A072233 Square array T(n,k) read by antidiagonals giving number of ways to distribute n indistinguishable objects in k indistinguishable containers; containers may be left empty.

1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 2, 1, 1, 0, 1, 2, 2, 1, 1, 0, 1, 3, 3, 2, 1, 1, 0, 1, 3, 4, 3, 2, 1, 1, 0, 1, 4, 5, 5, 3, 2, 1, 1, 0, 1, 4, 7, 6, 5, 3, 2, 1, 1, 0, 1, 5, 8, 9, 7, 5, 3, 2, 1, 1, 0, 1, 5, 10, 11, 10, 7, 5, 3, 2, 1, 1, 0, 1, 6, 12, 15, 13, 11, 7, 5, 3, 2, 1, 1, 0, 1, 6, 14, 18, 18, 14, 11, 7, 5, 3, 2, 1, 1

Offset: 0

Author

Martin Wohlgemuth (mail(AT)matroid.com), Jul 05 2002

Keywords

• easy
• nonn
• tabl

Comments

Regarded as a triangular table, this is another version of the number of partitions of n into k parts, A008284. - Franklin T. Adams-Watters, Dec 18 2006

From Gus Wiseman, Feb 10 2021: (Start)

T(n,k) is also the number of partitions of n with greatest part k, if we assume the greatest part of an empty partition to be 0. Row n = 9 counts the following partitions:

111111111 22221 333 432 54 63 72 81 9

222111 3222 441 522 621 711

2211111 3321 4221 531 6111

21111111 32211 4311 5211

33111 42111 51111

321111 411111

3111111

(End)

Examples

			Table begins (upper left corner = T(0,0)):
1 1 1  1  1  1  1  1  1 ...
0 1 1  1  1  1  1  1  1 ...
0 1 2  2  2  2  2  2  2 ...
0 1 2  3  3  3  3  3  3 ...
0 1 3  4  5  5  5  5  5 ...
0 1 3  5  6  7  7  7  7 ...
0 1 4  7  9 10 11 11 11 ...
0 1 4  8 11 13 14 15 15 ...
0 1 5 10 15 18 20 21 22 ...
There is 1 way to distribute 0 objects into k containers: T(0, k) = 1. The different ways for n=4, k=3 are: (oooo)()(), (ooo)(o)(), (oo)(oo)(), (oo)(o)(o), so T(4, 3) = 4.
From _Wolfdieter Lang_, Dec 03 2012 (Start)
The triangle a(n,k) = T(n-k,k) begins:
n\k  0  1  2  3  4  5  6  7  8  9 10 ...
00   1
01   0  1
02   0  1  1
03   0  1  1  1
04   0  1  2  1  1
05   0  1  2  2  1  1
06   0  1  3  3  2  1  1
07   0  1  3  4  3  2  1  1
08   0  1  4  5  5  3  2  1  1
09   0  1  4  7  6  5  3  2  1  1
10   0  1  5  8  9  7  5  3  2  1  1
...
Row n=5 is, for k=1..5, [1,2,2,1,1] which gives the number of partitions of n=5 with k parts. See A008284 and the Franklin T. Adams-Watters comment above. (End)
From _Gus Wiseman_, Feb 10 2021: (Start)
Row n = 9 counts the following partitions:
  9  54  333  3222  22221  222111  2211111  21111111  111111111
     63  432  3321  32211  321111  3111111
     72  441  4221  33111  411111
     81  522  4311  42111
         531  5211  51111
         621  6111
         711
(End)
		

Links

• Robert G. Wilson v, Table of n, a(n) for n = 0..10010
• Combinatorial Object Server, Information on Numerical Partitions
• FindStat - Combinatorial Statistic Finder, The length of the partition.

Crossrefs

Sum of antidiagonal entries T(n, k) with n+k=m equals A000041(m).

Alternating row sums are A081362.

Cf. A008284.

The version for factorizations is A316439.

The version for set partitions is A048993/A080510.

The version for strict partitions is A008289/A059607.

A047993 counts balanced partitions, ranked by A106529.

A063995/A105806 count partitions by Dyson rank.

Cf. A006141, A064174, A096401, A117409, A168659, A215366.

Programs

Mathematica

Flatten[Table[Length[IntegerPartitions[n, {k}]], {n, 0, 20}, {k, 0, n}]] (* Emanuele Munarini, Feb 24 2014 *)

Sage

from sage.combinat.partition import number_of_partitions_length
[[number_of_partitions_length(n, k) for k in (0..n)] for n in (0..10)] # Peter Luschny, Aug 01 2015

Formula

T(0, k) = 1, T(n, 0) = 0 (n>0), T(1, k) = 1 (k>0), T(n, 1) = 1 (n>0), T(n, k) = 0 for n < 0, T(n, k) = Sum[ T(n-k+i, k-i), i=0...k-1] Or, T(n, 1) = T(n, n) = 1, T(n, k) = 0 (k>n), T(n, k) = T(n-1, k-1) + T(n-k, k).

G.f. Product_{j=0..infinity} 1/(1-xy^j). Regarded as a triangular array, g.f. Product_{j=1..infinity} 1/(1-xy^j). - Franklin T. Adams-Watters, Dec 18 2006

O.g.f. of column No. k of the triangle a(n,k) is x^k/product(1-x^j,j=1..k), k>=0 (the undefined product for k=0 is put to 1). - Wolfdieter Lang, Dec 03 2012

Extensions

Corrected by Franklin T. Adams-Watters, Dec 18 2006