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Number of refinement-ordered pairs of factorizations of n into factors > 1.
25

%I #13 Aug 02 2018 08:45:48

%S 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,

%T 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,

%U 55,3,12,1,9,3,12,1,82,1,3,9,9,3,12,1,56,14

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

%C 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.

%C 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

%H Gus Wiseman, <a href="/A317144/a317144.png">The poset of factorizations of 36, ordered by refinement.</a>

%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>

%F a(n) >= A001055(n) + floor(A000005(n) / 2) - 1. - _David A. Corneth_, Jul 30 2018

%e The a(12) = 9 ordered pairs:

%e (2*2*3) <= (12)

%e (2*2*3) <= (2*6)

%e (2*2*3) <= (3*4)

%e (2*2*3) <= (2*2*3)

%e (2*6) <= (12)

%e (2*6) <= (2*6)

%e (3*4) <= (12)

%e (3*4) <= (3*4)

%e (12) <= (12)

%t sps[{}]:={{}};sps[set:{i_,___}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,___}];

%t mps[set_]:=Union[Sort[Sort/@(#/.x_Integer:>set[[x]])]&/@sps[Range[Length[set]]]];

%t facs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];

%t faccaps[fac_]:=Union[Sort/@Apply[Times,mps[fac],{2}]];

%t Table[Sum[Length[faccaps[fac]],{fac,facs[n]}],{n,100}]

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

%K nonn

%O 1,4

%A _Gus Wiseman_, Jul 22 2018