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A258119 Triangle T(n,k) in which the n-th row lists in increasing order the Heinz numbers of all perfect partitions of n.

%I A258119 #27 Feb 21 2023 16:04:54
%S A258119 1,2,4,6,8,16,18,20,32,64,42,54,56,128,100,256,162,176,512,1024,234,
%T A258119 260,294,392,416,486,500,2048,4096,1088,1458,8192,1936,2500,16384,798,
%U A258119 1026,1064,2058,2432,2744,4374,32768,65536,2300,3042,3380,5408,5888,12500,13122,131072
%N A258119 Triangle T(n,k) in which the n-th row lists in increasing order the Heinz numbers of all perfect partitions of n.
%C A258119 A partition of n is perfect if it contains just one partition of every number less than n when repeated parts are regarded as indistinguishable.
%C A258119 The Heinz number of a partition p = [p_1, p_2, ..., p_r] is defined as Product(p_j-th prime, j=1...r) (concept used by _Alois P. Heinz_ in A215366 as an "encoding" of a partition). For example, for the partition [1, 1, 1, 4] we get 2*2*2*7 = 56. It is in the sequence because the partition [1,1,1,4] is perfect.
%C A258119 Number of terms in row n is A002033(n). As a matter of fact, so far the triangle has been constructed by selecting those A002033(n) entries from row n of A215366 which correspond to perfect partitions. Last term in row n is 2^n.
%D A258119 J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p 123.
%H A258119 Alois P. Heinz, <a href="/A258119/b258119.txt">Rows n = 0..500, flattened</a>
%e A258119 54 = 2*3*3*3 is in the sequence because the partition [1,2,2,2] is perfect.
%e A258119 24 = 2*2*2*3 is not in the sequence because the partition [1,1,1,2] is not perfect (1+1+1=1+2; it is complete).
%e A258119 Triangle T(n,k) begins:
%e A258119    1;
%e A258119    2;
%e A258119    4;
%e A258119    6,  8;
%e A258119   16;
%e A258119   18, 20, 32;
%e A258119   64;
%e A258119   42, 54, 56, 128;
%e A258119   ...
%p A258119 with(numtheory):
%p A258119 T:= proc(m) local b, ll, p;
%p A258119       if m=0 then return 1 fi;
%p A258119       p:= proc(l) ll:=ll, 2^(l[1]-1)*mul(ithprime(
%p A258119       mul(l[j], j=1..i-1))^(l[i]-1), i=2..nops(l)) end:
%p A258119       b:= proc(n, l) `if`(n=1, p(l), seq(b(n/d, [l[], d])
%p A258119       , d=divisors(n) minus{1})) end:
%p A258119       ll:= NULL; b(m+1, []): sort([ll])[]
%p A258119     end:
%p A258119 seq(T(n), n=0..20);  # _Alois P. Heinz_, Jun 08 2015
%t A258119 T[0] = {1}; T[m_] := Module[{b, ll, p}, p[l_List] := (ll = Append[ll, 2^(l[[1]]-1)*Product[Prime[Product[l[[j]], {j, 1, i-1}]]^(l[[i]]-1), {i, 2, Length[l]}]]; 1); b[n_, l_List] := If[n == 1, p[l], Table[b[n/d, Append[l, d]], {d, Divisors[n] // Rest}]]; ll = {}; b[m+1, {}]; Sort[ll] ]; Table[T[n], {n, 0, 20}] // Flatten (* _Jean-François Alcover_, Jan 28 2016, after _Alois P. Heinz_ *)
%Y A258119 Cf. A000079, A215366, A002033, A258118.
%Y A258119 Column k=1 gives A259939.
%Y A258119 Row sums give A360713.
%K A258119 nonn,look,tabf
%O A258119 0,2
%A A258119 _Emeric Deutsch_, Jun 07 2015