A277624 Composite numbers which have a dominant prime factor. A prime factor p of n is dominant if floor(sqrt(p)) > (n/p).

22, 26, 34, 38, 46, 51, 57, 58, 62, 69, 74, 82, 86, 87, 93, 94, 106, 111, 116, 118, 122, 123, 124, 129, 134, 141, 142, 146, 148, 158, 159, 164, 166, 172, 177, 178, 183, 185, 188, 194, 201, 202, 205, 206, 212, 213, 214, 215, 218, 219, 226, 235, 236, 237, 244

Offset: 1

Author

Peter Luschny, Oct 24 2016

Keywords

• nonn
• easy

Comments

From David A. Corneth, Oct 26 2016 and Oct 27 2016: (Start)

Numbers of the form k * p where p > (k + 1)^2 and p prime and k > 1.

If n has a dominant prime factor, it's A006530(n).

All primes p > 4 have the property that floor(sqrt(A006530(p))) = floor(sqrt(p)) > (p/A006530(p)) = 1.

A063763 is a supersequence.

(End)

Examples

			133230 is in this sequence because 133230 = 2*3*5*4441 and 2*3*5 = 30 < 66 = floor(sqrt(4441)).
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A006530, A063763.

Programs

Maple

is_a := proc(n) max(numtheory:-factorset(n)):
not isprime(n) and floor(sqrt(%)) > (n/%) end:
select(is_a, [$1..244]);

Mathematica

Select[Select[Range@ 244, CompositeQ], Function[n, Total@ Boole@ Map[Function[p, Floor@ Sqrt@ p > n/p], FactorInteger[n][[All, 1]]] > 0]] (* Michael De Vlieger, Oct 27 2016 *)

PARI

upto(n) = my(l=List()); for(k=2, sqrtnint(n, 3), forprime(p=(k+1)^2, n\k, listput(l,k*p))); listsort(l); l
is(n) = if(!isprime(n)&&n>1, f=factor(n)[, 1];sqrtint(f[#f]) > n/f[#f], 0) \\ David A. Corneth, Oct 26 2016

Python

from sympy import primefactors
from gmpy2 import is_prime, isqrt
A277624_list = []
for n in range(2,10**3):
    if not is_prime(n):
        for p in primefactors(n):
            if isqrt(p)*p > n:
                A277624_list.append(n)
                break # Chai Wah Wu, Oct 25 2016

Formula

Conjecture: A092487(a(n)) = A006530(a(n)). - David A. Corneth, Oct 27 2016