%I #8 Jan 19 2023 02:04:24
%S 1,0,1,0,0,1,0,0,1,1,0,0,1,0,1,0,0,1,0,2,0,0,0,1,2,1,0,0,0,0,1,0,3,1,
%T 0,0,0,0,1,0,3,2,0,0,0,0,0,1,1,3,3,0,0,0,0,0,0,1,1,5,3,0,0,0,0,0,0,0,
%U 1,0,8,3,0,0,0,0,0,0,0,0,1,2,6,6,0,0,0
%N Triangle read by rows where T(n,k) is the number of integer partitions of n with adjusted frequency depth k whose parts cover an initial interval of positive integers.
%C The adjusted frequency depth of an integer partition (A325280) is 0 if the partition is empty, and otherwise it is 1 plus the number of times one must take the multiset of multiplicities to reach a singleton. For example, the partition (32211) has adjusted frequency depth 5 because we have: (32211) -> (221) -> (21) -> (11) -> (2).
%H Andrew Howroyd, <a href="/A325336/b325336.txt">Table of n, a(n) for n = 0..1325</a> (rows 0..50)
%e Triangle begins:
%e 1
%e 0 1
%e 0 0 1
%e 0 0 1 1
%e 0 0 1 0 1
%e 0 0 1 0 2 0
%e 0 0 1 2 1 0 0
%e 0 0 1 0 3 1 0 0
%e 0 0 1 0 3 2 0 0 0
%e 0 0 1 1 3 3 0 0 0 0
%e 0 0 1 1 5 3 0 0 0 0 0
%e 0 0 1 0 8 3 0 0 0 0 0 0
%e 0 0 1 2 6 6 0 0 0 0 0 0 0
%e 0 0 1 0 13 4 0 0 0 0 0 0 0 0
%e 0 0 1 0 12 8 1 0 0 0 0 0 0 0 0
%e 0 0 1 2 14 7 3 0 0 0 0 0 0 0 0 0
%e 0 0 1 0 17 11 3 0 0 0 0 0 0 0 0 0 0
%e 0 0 1 0 22 7 8 0 0 0 0 0 0 0 0 0 0 0
%e 0 0 1 2 17 16 10 0 0 0 0 0 0 0 0 0 0 0 0
%e 0 0 1 0 28 10 15 0 0 0 0 0 0 0 0 0 0 0 0 0
%e 0 0 1 1 29 13 20 0 0 0 0 0 0 0 0 0 0 0 0 0 0
%e Row 15 counts the following partitions:
%e 111111111111111 54321 433221 333321 4322211
%e 2222211111 443211 3332211 4332111
%e 3322221 33222111 43221111
%e 22222221 322221111
%e 32222211 332211111
%e 33321111 432111111
%e 222222111 321111111111
%e 3222111111
%e 3321111111
%e 22221111111
%e 32211111111
%e 222111111111
%e 2211111111111
%e 21111111111111
%t normQ[m_]:=Or[m=={},Union[m]==Range[Max[m]]];
%t fdadj[ptn_List]:=If[ptn=={},0,Length[NestWhileList[Sort[Length/@Split[#1]]&,ptn,Length[#1]>1&]]];
%t Table[Length[Select[IntegerPartitions[n],normQ[#]&&fdadj[#]==k&]],{n,0,30},{k,0,n}]
%o (PARI)
%o depth(p)={if(!#p, 0, my(r=1); while(#p > 1, my(L=List(), k=0); for(i=1, #p, if(i==#p||p[i]<>p[i+1], listput(L,i-k); k=i)); listsort(L); p=L; r++); r)}
%o isok(p)={if(#p, for(i=1, #p, if(p[i]-1 > if(i>1, p[i-1], 0), return(0)))); 1}
%o row(n)={my(v=vector(1+n)); forpart(p=n, if(isok(p), v[1+depth(Vec(p))]++)); v}
%o { for(n=0, 10, print(row(n))) } \\ _Andrew Howroyd_, Jan 18 2023
%Y Row sums are A000009.
%Y Column k = 3 is A325334.
%Y Column k = 4 is A325335.
%Y Cf. A181819, A182850, A317246, A320348, A323014, A325280, A325372.
%K nonn,tabl
%O 0,20
%A _Gus Wiseman_, May 01 2019