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A324923
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Number of distinct factors in the factorization of n into factors q(i) = prime(i)/i, i > 0.
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33
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0, 1, 2, 1, 3, 2, 2, 1, 2, 3, 4, 2, 3, 2, 3, 1, 3, 2, 2, 3, 3, 4, 3, 2, 3, 3, 2, 2, 4, 3, 5, 1, 4, 3, 4, 2, 3, 2, 3, 3, 4, 3, 3, 4, 3, 3, 4, 2, 2, 3, 4, 3, 2, 2, 4, 2, 3, 4, 4, 3, 3, 5, 3, 1, 4, 4, 3, 3, 3, 4, 4, 2, 4, 3, 3, 2, 5, 3, 5, 3, 2, 4, 4, 3, 5, 3, 4, 4, 3, 3, 4, 3, 5, 4, 4, 2, 4, 2, 4, 3, 4, 4, 3, 3, 4, 2, 3, 2
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OFFSET
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1,3
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COMMENTS
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Also the number of distinct proper terminal subtrees of the rooted tree with Matula-Goebel number n. See illustrations in A061773.
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LINKS
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FORMULA
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EXAMPLE
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The factorization 22 = q(1)^2 q(2) q(3) q(5) has four distinct factors, so a(22) = 4.
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MATHEMATICA
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difac[n_]:=If[n==1, {}, With[{i=PrimePi[FactorInteger[n][[1, 1]]]}, Sort[Prepend[difac[n*i/Prime[i]], i]]]];
Table[Length[Union[difac[n]]], {n, 100}]
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PROG
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(PARI)
A006530(n) = if(1==n, n, my(f=factor(n)); f[#f~, 1]);
A324923(n) = { my(lista = List([]), gpf, i); while(n > 1, gpf=A006530(n); i = primepi(gpf); n /= gpf; n *= i; listput(lista, i)); #Set(lista); }; \\ Antti Karttunen, Oct 23 2023
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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