[SeqFan]: Proposal for table: partitions of n such that LCM(summands)=m.

[Antti Karttunen]

I realized that the following table would be useful:

A(n,m): How many such (integer) partitions of n there are that their LCM 
= m?
(This implies that only divisors of m occur as the summands.
However, the condition is necessary, but not sufficient.)

"Calculating" by hand (not with the brain, in any case), I get:

  1 2 3 4 5 6 7 8
1 1,0,0,0,0,0,0,0,...
2 1,1,0,0,0,0,...
3 1,1,1,0,0,0,...
4 1,2,1,1,0,0,...
5 1,2,1,1,1,1,0,...
6 1,3,2,2,1,2,0,...


(n grows down, m to the right).
Rows should sum up to A000041,
and the last non-zero terms occurs at A000793(n) (Landau's function),
i.e. for 7, the partition 3+4 produces 3*4 = 12,
so it doesn't fit nicely into a triangle.

The above table does not seem to be in OEIS
(neither of the two transpositions), but I may
have missed it.
Then I'm especially interested about "subsidiary sequences"
like T(A000108(n),n), T(A000108(n),n+1),
T(A000108(n),2n), T(A000108(n),3n), etc... :-)



Terveisin,

Antti