A072811 - OEIS

%I #31 Sep 24 2023 11:09:51

%S 1,1,1,1,1,2,1,1,2,1,3,1,1,2,2,3,3,4,1,1,2,2,3,1,6,4,1,6,5,1,1,2,2,3,

%T 2,6,4,3,3,12,5,4,10,6,1,1,2,2,3,2,6,4,1,6,3,12,5,3,6,12,20,6,1,10,15,

%U 7,1,1,2,2,3,2,6,4,2,6,3,12,5,3,6,12,12,20,6,1,12,10,4,30,30,7,5,20,21,8,1

%N T(n,k) = multiplicity of the k-th partition of n in Mathematica order, defined to be the count of its permutations (compositions).

%C The sum of row n equals A011782(n). The first and last columns equal 1. The number of integers per row equals the partition number p(n). Row n is a vector of weights or multiplicities relating counts of ordered versus unordered objects classified according to the partitions of n.

%C a(n) is the multinomial coefficient of the signature of the n-th partition. - _Franklin T. Adams-Watters_, Apr 08 2008

%C Let f(x)=1/(1-sum(j>=1, c[j]*x^j))=sum(n>=0, w(n)*x^n), then the coefficients of wn=Pn(c[1],...,c[n]), listed in reverse lexicographic order, give row n of T(n,k). - _Groux Roland_, Mar 08 2011

%H Alois P. Heinz, <a href="/A072811/b072811.txt">Rows n = 0..26, flattened</a>

%e The partitions of 4 are {4}, {3,1}, {2,2}, {2,1,1}, {1,1,1,1}, so the fourth row equals 1,2,1,3,1 since these are the counts of the permutations of these lists.

%e Triangle begins:

%e 1;

%e 1;

%e 1, 1;

%e 1, 2, 1;

%e 1, 2, 1, 3, 1;

%e 1, 2, 2, 3, 3, 4, 1;

%e 1, 2, 2, 3, 1, 6, 4, 1, 6, 5, 1;

%t mult[li:{__Integer}] := Apply[Multinomial, Length/@Split[ Sort[li] ] ]; Table[mult/@Partitions[n], {n, 12}]

%o (PARI) \\ here mulp(v) computes the multiplicity of the given partition.

%o mulp(v) = {my(p=(#v)!, k=1); for(i=2, #v, k=if(v[i]==v[i-1], k+1, p/=k!; 1)); p/k!}

%o Row(n)={apply(mulp, vecsort([Vecrev(p) | p<-partitions(n)], , 4))}

%o { for(n=0, 9, print(Row(n))) } \\ _Peter Dolland_, Nov 11 2019

%Y Cf. A080577, A080575, A115621, A102462.

%K easy,nonn,look,tabf

%O 0,6

%A _Wouter Meeussen_, Aug 09 2002