A317144 Number of refinement-ordered pairs of factorizations of n into factors > 1.

1, 1, 1, 3, 1, 3, 1, 6, 3, 3, 1, 9, 1, 3, 3, 14, 1, 9, 1, 9, 3, 3, 1, 23, 3, 3, 6, 9, 1, 12, 1, 26, 3, 3, 3, 31, 1, 3, 3, 23, 1, 12, 1, 9, 9, 3, 1, 56, 3, 9, 3, 9, 1, 23, 3, 23, 3, 3, 1, 41, 1, 3, 9, 55, 3, 12, 1, 9, 3, 12, 1, 82, 1, 3, 9, 9, 3, 12, 1, 56, 14

Offset: 1

Author

Gus Wiseman, Jul 22 2018

Keywords

• nonn

Comments

If x and y are factorizations of the same integer and it is possible to produce x by further factoring the factors of y, flattening, and sorting, then x <= y.

As this is a sequence computed from exponents in factorization of n, distinct values of a(n) in this sequence can be found by computing a(A025487(k)) for k >= 0. - David A. Corneth, Jul 30 2018

Examples

			The a(12) = 9 ordered pairs:
  (2*2*3) <= (12)
  (2*2*3) <= (2*6)
  (2*2*3) <= (3*4)
  (2*2*3) <= (2*2*3)
    (2*6) <= (12)
    (2*6) <= (2*6)
    (3*4) <= (12)
    (3*4) <= (3*4)
     (12) <= (12)
		

Links

• Gus Wiseman, The poset of factorizations of 36, ordered by refinement.
• Index entries for sequences computed from exponents in factorization of n

Crossrefs

Cf. A000005, A001055, A007716, A025487, A045778, A162247, A281113, A317142, A317145, A317146.

Programs

Mathematica

sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}];
mps[set_]:=Union[Sort[Sort/@(#/.x_Integer:>set[[x]])]&/@sps[Range[Length[set]]]];
facs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
faccaps[fac_]:=Union[Sort/@Apply[Times,mps[fac],{2}]];
Table[Sum[Length[faccaps[fac]],{fac,facs[n]}],{n,100}]

Formula

a(n) >= A001055(n) + floor(A000005(n) / 2) - 1. - David A. Corneth, Jul 30 2018