A320267 Number of balanced complete orderless tree-factorizations of n.

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, 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, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 3, 1, 1, 1

Offset: 1

Author

Gus Wiseman, Oct 08 2018

Keywords

• nonn

Comments

a(1) = 1 by convention.

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

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

a(n) depends only on the prime signature of n. - Andrew Howroyd, Nov 18 2018

Examples

			The a(96) = 5 balanced complete orderless tree-factorizations:
     (2*2*2*2*2*3)
   ((2*2)*(2*2*2*3))
   ((2*3)*(2*2*2*2))
   ((2*2*2)*(2*2*3))
  ((2*2)*(2*2)*(2*3))
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..10000

Crossrefs

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

Programs

Mathematica

facs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
oltfacs[n_]:=If[n<=1,{{}},Prepend[Union@@Function[q,Sort/@Tuples[oltfacs/@q]]/@DeleteCases[facs[n],{n}],n]];
Table[Length[Select[oltfacs[n],And[SameQ@@Length/@Position[#,_Integer],FreeQ[#,_Integer?(!PrimeQ[#]&)]]&]],{n,100}]

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}
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

Formula

a(p^n) = A120803(n) for prime p. - Andrew Howroyd, Nov 18 2018