%I #43 Feb 13 2021 14:37:15
%S 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,
%T 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,
%U 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
%N 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.
%C 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
%C From _Gus Wiseman_, Feb 10 2021: (Start)
%C 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:
%C 111111111 22221 333 432 54 63 72 81 9
%C 222111 3222 441 522 621 711
%C 2211111 3321 4221 531 6111
%C 21111111 32211 4311 5211
%C 33111 42111 51111
%C 321111 411111
%C 3111111
%C (End)
%H Robert G. Wilson v, <a href="/A072233/b072233.txt">Table of n, a(n) for n = 0..10010</a>
%H Combinatorial Object Server, <a href="http://combos.org/part">Information on Numerical Partitions</a>
%H FindStat - Combinatorial Statistic Finder, <a href="http://www.findstat.org/StatisticsDatabase/St000010">The length of the partition.</a>
%F 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).
%F 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
%F 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
%e Table begins (upper left corner = T(0,0)):
%e 1 1 1 1 1 1 1 1 1 ...
%e 0 1 1 1 1 1 1 1 1 ...
%e 0 1 2 2 2 2 2 2 2 ...
%e 0 1 2 3 3 3 3 3 3 ...
%e 0 1 3 4 5 5 5 5 5 ...
%e 0 1 3 5 6 7 7 7 7 ...
%e 0 1 4 7 9 10 11 11 11 ...
%e 0 1 4 8 11 13 14 15 15 ...
%e 0 1 5 10 15 18 20 21 22 ...
%e 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.
%e From _Wolfdieter Lang_, Dec 03 2012 (Start)
%e The triangle a(n,k) = T(n-k,k) begins:
%e n\k 0 1 2 3 4 5 6 7 8 9 10 ...
%e 00 1
%e 01 0 1
%e 02 0 1 1
%e 03 0 1 1 1
%e 04 0 1 2 1 1
%e 05 0 1 2 2 1 1
%e 06 0 1 3 3 2 1 1
%e 07 0 1 3 4 3 2 1 1
%e 08 0 1 4 5 5 3 2 1 1
%e 09 0 1 4 7 6 5 3 2 1 1
%e 10 0 1 5 8 9 7 5 3 2 1 1
%e ...
%e 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)
%e From _Gus Wiseman_, Feb 10 2021: (Start)
%e Row n = 9 counts the following partitions:
%e 9 54 333 3222 22221 222111 2211111 21111111 111111111
%e 63 432 3321 32211 321111 3111111
%e 72 441 4221 33111 411111
%e 81 522 4311 42111
%e 531 5211 51111
%e 621 6111
%e 711
%e (End)
%t Flatten[Table[Length[IntegerPartitions[n, {k}]], {n, 0, 20}, {k, 0, n}]] (* _Emanuele Munarini_, Feb 24 2014 *)
%o (Sage)
%o from sage.combinat.partition import number_of_partitions_length
%o [[number_of_partitions_length(n, k) for k in (0..n)] for n in (0..10)] # _Peter Luschny_, Aug 01 2015
%Y Sum of antidiagonal entries T(n, k) with n+k=m equals A000041(m).
%Y Alternating row sums are A081362.
%Y Cf. A008284.
%Y The version for factorizations is A316439.
%Y The version for set partitions is A048993/A080510.
%Y The version for strict partitions is A008289/A059607.
%Y A047993 counts balanced partitions, ranked by A106529.
%Y A063995/A105806 count partitions by Dyson rank.
%Y Cf. A006141, A064174, A096401, A117409, A168659, A215366.
%K easy,nonn,tabl
%O 0,13
%A Martin Wohlgemuth (mail(AT)matroid.com), Jul 05 2002
%E Corrected by _Franklin T. Adams-Watters_, Dec 18 2006