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Alois P. Heinz, <a href="/A322027/b322027.txt">Table of n, a(n) for n = 1..65536</a>
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Maximum order of primeness among the prime factors of n; a(1) = 0.
seq(a(n), n=1..120); # _Alois P. Heinz_, Nov 24 2018
with(numtheory):
p:= proc(n) option remember;
`if`(isprime(n), 1+p(pi(n)), 0)
end:
a:= n-> max(0, map(p, factorset(n))):
seq(a(n), n=1..120);
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allocated for Gus WisemanMaximum order of primeness among the prime factors of n.
0, 1, 2, 1, 3, 2, 1, 1, 2, 3, 4, 2, 1, 1, 3, 1, 2, 2, 1, 3, 2, 4, 1, 2, 3, 1, 2, 1, 1, 3, 5, 1, 4, 2, 3, 2, 1, 1, 2, 3, 2, 2, 1, 4, 3, 1, 1, 2, 1, 3, 2, 1, 1, 2, 4, 1, 2, 1, 3, 3, 1, 5, 2, 1, 3, 4, 2, 2, 2, 3, 1, 2, 1, 1, 3, 1, 4, 2, 1, 3, 2, 2, 2, 2, 3, 1, 2
1,3
N. Fernandez, <a href="http://www.borve.org/primeness/FOP.html">An order of primeness, F(p)</a>
N. Fernandez, <a href="/A006450/a006450.html">An order of primeness</a> [cached copy, included with permission of the author]
a(105) = 3 because the prime factor of 105 = 3*5*7 with maximum order of primeness is 5, with order 3.
Table[If[n==1, 0, Max@@(Length[NestWhileList[PrimePi, PrimePi[#], PrimeQ]]&/@FactorInteger[n][[All, 1]])], {n, 100}]
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nonn
Gus Wiseman, Nov 24 2018
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