%I
%S 1,5,1,55,15,1,935,220,75,30,1,21505,4675,2750,550,375,50,1,623645,
%T 129030,70125,30250,14025,16500,1875,1100,1125,75,1,21827575,4365515,
%U 2258025,1799875,451605,490875,211750,144375,32725,57750,13125,1925,2625,105,1,894930575
%N Partition number array, called M32(5), related to A013988(n,m)= S2(5;n,m) ( generalized Stirling triangle).
%C Each partition of n, ordered as in AbramowitzStegun (ASt order; for the reference see A134278), is mapped to a nonnegative integer a(n,k)=:M32(5;n,k) with the kth partition of n in ASt order.
%C The sequence of row lengths is A000041 (partition numbers) [1, 2, 3, 5, 7, 11, 15, 22, 30, 42, ...].
%C a(n,k) enumerates special unordered forests related to the kth partition of n in the ASt order. The kth partition of n is given by the exponents enk =(e(n,k,1),...,e(n,k,n)) of 1,2,...n. The number of parts is m = sum(e(n,k,j),j=1..n). The special (enk)forest is composed of m rooted increasing (r+4)ary trees if the outdegree is r >= 0.
%C If M32(5;n,k) is summed over those k with fixed number of parts m one obtains triangle A013988(n,m)= S2(5;n,m), a generalization of Stirling numbers of the second kind. For S2(K;n,m), K from the integers, see the reference under A035342.
%H W. Lang, <a href="/A144268/a144268.txt">First 10 rows of the array and more.</a>
%H W. Lang, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL12/Lang/lang.html">Combinatorial Interpretation of Generalized Stirling Numbers</a>, J. Int. Seqs. Vol. 12 (2009) 09.3.3.
%F a(n,k)= (n!/product(e(n,k,j)!*j!^(e(n,k,j),j=1..n))*product(S2(5,j,1)^e(n,k,j),j=1..n) = M3(n,k)*product(S2(5,j,1)^e(n,k,j),j=1..n), with S2(5,n,1)= A008543(n1) = (6*n7)(!^6) (6factorials) for n>=2 and 1 if n=1 and the exponent e(n,k,j) of j in the kth partition of n in the ASt ordering of the partitions of n. Exponents 0 can be omitted due to 0!=1. M3(n,k):= A036040(n,k), k=1..p(n), p(n):= A000041(n).
%e a(4,3)=75. The relevant partition of 4 is (2^2). The 75 unordered (0,2,0,0)forests are composed of the following 2 rooted increasing trees 12,34; 13,24 and 14,23. The trees are 5ary because r=1 vertices are 5ary and for the leaves (r=0) the arity does not matter. Each of the three differently labeled forests comes therefore in 5^2=25 versions due to the two 5ary root vertices.
%Y Cf. A144267 (M32(4) array).
%K nonn,easy,tabf
%O 1,2
%A _Wolfdieter Lang_, Oct 09 2008
