login
A000140
Kendall-Mann numbers: the most common number of inversions in a permutation on n letters is floor(n*(n-1)/4); a(n) is the number of permutations with this many inversions.
(Formerly M1665 N0655)
21
1, 1, 2, 6, 22, 101, 573, 3836, 29228, 250749, 2409581, 25598186, 296643390, 3727542188, 50626553988, 738680521142, 11501573822788, 190418421447330, 3344822488498265, 62119523114983224, 1214967840930909302, 24965661442811799655, 538134522243713149122
OFFSET
1,3
COMMENTS
Row maxima of A008302, see example.
The term a(0) would be 1: the empty product is one and there is just one coefficient 1=x^0, corresponding to the 1 empty permutation (which has 0 inversions).
From Ryen Lapham and Anant Godbole, Dec 12 2006: (Start)
Also, the number of permutations on {1,2,...,n} for which the number A of monotone increasing subsequences of length 2 and the number D of monotone decreasing 2-subsequences are as close to each other as possible, i.e., 0 or 1. We call such permutations 2-balanced.
If 4|n(n-1) then (with A and D as above) the feasible values of A-D are C(n,2), C(n,2)-2,...,2,0,-2,...,-C(n,2), whereas if 4 does not divide n(n-1), A-D may equal C(n,2), C(n,2)-2,...,1,-1,...,-C(n,2). Let a_n(i) equal the number of permutations with A-D the i-th highest feasible value.
The sequence in question gives the number of permutations for which A-D=0 or A-D=1, i.e., it equals A_n(j) where j = floor((binomial(n,2)+2)/2). Here is the recursion: a_n(i) = a_n(i-1) + a_{n-1}(i) for 1 <= i <= n and a_n(n+k) = a_n(n+k-1) + a_{n-1}(n+k) - a_n(k) for k >= 1. (End)
The only two primes found < 301 are for n = 3 and 6.
Define an ordered list to have n terms with terms t(k) for k=1..n. Specify that t(k) ranges from 1 to k, hence the third term t(3) can be 1, 2, or 3. Find all sums of the terms for all n! allowable arrangements to obtain a maximum sum for the greatest number of arrangements. This number is a(n). For n=4, the maximum sum 7 appears in 6 arrangements: 1114, 1123, 1213, 1222, 1231, and 1132. - J. M. Bergot, May 14 2015
Named after the British statistician Maurice George Kendall (1907-1983) and the Austrian-American mathematician Henry Berthold Mann (1905-2000). - Amiram Eldar, Apr 07 2023
REFERENCES
F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 241.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Robert Israel, Robert G. Wilson v and N. J. A. Sloane Table of n, a(n) for n = 1..400 (a(1) to a(61) from Sloane, a(62) to a(350) from Wilson).
Dorin Andrica and Ovidiu Bagdasar, On some results concerning the polygonal polynomials, Carpathian Journal of Mathematics (2019) Vol. 35, No. 1, 1-11.
Dominique Foata, Distributions eulériennes et mahoniennes sur le groupe des permutations, pp. 27-49 of M. Aigner, editor, Higher Combinatorics, Reidel, Dordrecht, Holland, 1977.
Mikhail Gaichenkov, The property of Kendall-Mann numbers, MathOverflow, 2010.
A. Waksman, On the complexity of inversions, IEEE Trans. Computers, 19 (1970), 1225-1226.
Wikipedia, Inversion.
Wikipedia, q-Pochhammer symbol. - Paul Muljadi, Jan 18 2011
FORMULA
Largest coefficient of (1)(x+1)(x^2+x+1)...(x^(n-1) + ... + x + 1). - David W. Wilson
The number of terms is given in A000124.
a(n+1)/a(n) = n - 1/2 + O(1/n^{1-epsilon}) as n --> infinity (compare with A008302, A181609, A001147). - Mikhail Gaichenkov, Apr 11 2014
Asymptotics (Mikhail Gaichenkov, 2010): a(n) ~ 6 * n^(n-1) / exp(n). - Vaclav Kotesovec, May 17 2015
EXAMPLE
From Joerg Arndt, Jan 16 2011: (Start)
a(4) = 6 because the among the permutations of 4 elements those with 3 inversions are the most frequent and appear 6 times:
[inv. table] [permutation] number of inversions
0: [ 0 0 0 ] [ 0 1 2 3 ] 0
1: [ 1 0 0 ] [ 1 0 2 3 ] 1
2: [ 0 1 0 ] [ 0 2 1 3 ] 1
3: [ 1 1 0 ] [ 2 0 1 3 ] 2
4: [ 0 2 0 ] [ 1 2 0 3 ] 2
5: [ 1 2 0 ] [ 2 1 0 3 ] 3 (*)
6: [ 0 0 1 ] [ 0 1 3 2 ] 1
7: [ 1 0 1 ] [ 1 0 3 2 ] 2
8: [ 0 1 1 ] [ 0 3 1 2 ] 2
9: [ 1 1 1 ] [ 3 0 1 2 ] 3 (*)
10: [ 0 2 1 ] [ 1 3 0 2 ] 3 (*)
11: [ 1 2 1 ] [ 3 1 0 2 ] 4
12: [ 0 0 2 ] [ 0 2 3 1 ] 2
13: [ 1 0 2 ] [ 2 0 3 1 ] 3 (*)
14: [ 0 1 2 ] [ 0 3 2 1 ] 3 (*)
15: [ 1 1 2 ] [ 3 0 2 1 ] 4
16: [ 0 2 2 ] [ 2 3 0 1 ] 4
17: [ 1 2 2 ] [ 3 2 0 1 ] 5
18: [ 0 0 3 ] [ 1 2 3 0 ] 3 (*)
19: [ 1 0 3 ] [ 2 1 3 0 ] 4
20: [ 0 1 3 ] [ 1 3 2 0 ] 4
21: [ 1 1 3 ] [ 3 1 2 0 ] 5
22: [ 0 2 3 ] [ 2 3 1 0 ] 5
23: [ 1 2 3 ] [ 3 2 1 0 ] 6
The statistics are reflected by the coefficients of the polynomial
(1+x)*(1+x+x^2)*(1+x+x^2+x^3) ==
x^6 + 3*x^5 + 5*x^4 + 6*x^3 + 5*x^2 + 3*x^1 + 1*x^0
There is 1 permutation (the identity) with 0 inversions,
3 permutations with 1 inversion, 5 with 2 inversions,
6 with 3 inversions (the most frequent, marked with (*) ), 5 with 4 inversions, 3 with 5 inversions, and one with 6 inversions. (End)
G.f. = x + x^2 + 2*x^3 + 6*x^4 + 22*x^5 + 101*x^6 + 573*x^7 + 3836*x^8 + ...
MAPLE
f := 1: for n from 0 to 40 do f := f*add(x^i, i=0..n): s := series(f, x, n*(n+1)/2+1): m := max(coeff(s, x, j) $ j=0..n*(n+1)/2): printf(`%d, `, m) od: # James A. Sellers, Dec 07 2000 [offset is off by 1 - N. J. A. Sloane, May 23 2006]
P:= [1]: a[1]:= 1:
for n from 2 to 100 do
P:= expand(P * add(x^j, j=0..n-1));
a[n]:= max(eval(convert(P, list), x=1));
od:
seq(a[i], i=1..100); # Robert Israel, Dec 14 2014
MATHEMATICA
f[n_] := Max@ CoefficientList[ Expand@ Product[ Sum[x^i, {i, 0, j}], {j, n-1}], x]; Array[f, 20]
Flatten[{1, 1, Table[Coefficient[Expand[Product[Sum[x^k, {k, 0, m-1}], {m, 1, n}]], x^Floor[n*(n-1)/4]], {n, 3, 20}]}] (* Vaclav Kotesovec, May 13 2016 *)
Table[SeriesCoefficient[QPochhammer[x, x, n]/(1-x)^n, {x, 0, Floor[n*(n-1)/4]}], {n, 1, 20}] (* Vaclav Kotesovec, May 13 2016 *)
PROG
(Magma) /* based on David W. Wilson's formula */ PS<x>:=PowerSeriesRing(Integers()); [ Max(Coefficients(&*[&+[ x^i: i in [0..j] ]: j in [0..n-1] ])): n in [1..21] ]; // Klaus Brockhaus, Jan 18 2011
(PARI) {a(n) = if( n<0, 0, vecmax( Vec( prod(k=1, n, 1 - x^k) / (1 - x)^n)))}; /* Michael Somos, Apr 21 2014 */
(Python)
from math import prod
from sympy import Poly
from sympy.abc import x
def A000140(n): return 1 if n == 1 else max(Poly(prod(sum(x**i for i in range(j+1)) for j in range(n))).all_coeffs()) # Chai Wah Wu, Feb 02 2022
CROSSREFS
Row maxima of A008302.
Odd terms are A186888.
Sequence in context: A012268 A009655 A002772 * A079263 A129815 A345194
KEYWORD
nonn,easy,core,nice
EXTENSIONS
Edited by N. J. A. Sloane, Mar 05 2011
STATUS
approved