%I #50 Nov 07 2022 14:52:10
%S 0,0,0,0,0,0,0,0,0,0,0,0,1,2,0,1,1,0,1,1,3,1,1,1,1,3,2,7,3,2,5,4,3,5,
%T 9,2,5,6,9,5,9,14,9,7,5,10,10,11,18,7,11,16,14,12,12,23,19,13,18,11,
%U 20,19,32,17,21,18,25,19,21,27,22,21,31,27,24,28,42,34,33,21,28,31,35,47
%N Number of m for which a+b+c = n; abc = m has at least two distinct solutions (a,b,c) with 1 <= a <= b <= c.
%C A triple (a,b,c) as described in the name cannot have c prime. - _David A. Corneth_, Aug 01 2018
%H Alois P. Heinz, <a href="/A060277/b060277.txt">Table of n, a(n) for n = 1..10000</a> (first 1500 terms from Hugo Pfoertner)
%H IBM Research, <a href="https://www.research.ibm.com/haifa/ponderthis/challenges/July2018.html">Ponder This Challenge - July 2018</a>.
%H John B. Kelly, <a href="http://dx.doi.org/10.1090/S0002-9939-1964-0168542-2">Partitions with equal products</a>, Proc. Amer. Math. Soc. 15 (1964), 987-990
%H John B. Kelly, <a href="http://dx.doi.org/10.1090/S0002-9939-1989-0984800-0">Partitions with equal products. II</a>, Proc. Amer. Math. Soc. 107 (1989), 887-893
%F a(n) = Sum_{k>=2} A317578(n,k). - _Alois P. Heinz_, Aug 01 2018
%e (14 = 6+6+2 = 8+3+3, 72 = 6*6*2 = 8*3*3); (14 = 8+5+1 = 10+2+2, 40 = 8*5*1 = 10*2*2); 14 has two "m" variables. so a(14)=2.
%t a[n_] := Count[ Tally[ Times @@@ IntegerPartitions[n, {3}]], {m_,c_} /; c>1]; Array[a, 84] (* _Giovanni Resta_, Jul 27 2018 *)
%o (PARI) a(n)={my(M=Map()); for(i=n\3, n, for(j=(n-i+1)\2, min(n-1-i, i), my(k=n-i-j); my(m=i*j*k); my(z); mapput(M, m, if(mapisdefined(M, m, &z), z + 1, 1)))); #select(z->z>=2, if(#M, Mat(M)[, 2], []))} \\ _Andrew Howroyd_, Jul 27 2018
%Y Cf. A001399, A060275, A069905, A103277, A103278, A317388, A317578.
%K nonn
%O 1,14
%A _Naohiro Nomoto_, Mar 23 2001
%E Description revised by _David W. Wilson_ and _Don Reble_, Jun 04 2002