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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. 87
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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
0,13
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)
LINKS
Combinatorial Object Server, Information on Numerical Partitions
FindStat - Combinatorial Statistic Finder, The length of the partition.
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
EXAMPLE
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)
MATHEMATICA
Flatten[Table[Length[IntegerPartitions[n, {k}]], {n, 0, 20}, {k, 0, n}]] (* Emanuele Munarini, Feb 24 2014 *)
PROG
(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
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.
Sequence in context: A344612 A069713 A319453 * A264391 A116598 A244925
KEYWORD
easy,nonn,tabl
AUTHOR
Martin Wohlgemuth (mail(AT)matroid.com), Jul 05 2002
EXTENSIONS
Corrected by Franklin T. Adams-Watters, Dec 18 2006
STATUS
approved

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Last modified March 19 02:58 EDT 2024. Contains 370952 sequences. (Running on oeis4.)