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A277624
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Composite numbers which have a dominant prime factor. A prime factor p of n is dominant if floor(sqrt(p)) > (n/p).
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3
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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
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OFFSET
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1,1
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COMMENTS
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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.
(End)
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LINKS
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FORMULA
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EXAMPLE
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133230 is in this sequence because 133230 = 2*3*5*4441 and 2*3*5 = 30 < 66 = floor(sqrt(4441)).
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MAPLE
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is_a := proc(n) max(numtheory:-factorset(n)):
not isprime(n) and floor(sqrt(%)) > (n/%) end:
select(is_a, [$1..244]);
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MATHEMATICA
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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 *)
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PROG
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(Python)
from sympy import primefactors
from gmpy2 import is_prime, isqrt
for n in range(2, 10**3):
if not is_prime(n):
for p in primefactors(n):
if isqrt(p)*p > n:
(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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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