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A071626
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Number of distinct exponents in the prime factorization of n!.
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40
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0, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 8, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11
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OFFSET
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1,4
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COMMENTS
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Erdős proved that there exist two constants c1, c2 > 0 such that c1 (n / log(n))^(1/2) < a(n) < c2 (n / log(n))^(1/2). - Carlo Sanna, May 28 2019
R. Heyman and R. Miraj proved that the cardinality of the set { floor(n/p) : p <= n, p prime } is same as the number of distinct exponents in the prime factorization of n!. - Md Rahil Miraj, Apr 05 2024
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LINKS
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P. Erdős, Miscellaneous problems in number theory, Proceedings of the Eleventh Manitoba Conference on Numerical Mathematics and Computing (Winnipeg, Man., 1981), Congressus Numerantium 34 (1982), 25-45.
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FORMULA
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EXAMPLE
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n=7: 7! = 5040 = 2*2*2*2*3*3*5*7; three different exponents arise: 4, 2 and 1; a(7)=3.
n=7: { floor(7/p) : p <= 7, p prime } = {3,2,1}. So, its cardinality is 3. - Md Rahil Miraj, Apr 05 2024
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MATHEMATICA
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ffi[x_] := Flatten[FactorInteger[x]] lf[x_] := Length[FactorInteger[x]] ep[x_] := Table[Part[ffi[x], 2*w], {w, 1, lf[x]}] Table[Length[Union[ep[w! ]]], {w, 1, 100}]
Table[Length[Union[Last/@If[n==1, {}, FactorInteger[n!]]]], {n, 30}] (* Gus Wiseman, May 15 2019 *)
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PROG
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(PARI) a(n) = #Set(factor(n!)[, 2]); \\ Michel Marcus, Sep 05 2017
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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