A027854 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.

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

Offset: 1

Author

Leroy Quet

Keywords

• nonn

Comments

Numbers n > 1 such that n/A053585(n) > A006530(n). - Michael De Vlieger, Jul 13 2017

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

This sequence is a subsequence and subsemigroup of A289484. - Richard Peterson, Oct 29 2022

Examples

			From _Michael De Vlieger_, Jul 13 2017: (Start)
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)
		

Links

• Giovanni Resta, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A006530, A027855, A053585.

Programs

Mathematica

Select[Range@ 280, Function[n, (n/Apply[Power, Last@ #]) > #[[-1, 1]] &@ FactorInteger[n]]] (* Michael De Vlieger, Jul 13 2017 *)

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

Extensions

Extended by Ray Chandler, Nov 17 2008

Offset changed to 1 by Michel Marcus, Jan 16 2014