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A027854
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Mutinous numbers: n > 1 such that n/p^k > p, where p is the largest prime dividing n and p^k is the highest power of p dividing n.
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8
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12, 24, 30, 36, 40, 45, 48, 56, 60, 63, 70, 72, 80, 84, 90, 96, 105, 108, 112, 120, 126, 132, 135, 140, 144, 150, 154, 160, 165, 168, 175, 176, 180, 182, 189, 192, 195, 198, 200, 208, 210, 216, 220, 224, 225, 231, 234, 240, 252, 260, 264, 270, 273, 275, 280
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OFFSET
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1,1
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COMMENTS
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If p = A006530(a(n)) then p * a(n) is in the sequence. E.g., as 12 is in the sequence with gpf(12) = A006530(12) = 3, 12*3^k is in the sequence for k > 0. Conjecture: if m is in the sequence then so is A003961(m). - David A. Corneth, Jul 13 2017
At present this and A027855 are complements in the set of integers >= 2. If a 1 were inserted at the start, then this and A027855 are complements in the set of positive integers. - Harry Richman, Sep 08 2019
The sequence is closed under multiplication (a semigroup). For, suppose x = p^i*m1, y = q^j*m2 are in the sequence, with p, q, p^i, p^j as given, with m1 > p and m2 > q, and suppose q >= p. If q = p then xy/q^(i+j) = m1*m2 > q. If q > p, then xy/q^j = p^i*m1*m2 > q (since q > p and p is greater than all primes in m1). - Richard Peterson, May 29 2022
There are subsequences that constitute subsemigroups: Consider as a subsequence all terms x such that x/p^k > a*p^b, with p,k as specified in the definition and a,b fixed real numbers greater than or equal to 1. Each pair (a,b) determines a subsequence that is also a subsemigroup of the original (1,1) semigroup that constitutes the whole sequence. The proof of closure is similar. To see that such proposed subsemigroups are nonempty, choose any prime p greater than 2 and multiply p by a sufficiently large power of 2. - Richard Peterson, May 29 2022
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LINKS
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EXAMPLE
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12 is a term since 12/A053585(12) = 12/3 = 4, A006530(12) = 3, and 4 > 3.
30 is a term since 30/A053585(30) = 30/5 = 6, A006530(30) = 5, and 6 > 5.
(End)
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MATHEMATICA
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Select[Range@ 280, Function[n, (n/Apply[Power, Last@ #]) > #[[-1, 1]] &@ FactorInteger[n]]] (* Michael De Vlieger, Jul 13 2017 *)
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PROG
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(PARI) isok(n) = {my(f = factor(n)); my(maxf = #f~); my(p = f[maxf, 1]); my(pk = f[maxf, 2]); (n/p^pk) > p; } \\ Michel Marcus, Jan 16 2014
(Python)
from sympy import factorint, primefactors
def a053585(n):
if n==1: return 1
p = primefactors(n)[-1]
return p**factorint(n)[p]
print([n for n in range(2, 301) if n>a053585(n)*primefactors(n)[-1]]) # Indranil Ghosh, Jul 13 2017
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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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