A027855 Antimutinous 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.

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 37, 38, 39, 41, 42, 43, 44, 46, 47, 49, 50, 51, 52, 53, 54, 55, 57, 58, 59, 61, 62, 64, 65, 66, 67, 68, 69, 71, 73, 74, 75, 76, 77, 78, 79, 81, 82, 83, 85, 86, 87

Offset: 1

Author

Leroy Quet

Keywords

• nonn

Comments

Numbers which can be expressed as m*p^k, for p prime and m < p and k > 0. List contains n if A006530(n) > A051119(n). - Harry Richman, Aug 19 2019

Links

• Ivan Neretin, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A006530, A027854, A051119.

Programs

Maple

isA027855 := proc(n) local p,k,pk; if n <= 1 then false; else p := A006530(n) ; pk := p ; while n mod ( pk*p) = 0 do pk := pk*p ; od: if n< p*pk then true ; else false ; fi ; fi ; end proc:
for n from 2 to 120 do if isA027855(n) then printf("%d, ",n) ; fi ; od: # R. J. Mathar, Dec 02 2007

Mathematica

Select[Range@100, #1^(#2 + 1) & @@ FactorInteger[#][[-1]] > # &] (* Ivan Neretin, Jul 09 2015 *)

PARI

is(n) = my(f = factor(n)); c = n\f[#f~, 1]^f[#f~, 2]; c < f[#f~, 1] \\ David A. Corneth, Aug 19 2019

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)Indranil Ghosh, Jul 13 2017

Extensions

More terms from R. J. Mathar, Dec 02 2007