A318160 -id:A318160 - cp's OEIS frontend

A180281 Triangle read by rows: T(n,k) = number of arrangements of n indistinguishable balls in n boxes with the maximum number of balls in any box equal to k.

1, 1, 2, 1, 6, 3, 1, 18, 12, 4, 1, 50, 50, 20, 5, 1, 140, 195, 90, 30, 6, 1, 392, 735, 392, 147, 42, 7, 1, 1106, 2716, 1652, 672, 224, 56, 8, 1, 3138, 9912, 6804, 2970, 1080, 324, 72, 9, 1, 8952, 35850, 27600, 12825, 4950, 1650, 450, 90, 10, 1, 25652, 128865, 110715, 54450, 22022, 7865, 2420, 605, 110, 11

Offset: 1

R. H. Hardin, Aug 24 2010

To clarify a slight ambiguity in the definition, the heaviest box in such an arrangement should contain exactly k balls. - Gus Wiseman, Sep 22 2016

			The T(4,2)=18 arrangements are {0022, 0112, 0121, 0202, 0211, 0220, 1012, 1021, 1102, 1120, 1201, 1210, 2002, 2011, 2020, 2101, 2110, 2200}.
Triangle starts
  1
  1   2
  1   6   3
  1  18  12  4
  1  50  50 20  5
  1 140 195 90 30 6
  ...

Alois P. Heinz, Rows n = 1..200, flattened (first 59 rows from R. H. Hardin)

Cf. A097861, A180282, A180291.
Row sums give A088218.
T(n,ceiling(n/2)) gives A318160.
T(2n,n) gives A318161.
T(2n-1,n) gives A318161.

b:= proc(n, i, k) option remember; `if`(n=0, 1,
      `if`(i=0, 0, add(b(n-j, i-1, k), j=0..min(n, k))))
    end:
T:= (n, k)-> b(n$2, k)-b(n$2, k-1):
seq(seq(T(n,k), k=1..n), n=1..12);  # Alois P. Heinz, Aug 16 2018
# second Maple program:
T:= (n, k)-> coeff(series(((x^(k+1)-1)/(x-1))^n
             -((x^k-1)/(x-1))^n, x, n+1), x, n):
seq(seq(T(n, k), k=1..n), n=1..12);  # Alois P. Heinz, Aug 17 2018

T[n_,k_]:=Select[Tuples[Range[0,k],n],And[Max[#]===k,Total[#]===n]&]; (* Gus Wiseman, Sep 22 2016 *)
SequenceForm@@@T[4,2] (* example *)
Join@@Table[Length[T[n,k]],{n,1,6},{k,1,n}] (* sequence *)
(* Second program: *)
b[n_, i_, k_] := b[n, i, k] = If[n == 0, 1, If[i == 0, 0, Sum[b[n-j, i-1, k], {j, 0, Min[n, k]}]]];
T[n_, k_] := b[n, n, k] - b[n, n, k-1];
Table[Table[T[n, k], {k, 1, n}], {n, 1, 12}] // Flatten (* Jean-François Alcover, Aug 28 2022, after Alois P. Heinz *)

Empirical: right half of table, T(n,k) = n*binomial(2*n-k-2,n-2) for 2*k > n; also, T(n,2) = Sum_{j=1..n} binomial(n,j)*binomial(n-j,j) = 2*A097861(n). - Robert Gerbicz in the Sequence Fans Mailing List
From Alois P. Heinz, Aug 17 2018: (Start)
T(n,k) = [x^n] ((x^(k+1)-1)/(x-1))^n - ((x^k-1)/(x-1))^n.
T(n,k) = A305161(n,k) - A305161(n,k-1). (End)

A318161 Number of compositions of 2n into exactly 2n nonnegative parts with largest part n.

Original entry on oeis.org

1, 1, 18, 195, 1652, 12825, 96030, 705341, 5116200, 36773397, 262462010, 1862790699, 13160496684, 92624149475, 649794035142, 4545979700445, 31727803153232, 220975193536845, 1536191185018770, 10661898343645847, 73890140441316420, 511405029708269529

Offset: 0

Alois P. Heinz, Aug 19 2018

nonn

			a(2) = 18: 0022, 0112, 0121, 0202, 0211, 0220, 1012, 1021, 1102, 1120, 1201, 1210, 2002, 2011, 2020, 2101, 2110, 2200.

Alois P. Heinz, Table of n, a(n) for n = 0..1204

Bisection of A318160 (even part).

Cf. A180281.

Flatten[{1, Table[n - 2*n^2 + 2*n*Binomial[3*n - 2, n], {n, 1, 20}]}] (* Vaclav Kotesovec, Sep 20 2019 *)

a(n) = A180281(2n,n).

For n>0, a(n) = n - 2*n^2 + 2*n*binomial(3*n - 2, n). - Vaclav Kotesovec, Sep 20 2019

A318162 Number of compositions of 2n-1 into exactly 2n-1 nonnegative parts with largest part n.

Original entry on oeis.org

1, 6, 50, 392, 2970, 22022, 160888, 1162800, 8335338, 59366450, 420630210, 2967563040, 20861295000, 146203657992, 1021964428880, 7127260128736, 49606676100234, 344658278690250, 2390849931605590, 16561583202364200, 114577083158683530, 791757148201073670

Offset: 1

Alois P. Heinz, Aug 19 2018

nonn

			a(1) = 1: 1.
a(2) = 6: 012, 021, 102, 120, 201, 210.
a(3) = 50: 00023, 00032, 00113, 00131, 00203, 00230, 00302, 00311, 00320, 01013, 01031, 01103, 01130, 01301, 01310, 02003, 02030, 02300, 03002, 03011, 03020, 03101, 03110, 03200, 10013, 10031, 10103, 10130, 10301, 10310, 11003, 11030, 11300, 13001, 13010, 13100, 20003, 20030, 20300, 23000, 30002, 30011, 30020, 30101, 30110, 30200, 31001, 31010, 31100, 32000.

Alois P. Heinz, Table of n, a(n) for n = 1..1205

Bisection of A318160 (odd part).

Cf. A180281.

a:= proc(n) option remember; (2*n-1)*`if`(n<3, n,
      3*(3*n-4)*(3*n-5)*a(n-1)/(2*(n-1)*(2*n-3)^2))
    end:
seq(a(n), n=1..30);

Flatten[{1, Table[2*(2*n - 1)*Binomial[3*n - 4, n-2], {n, 2, 20}]}] (* Vaclav Kotesovec, Sep 20 2019 *)

a(n) = A180281(2n-1,n).

For n>1, a(n) = 2*(2*n - 1) * binomial(3*n - 4, n-2). - Vaclav Kotesovec, Sep 20 2019

cp's OEIS Frontend

A180281 Triangle read by rows: T(n,k) = number of arrangements of n indistinguishable balls in n boxes with the maximum number of balls in any box equal to k.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A318161 Number of compositions of 2n into exactly 2n nonnegative parts with largest part n.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A318162 Number of compositions of 2n-1 into exactly 2n-1 nonnegative parts with largest part n.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula