OFFSET
1,73
COMMENTS
Counts finite sets of positive integers by sum and product.
EXAMPLE
Array begins:
k=1 k=2 k=3 k=4 k=5 k=6 k=7 k=8 k=9 k10 k11 k12
-----------------------------------------------
n=0: 1 0 0 0 0 0 0 0 0 0 0 0
n=1: 1 0 0 0 0 0 0 0 0 0 0 0
n=2: 0 1 0 0 0 0 0 0 0 0 0 0
n=3: 0 1 1 0 0 0 0 0 0 0 0 0
n=4: 0 0 1 1 0 0 0 0 0 0 0 0
n=5: 0 0 0 1 1 1 0 0 0 0 0 0
n=6: 0 0 0 0 1 2 0 1 0 0 0 0
n=7: 0 0 0 0 0 1 1 1 0 1 0 1
n=8: 0 0 0 0 0 0 1 1 0 1 0 2
n=9: 0 0 0 0 0 0 0 1 1 0 0 1
n=10: 0 0 0 0 0 0 0 0 1 1 0 0
n=11: 0 0 0 0 0 0 0 0 0 1 1 0
n=12: 0 0 0 0 0 0 0 0 0 0 1 1
The A(8,12) = 2 sets are: {2,6}, {1,3,4}.
The A(14,40) = 2 sets are: {4,10}, {1,5,8}.
Antidiagonals begin:
n+k=1: 1
n+k=2: 0 1
n+k=3: 0 0 0
n+k=4: 0 0 1 0
n+k=5: 0 0 0 1 0
n+k=6: 0 0 0 1 0 0
n+k=7: 0 0 0 0 1 0 0
n+k=8: 0 0 0 0 1 0 0 0
n+k=9: 0 0 0 0 0 1 0 0 0
n+k=10: 0 0 0 0 0 1 0 0 0 0
n+k=11: 0 0 0 0 0 1 1 0 0 0 0
n+k=12: 0 0 0 0 0 0 2 0 0 0 0 0
n+k=13: 0 0 0 0 0 0 0 1 0 0 0 0 0
n+k=14: 0 0 0 0 0 0 1 1 0 0 0 0 0 0
n+k=15: 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0
n+k=16: 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
For example, antidiagonal n+k=11 counts the following sets:
n=5: {2,3}
n=6: {1,5}
so the 11th antidiagonal is: (0,0,0,0,0,1,1,0,0,0,0).
MATHEMATICA
nn=12;
tt=Table[Length[Select[IntegerPartitions[n], UnsameQ@@#&&Times@@#==k&]], {n, 0, nn}, {k, 1, nn}] (* array *)
tr=Table[tt[[j, i-j]], {i, 2, nn}, {j, i-1}] (* antidiagonals *)
Join@@tr (* sequence *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Gus Wiseman, Jan 01 2025
STATUS
approved