, 25472408256000, 4002577182720000, 728259627506688000, 152076300005855232000, 36154290403284480000000, 9714114050265240698880000, 2930311008702414783774720000, 986466808456816565267988480000, 368586443487759607372452986880000
Offset: 0

Views

Author

Petros Hadjicostas, May 16 2020

Keywords

Crossrefs

Programs

  • 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); }

Formula

a(n) = ((n!)^2/8)*A330895(n)/A330907(n).
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).
Showing 1-4 of 4 results.