OFFSET
1
COMMENTS
Counts finite sets of positive integers > 1 by sum and product. Compare to the triangle A379737.
FORMULA
For k >= n we have A(n,k) = A379737(k,n).
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: 0 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 0 1 0 0 0 0 0 0 0 0 0
n=4: 0 0 0 1 0 0 0 0 0 0 0 0
n=5: 0 0 0 0 1 1 0 0 0 0 0 0
n=6: 0 0 0 0 0 1 0 1 0 0 0 0
n=7: 0 0 0 0 0 0 1 0 0 1 0 1
n=8: 0 0 0 0 0 0 0 1 0 0 0 1
n=9: 0 0 0 0 0 0 0 0 1 0 0 0
n=10: 0 0 0 0 0 0 0 0 0 1 0 0
n=11: 0 0 0 0 0 0 0 0 0 0 1 0
n=12: 0 0 0 0 0 0 0 0 0 0 0 1
For example, the: A(17,72) = 2 sets are: {2,3,12}, {8,9}.
Antidiagonals begin:
n+k=1: 1
n+k=2: 0 0
n+k=3: 0 0 0
n+k=4: 0 0 1 0
n+k=5: 0 0 0 0 0
n+k=6: 0 0 0 1 0 0
n+k=7: 0 0 0 0 0 0 0
n+k=8: 0 0 0 0 1 0 0 0
n+k=9: 0 0 0 0 0 0 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 0 0 0 0 0
n+k=12: 0 0 0 0 0 0 1 0 0 0 0 0
n+k=13: 0 0 0 0 0 0 0 0 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 0 0 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=14 counts the following sets:
n=6: {2,4}
n=7: {7}
so the 14th antidiagonal is: (0,0,0,0,0,0,1,1,0,0,0,0,0,0).
Antidiagonal n+k=89 counts the following sets:
n=17: {2,3,12}, {8,9}
n=19: {5,14}
n=21: {4,17}
n=31: {2,29}
so the 89th antidiagonal is: (...,0,0,0,2,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,...).
MATHEMATICA
nn=12;
tt=Table[Length[Select[IntegerPartitions[n], FreeQ[#, 1]&&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