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Number of balanced complete orderless tree-factorizations of n.
2

%I #8 Nov 18 2018 14:59:36

%S 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,1,1,1,1,1,1,2,1,1,

%T 1,3,1,1,1,2,1,1,1,1,1,1,1,3,1,1,1,1,1,2,1,2,1,1,1,3,1,1,1,4,1,1,1,1,

%U 1,1,1,4,1,1,1,1,1,1,1,3,2,1,1,3,1,1,1

%N Number of balanced complete orderless tree-factorizations of n.

%C a(1) = 1 by convention.

%C A rooted tree is balanced if all leaves are the same distance from the root.

%C An orderless tree-factorization (see A292504 for definition) is complete if all leaves are prime numbers.

%C a(n) depends only on the prime signature of n. - _Andrew Howroyd_, Nov 18 2018

%H Andrew Howroyd, <a href="/A320267/b320267.txt">Table of n, a(n) for n = 1..10000</a>

%F a(p^n) = A120803(n) for prime p. - _Andrew Howroyd_, Nov 18 2018

%e The a(96) = 5 balanced complete orderless tree-factorizations:

%e (2*2*2*2*2*3)

%e ((2*2)*(2*2*2*3))

%e ((2*3)*(2*2*2*2))

%e ((2*2*2)*(2*2*3))

%e ((2*2)*(2*2)*(2*3))

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

%t oltfacs[n_]:=If[n<=1,{{}},Prepend[Union@@Function[q,Sort/@Tuples[oltfacs/@q]]/@DeleteCases[facs[n],{n}],n]];

%t Table[Length[Select[oltfacs[n],And[SameQ@@Length/@Position[#,_Integer],FreeQ[#,_Integer?(!PrimeQ[#]&)]]&]],{n,100}]

%o (PARI) MultEulerT(u)={my(v=vector(#u)); v[1]=1; for(k=2, #u, forstep(j=#v\k*k, k, -k, my(i=j, e=0); while(i%k==0, i/=k; e++; v[j]+=binomial(e+u[k]-1, e)*v[i]))); v}

%o seq(n)={my(u=vector(n, i, i==1 || isprime(i)), v=vector(n)); while(u, v+=u; u[1]=1; u=MultEulerT(u)-u); v} \\ _Andrew Howroyd_, Nov 18 2018

%Y Cf. A001055, A048816, A050336, A281118, A292505, A119262, A120803, A292504, A320160, A320266.

%K nonn

%O 1,16

%A _Gus Wiseman_, Oct 08 2018