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%I #100 Mar 04 2023 20:54:41
%S 12,24,30,36,40,45,48,56,60,63,70,72,80,84,90,96,105,108,112,120,126,
%T 132,135,140,144,150,154,160,165,168,175,176,180,182,189,192,195,198,
%U 200,208,210,216,220,224,225,231,234,240,252,260,264,270,273,275,280
%N 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.
%C Numbers n > 1 such that n/A053585(n) > A006530(n). - _Michael De Vlieger_, Jul 13 2017
%C 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
%C 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
%C 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
%C 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
%C This sequence is a subsequence and subsemigroup of A289484. - _Richard Peterson_, Oct 29 2022
%H Giovanni Resta, <a href="/A027854/b027854.txt">Table of n, a(n) for n = 1..10000</a>
%e From _Michael De Vlieger_, Jul 13 2017: (Start)
%e 12 is a term since 12/A053585(12) = 12/3 = 4, A006530(12) = 3, and 4 > 3.
%e 30 is a term since 30/A053585(30) = 30/5 = 6, A006530(30) = 5, and 6 > 5.
%e (End)
%t Select[Range@ 280, Function[n, (n/Apply[Power, Last@ #]) > #[[-1, 1]] &@ FactorInteger[n]]] (* _Michael De Vlieger_, Jul 13 2017 *)
%o (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
%o (Python)
%o from sympy import factorint, primefactors
%o def a053585(n):
%o if n==1: return 1
%o p = primefactors(n)[-1]
%o return p**factorint(n)[p]
%o print([n for n in range(2, 301) if n>a053585(n)*primefactors(n)[-1]]) # _Indranil Ghosh_, Jul 13 2017
%Y Cf. A006530, A027855, A053585.
%K nonn
%O 1,1
%A _Leroy Quet_
%E Extended by _Ray Chandler_, Nov 17 2008
%E Offset changed to 1 by _Michel Marcus_, Jan 16 2014
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