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A295281
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Number of complete strict tree-factorizations of n > 1.
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7
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1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 4, 1, 0, 1, 1, 1, 3, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 0, 1, 4, 1, 1, 1, 4, 1, 6, 1, 1, 1, 1, 1, 4, 1, 1, 0, 1, 1, 9, 1, 1, 1, 1, 1, 9, 1
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OFFSET
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2,29
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COMMENTS
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A strict tree-factorization (see A295279 for definition) is complete if its leaves are all prime numbers.
a(n) depends only on the prime signature of n.
This sequence is very similar but not identical to the number of complete orderless identity tree-factorizations of n. The first difference is at n=900 (square of three primes). Here a(n) = 191 whereas the other sequence would have 197. (End)
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LINKS
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FORMULA
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a(product of n distinct primes) = A000311(n).
Positions of zeros are proper prime powers A025475. Positions of nonzero entries are A085971.
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EXAMPLE
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The a(72) = 6 complete strict tree-factorizations are: 2*3*(2*(2*3)), 2*(2*3*(2*3)), 2*(2*(3*(2*3))), 2*(3*(2*(2*3))), 3*(2*(2*(2*3))), (2*3)*(2*(2*3)).
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MATHEMATICA
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postfacs[n_]:=If[n<=1, {{}}, Join@@Table[Map[Prepend[#, d]&, Select[postfacs[n/d], Min@@#>=d&]], {d, Rest[Divisors[n]]}]];
sftc[n_]:=Prepend[Join@@Function[fac, Tuples[sftc/@fac]]/@Select[postfacs[n], And[Length[#]>1, UnsameQ@@#]&], n];
Table[Length[Select[sftc[n], FreeQ[#, _Integer?(!PrimeQ[#]&)]&]], {n, 2, 100}]
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PROG
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(PARI) seq(n)={my(v=vector(n), w=vector(n)); v[1]=1; for(k=2, n, w[k]=v[k]+isprime(k); forstep(j=n\k*k, k, -k, v[j]+=w[k]*v[j/k])); w[2..n]} \\ Andrew Howroyd, Nov 18 2018
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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