OFFSET
2,4
COMMENTS
Each row of triangle has 2^(n-1) terms.
The n-th row is built by multiplying the elements of the nonempty subsets of [n-1], sorting, removing the duplicates, and appending the reversed version. Equivalently, the n-th row is also given by the positive divisors of (n-1)! listed in increasing order (A079210), appending the reversed version. - Stefano Spezia, Aug 17 2022
FORMULA
A closed form for the sequence is: S_n =(1/n!) Sum {(n-1)!/{(n-1)^i(n-2)^j...2^q 1^r} (ij...qr = 00...00 to 11...11) where ij...qr is a base-2 number whose n-1 digits appear as exponents in the sum.
EXAMPLE
The first five rows are:
n=2: 1 1
n=3: 1 2 2 1
n=4: 1 2 3 6 6 3 2 1
n=5: 1 2 3 4 6 8 12 24 24 12 8 6 4 3 2 1
MATHEMATICA
Multiply[s_]:=Product[Part[s, i], {i, Length[s]}]; nmax=6; a={}; For[n=2, n<=nmax, n++, b=DeleteDuplicates[Sort[Table[Multiply[Part[Subsets[Drop[Range[n-1], 1]], i]], {i, 2^Length[Range[n-2]]}]]]; AppendTo[a, Join[b, Reverse[b]]]]; a (* or *)
nmax=6; a={}; For[n=2, n<=nmax, n++, b=Divisors[(n-1)!]; AppendTo[a, Join[b, Reverse[b]]]]; a (* Stefano Spezia, Aug 17 2022 *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Daniel Tisdale, Nov 03 2011
STATUS
approved