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A071626
Number of distinct exponents in the prime factorization of n!.
40
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
OFFSET
1,4
COMMENTS
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
LINKS
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.
Randell Heyman and Md Rahil Miraj, On some floor function sets, arXiv:2309.16072 [math.NT], 2023-2024.
FORMULA
a(n) = A071625(n!) = A323023(n!,3). - Gus Wiseman, May 15 2019
EXAMPLE
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
MATHEMATICA
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 *)
PROG
(PARI) a(n) = #Set(factor(n!)[, 2]); \\ Michel Marcus, Sep 05 2017
KEYWORD
nonn,easy
AUTHOR
Labos Elemer, May 29 2002
STATUS
approved