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%I #32 Feb 21 2023 16:06:29
%S 1,2,4,6,8,12,16,18,20,24,32,30,36,40,48,64,42,54,56,60,72,80,96,128,
%T 84,90,100,108,112,120,144,160,192,256,126,132,140,150,162,168,176,
%U 180,200,216,224,240,288,320,384,512,198,210,220,252,264,270,280,300,324,336,352,360,400,432,448,480,576,640,768,1024
%N Triangle T(n,k) in which the n-th row lists in increasing order the Heinz numbers of all complete partitions of n.
%C A partition of n is complete if every number from 1 to n can be represented as a sum of parts of the partition.
%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 complete.
%C Except for a(0)=1, there are no odd numbers in the sequence. Indeed, a partition having an odd Heinz number does not have 1 as a part and, consequently, it cannot be complete.
%C Number of terms in row n is A126796(n). As a matter of fact, so far, the triangle has been constructed by selecting those A126796(n) entries from row n of A215366 which correspond to complete partitions. Last term in row n is 2^n.
%H Alois P. Heinz, <a href="/A258118/b258118.txt">Rows n = 0..30, flattened</a>
%H SeungKyung Park, <a href="http://www.fq.math.ca/Scanned/36-4/park.pdf">Complete Partitions</a>, Fibonacci Quarterly, Vol. 36 (1998), pp. 354-360.
%e 54 = 2*3*3*3 is in the sequence because the partition [1,2,2,2] is complete.
%e 28 = 2*2*7 is not in the sequence because the partition [1,1,4] is not complete.
%e Triangle T(n,k) begins:
%e 1;
%e 2;
%e 4;
%e 6, 8;
%e 12, 16;
%e 18, 20, 24, 32;
%e 30, 36, 40, 48, 64;
%e 42, 54, 56, 60, 72, 80, 96, 128;
%e 84, 90, 100, 108, 112, 120, 144, 160, 192, 256;
%e ...
%p T:= proc(m) local b, ll, p;
%p p:= proc(l) ll:=ll, (mul(ithprime(j), j=l)); 1 end:
%p b:= proc(n, i, l) `if`(i<2, p([l[], 1$n]), `if`(n<2*i-1,
%p b(n, iquo(n+1, 2), l), b(n, i-1, l)+b(n-i, i, [l[], i])))
%p end: ll:= NULL; b(m, iquo(m+1, 2), []): sort([ll])[]
%p end:
%p seq(T(n), n=0..12); # _Alois P. Heinz_, Jun 07 2015
%t T[m_] := Module[{b, ll, p}, p[l_List] := (ll = Append[ll, Product[Prime[j], {j, l}]]; 1); b[n_, i_, l_List] := If[i<2, p[Join[l, Array[1&, n]]], If[n < 2*i-1, b[n, Quotient[n+1, 2], l], b[n, i-1, l] + b[n-i, i, Append[l, i] ]]]; ll = {}; b[m, Quotient[m+1, 2], {}]; Sort[ll]]; Table[T[n], {n, 0, 12}] // Flatten (* _Jean-François Alcover_, Jan 28 2016, after _Alois P. Heinz_ *)
%Y Cf. A000079, A215366, A126796, A258119.
%Y Column k=1 gives A259941.
%Y Row sums give A360791.
%K nonn,look,tabf
%O 0,2
%A _Emeric Deutsch_, Jun 07 2015
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