A258119 - OEIS

%I #27 Feb 21 2023 16:04:54

%S 1,2,4,6,8,16,18,20,32,64,42,54,56,128,100,256,162,176,512,1024,234,

%T 260,294,392,416,486,500,2048,4096,1088,1458,8192,1936,2500,16384,798,

%U 1026,1064,2058,2432,2744,4374,32768,65536,2300,3042,3380,5408,5888,12500,13122,131072

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

%C 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 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 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 J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p 123.

%H Alois P. Heinz, <a href="/A258119/b258119.txt">Rows n = 0..500, flattened</a>

%e 54 = 2*3*3*3 is in the sequence because the partition [1,2,2,2] is perfect.

%e 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 Triangle T(n,k) begins:

%e    1;

%e    2;

%e    4;

%e    6,  8;

%e   16;

%e   18, 20, 32;

%e   64;

%e   42, 54, 56, 128;

%e   ...

%p with(numtheory):

%p T:= proc(m) local b, ll, p;

%p       if m=0 then return 1 fi;

%p       p:= proc(l) ll:=ll, 2^(l[1]-1)*mul(ithprime(

%p       mul(l[j], j=1..i-1))^(l[i]-1), i=2..nops(l)) end:

%p       b:= proc(n, l) `if`(n=1, p(l), seq(b(n/d, [l[], d])

%p       , d=divisors(n) minus{1})) end:

%p       ll:= NULL; b(m+1, []): sort([ll])[]

%p     end:

%p seq(T(n), n=0..20);  # _Alois P. Heinz_, Jun 08 2015

%t 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 Cf. A000079, A215366, A002033, A258118.

%Y Column k=1 gives A259939.

%Y Row sums give A360713.

%K nonn,look,tabf

%O 0,2

%A _Emeric Deutsch_, Jun 07 2015