A128566 -id:A128566 - cp's OEIS frontend

A281425 a(n) = [q^n] (1 - q)^n / Product_{j=1..n} (1 - q^j).

1, 0, 1, -1, 2, -4, 9, -21, 49, -112, 249, -539, 1143, -2396, 5013, -10550, 22420, -48086, 103703, -223806, 481388, -1029507, 2187944, -4625058, 9742223, -20490753, 43111808, -90840465, 191773014, -405523635, 858378825, -1817304609, 3845492204, -8129023694, 17162802918, -36191083386

Offset: 0

Ilya Gutkovskiy, Oct 05 2017

sign

a(n) is n-th term of the Euler transform of -n + 1, 1, 1, 1, ...

Inverse zero-based binomial transform of A000041. The version for strict partitions is A380412, or A293467 up to sign. - Gus Wiseman, Feb 06 2025

Vaclav Kotesovec, Table of n, a(n) for n = 0..3000
A. M. Odlyzko, Differences of the partition function, Acta Arithmetica 49.3 (1988): 237-254.
Dennis Stanton and Doron Zeilberger, The Odlyzko conjecture and O’Hara’s unimodality proof, Proceedings of the American Mathematical Society 107.1 (1989): 39-42.

Cf. A128566, A292463, A292541, A292613.
Cf. A000041, A002865, A053445, A275638, A275639, A275640, A275641, A275642, A275643, A275644.
Cf. A218481, A294466, A095051.
Cf. A266232, A294467, A293467, A294468.
Row n=0 of A175804 (row n=1 is A320590 except first term).
Cf. A000009, A008284, A087897, A116608, A129519, A225486, A320591, A377056, A378622.

b:= proc(n, k) option remember; `if`(k=0,
      combinat[numbpart](n), b(n, k-1)-b(n-1, k-1))
    end:
a:= n-> b(n$2):
seq(a(n), n=0..35);  # Alois P. Heinz, Dec 21 2024

Table[SeriesCoefficient[(1 - q)^n / Product[(1 - q^j), {j, 1, n}], {q, 0, n}], {n, 0, 35}]
Table[SeriesCoefficient[(1 - q)^n QPochhammer[q^(1 + n), q]/QPochhammer[q, q], {q, 0, n}], {n, 0, 35}]
Table[SeriesCoefficient[1/QFactorial[n, q], {q, 0, n}], {n, 0, 35}]
Table[Differences[PartitionsP[Range[0, n]], n], {n, 0, 35}] // Flatten
Table[Sum[(-1)^j*Binomial[n, j]*PartitionsP[n-j], {j, 0, n}], {n, 0, 30}] (* Vaclav Kotesovec, Oct 06 2017 *)

a(n) = [q^n] 1/((1 + q)*(1 + q + q^2)*...*(1 + q + ... + q^(n-1))).
a(n) = Sum_{j=0..n} (-1)^j * binomial(n, j) * A000041(n-j). - Vaclav Kotesovec, Oct 06 2017
a(n) ~ (-1)^n * 2^(n - 3/2) * exp(Pi*sqrt(n/12) + Pi^2/96) / (sqrt(3)*n). - Vaclav Kotesovec, May 07 2018

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.

Original entry on oeis.org

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

N. J. A. Sloane

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

			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 + ...

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).

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.
Richard K. Guy, Letter to N. J. A. Sloane with attachment, Mar 1988.
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
Index entries for "core" sequences.

Row maxima of A008302.
Odd terms are A186888.
Cf. A000124, A001147, A128566, A181609.

/* based on David W. Wilson's formula */ PS:=PowerSeriesRing(Integers()); [ Max(Coefficients(&*[&+[ x^i: i in [0..j] ]: j in [0..n-1] ])): n in [1..21] ]; // Klaus Brockhaus, Jan 18 2011

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 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

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 *)

{a(n) = if( n<0, 0, vecmax( Vec( prod(k=1, n, 1 - x^k) / (1 - x)^n)))}; /* Michael Somos, Apr 21 2014 */

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

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

Edited by N. J. A. Sloane, Mar 05 2011

A128564 Triangle, read by rows, where T(n,k) equals the number of permutations of {1..n+1} with [(nk+k)/2] inversions for n>=k>=0.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 5, 5, 1, 1, 9, 22, 15, 1, 1, 29, 90, 90, 29, 1, 1, 49, 359, 573, 359, 98, 1, 1, 174, 1415, 3450, 3450, 1415, 174, 1, 1, 285, 5545, 17957, 29228, 21450, 5545, 628, 1, 1, 1068, 21670, 110010, 230131, 230131, 110010, 21670, 1068, 1

Offset: 0

Paul D. Hanna, Mar 12 2007

Row sums equal 2*n! for n>0.

			Row sums equal 2*n! for n>0:
[1, 2, 4, 12, 48, 240, 1440, 10080, 80640, ..., 2*n!,...].
Triangle begins:
  1;
  1,    1;
  1,    2,     1;
  1,    5,     5,      1;
  1,    9,    22,     15,       1;
  1,   29,    90,     90,      29,       1;
  1,   49,   359,    573,     359,      98,       1;
  1,  174,  1415,   3450,    3450,    1415,     174,      1;
  1,  285,  5545,  17957,   29228,   21450,    5545,    628,     1;
  1, 1068, 21670, 110010,  230131,  230131,  110010,  21670,  1068,    1;
  1, 1717, 84591, 526724, 1729808, 2409581, 1729808, 686763, 84591, 4015, 1;
  ...

Alois P. Heinz, Rows n = 0..140, flattened

Cf. A008302 (Mahonian numbers); A128565 (column 1), A128566 (column 2).

Row sums give A098558.

b:= proc(u, o) option remember; expand(`if`(u+o=0, 1,
       add(b(u+j-1, o-j)*x^(u+j-1), j=1..o)+
       add(b(u-j, o+j-1)*x^(u-j), j=1..u)))
    end:
T:= (n, k)-> coeff(b(n+1, 0), x, iquo((n+1)*k, 2)):
seq(seq(T(n,k), k=0..n), n=0..10);  # Alois P. Heinz, May 02 2017

b[u_, o_] := b[u, o] = Expand[If[u + o == 0, 1, Sum[b[u + j - 1, o - j]* x^(u+j-1), {j, 1, o}] + Sum[b[u-j, o+j-1]*x^(u-j), {j, 1, u}]]];
T[n_, k_] := Coefficient[b[n+1, 0], x, Quotient[(n+1)*k, 2]];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* Jean-François Alcover, Dec 06 2019, after Alois P. Heinz *)

{T(n,k)=local(faq=prod(j=1, n+1, (1-q^j)/(1-q))); polcoeff(faq, (n*k+k)\2, q)}

T(n,k) = A008302(n+1, [(nk+k)/2]) = coefficient of q^[(nk+k)/2] in the q-factorial of n+1 for n>=0.

A214086 Number of involutions of length n having exactly n inversions.

Original entry on oeis.org

1, 0, 0, 1, 1, 2, 10, 17, 39, 119, 254, 613, 1623, 3791, 9281, 23469, 56823, 140035, 349167, 857868, 2122297, 5274213, 13044870, 32357838, 80421284, 199613489, 496144310, 1234420850, 3070773600, 7644879181, 19044266176, 47450853932, 118290412077, 295019269801

Offset: 0

Geoffrey Critzer and Alois P. Heinz, Mar 06 2013

nonn

			a(0) = 1: (), the empty involution.
a(3) = 1: (3,2,1); inversions are (3,2), (3,1), (2,1).
a(4) = 1: (3,4,1,2); inversions are (3,1), (3,2), (4,1), (4,2).
a(5) = 2: (1,5,3,4,2), (4,2,3,1,5).
a(6) = 10: (1,2,6,5,4,3), (1,4,6,2,5,3), (1,5,3,6,2,4), (1,5,4,3,2,6), (2,1,6,4,5,3), (3,2,1,6,5,4), (3,5,1,4,2,6), (4,2,3,1,6,5), (4,2,5,1,3,6), (4,3,2,1,5,6).

Alois P. Heinz, Table of n, a(n) for n = 0..2000
Wikipedia, Inversion
Wikipedia, Involution

Diagonal of A213910.

Cf. A128566.

T:= proc(n, k) option remember; `if`(n<0 or k<0, 0, `if`(k=0, 1,
      T(n-1, k) + add(T(n-2, k-2*(n-j)+1), j=1..n-1)))
    end:
a:= n-> T(n$2):
seq(a(n), n=0..50);

Needs["Combinatorica`"];
Table[Count[Map[Inversions,Involutions[n]],n],{n,0,12}]
(* Second program: *)
T[n_, k_] := T[n, k] = If[n < 0 || k < 0, 0, If[k == 0, 1, T[n-1, k] + Sum[T[n-2, k-2*(n-j)+1], {j, 1, n-1}]]];
a[n_] := T[n, n];
Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Dec 04 2023, after Maple code *)

a(n) = T(n,n) with T(n,k) = T(n-1,k) + Sum_{j=1..n-1} T(n-2,k-2*(n-j)+1) for n>=0, k>0; T(n,k) = 0 for n<0 or k<0; T(n,0) = 1 for n>=0.

a(n) ~ c * d^n / sqrt(n), where d = 2.529010713480526199368... is the root of the equation 4 - 23*d - 36*d^2 - 8*d^3 + 4*d^5 = 0, c = 0.08933570507578447270759... . - Vaclav Kotesovec, Sep 07 2014

A128565 Column 1 of triangle A128564; a(n) equals the number of permutations of {1..n+2} with [n/2+1] inversions for n>=0.

Original entry on oeis.org

1, 2, 5, 9, 29, 49, 174, 285, 1068, 1717, 6655, 10569, 41926, 66013, 266338, 416687, 1703027, 2651355, 10947079, 16976806, 70673825, 109256095, 457927079, 706071989, 2976282415, 4579020513, 19395654894, 29784426945, 126688273871, 194231327451, 829176461458

Offset: 0

Paul D. Hanna, Mar 12 2007

nonn

Cf. A008302 (Mahonian numbers); A128564 (triangle), A128566 (column 2).

{a(n)=polcoeff(prod(j=1, n+2, (1-q^j)/(1-q)),(n+2)\2,q)}

a(n) = A008302(n+2,[n/2+1]) = coefficient of q^[n/2+1] in the q-factorial of n+2 for n>=0.

A384368 Number of permutations of [2n] with n inversions.

Original entry on oeis.org

1, 1, 5, 29, 174, 1068, 6655, 41926, 266338, 1703027, 10947079, 70673825, 457927079, 2976282415, 19395654894, 126688273871, 829176461458, 5436687172806, 35703722618623, 234807844921153, 1546217013188447, 10193761267335877, 67275841673522196, 444431529264364506

Offset: 0

Alois P. Heinz, May 27 2025

nonn

			a(0) = 1: the empty permutation.
a(1) = 1: 21.
a(2) = 5: 1342, 1423, 2143, 2314, 3124.
a(3) = 29: 123654, 124563, 124635, 125364, 125436, 126345, 132564, 132645, 134265, 134526, 135246, 142365, 142536, 143256, 152346, 213564, 213645, 214365, 214536, 215346, 231465, 231546, 234156, 241356, 312465, 312546, 314256, 321456, 412356.

Alois P. Heinz, Table of n, a(n) for n = 0..1208

Cf. A008302, A128566.

Cf. A100220.

a:= n-> coeff(series(mul((1-q^j)/(1-q), j=1..2*n), q, n+1), q, n):
seq(a(n), n=0..23);

a(n) = A008302(2n,n).

a(n) ~ c * 3^(3*n - 1/2) / (sqrt(Pi*n) * 2^(2*n)), where c = QPochhammer(1/3) = A100220 = 0.5601260779279489449697922433141400143797363337983... - Vaclav Kotesovec, Jun 09 2025

A304781 a(n) = [x^n] (1/(1 - x)^n)*Product_{k>=1} (1 + x^k).

Original entry on oeis.org

1, 2, 6, 21, 75, 274, 1016, 3807, 14377, 54627, 208584, 799669, 3076167, 11867511, 45897145, 177888715, 690770763, 2686879415, 10466761637, 40828165464, 159453481037, 623427464093, 2439907421914, 9557831470082, 37472409664888, 147028505564603, 577302980976146

Offset: 0

Ilya Gutkovskiy, May 18 2018

nonn

Number of partitions of n into odd parts with n + 1 kinds of 1.

Index entries for sequences related to partitions

Cf. A000009, A036469, A095944, A128566, A128593, A292463, A292613, A293467.

Table[SeriesCoefficient[1/(1 - x)^n Product[(1 + x^k), {k, 1, n}], {x, 0, n}], {n, 0, 26}]
Table[SeriesCoefficient[1/(1 - x)^n Product[1/(1 - x^(2 k - 1)), {k, 1, n}], {x, 0, n}], {n, 0, 26}]
Table[SeriesCoefficient[1/(1 - x)^n Exp[Sum[(-1)^(k + 1) x^k/(k (1 - x^k)), {k, 1, n}]], {x, 0, n}], {n, 0, 26}]
Table[SeriesCoefficient[QPochhammer[-1, x]/(2 (1 - x)^n), {x, 0, n}], {n, 0, 26}]

a(n) = [x^n] (1/(1 - x)^n)*Product_{k>=1} 1/(1 - x^(2*k-1)).
a(n) = [x^n] (1/(1 - x)^n)*exp(Sum_{k>=1} (-1)^(k+1)*x^k/(k*(1 - x^k))).
a(n) ~ QPochhammer[-1, 1/2] * 4^(n-1) / sqrt(Pi*n). - Vaclav Kotesovec, May 18 2018

cp's OEIS Frontend

A281425 a(n) = [q^n] (1 - q)^n / Product_{j=1..n} (1 - q^j).

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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.

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A128564 Triangle, read by rows, where T(n,k) equals the number of permutations of {1..n+1} with [(nk+k)/2] inversions for n>=k>=0.

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A214086 Number of involutions of length n having exactly n inversions.

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A128565 Column 1 of triangle A128564; a(n) equals the number of permutations of {1..n+2} with [n/2+1] inversions for n>=0.

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A384368 Number of permutations of [2n] with n inversions.

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A304781 a(n) = [x^n] (1/(1 - x)^n)*Product_{k>=1} (1 + x^k).

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