|
|
A060277
|
|
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.
|
|
6
|
|
|
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, 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, 20, 19, 32, 17, 21, 18, 25, 19, 21, 27, 22, 21, 31, 27, 24, 28, 42, 34, 33, 21, 28, 31, 35, 47
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,14
|
|
COMMENTS
|
A triple (a,b,c) as described in the name cannot have c prime. - David A. Corneth, Aug 01 2018
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
(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.
|
|
MATHEMATICA
|
a[n_] := Count[ Tally[ Times @@@ IntegerPartitions[n, {3}]], {m_, c_} /; c>1]; Array[a, 84] (* Giovanni Resta, Jul 27 2018 *)
|
|
PROG
|
(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
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|