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A229094 Composite squarefree numbers k such that the arithmetic mean of the distinct prime factors of k is a prime p, and p divides k. 3
105, 231, 627, 897, 935, 1365, 1581, 1729, 2465, 2967, 4123, 4301, 4715, 5313, 5487, 6045, 7293, 7685, 7881, 7917, 9717, 10707, 10965, 11339, 12597, 14637, 14993, 16377, 16445, 17353, 18753, 20213, 20757, 20915, 21045, 23779, 25327, 26331, 26765, 26961 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
Let A(x) be the set of terms <= x. The estimates x/(exp((2 + o(1))*sqrt(log x log log x)) <= #A(x) <= x/(exp((1/sqrt(2) + o(1))*sqrt(log x log log x)) hold as x -> infinity.
LINKS
Florian Luca and Francesco Pappalardi, Composite positive integers with an average prime factor, Acta Arithmetica, Vol. 129, No. 2 (2007), pp. 197-201.
FORMULA
omega(a(n)) > 2. - David A. Corneth, May 01 2017
EXAMPLE
935 is in the list for the following reasons. First, 935 is squarefree and composite. Secondly the distinct prime factors of 935 are 5, 11, and 17, and the average of these three prime factors is 11, which is also prime. Finally, 935 is divisible by 11 (the prime average of the distinct prime factors).
Similarly, 1365 is in the list since it is composite, squarefree, and its distinct prime factors are 3, 5, 7, and 13. The average of the prime factors is 28/4=7, 7 is prime, and 7 divides 1365. - Tom Edgar, Oct 21 2014
MATHEMATICA
Reap[For[k = 6, k < 10^5, k++, If[SquareFreeQ[k] && CompositeQ[k], m = Mean[FactorInteger[k][[All, 1]]]; If[IntegerQ[m] && PrimeQ[m] && Mod[k, m] == 0, Print[k]; Sow[k]]]]][[2, 1]] (* Jean-François Alcover, May 01 2017 *)
PROG
(PARI) for(n=2, 26961, if(issquarefree(n)&&!isprime(n), o=omega(n); s=sum(i=1, o, factor(n)[, 1][i]); a=s/o; if(!frac(a)&&isprime(a)&&!Mod(n, a), print1(n, ", "))));
CROSSREFS
Subsequence of A120944.
Cf. A185642.
Sequence in context: A176878 A088595 A308643 * A307108 A262723 A250757
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified March 19 06:47 EDT 2024. Contains 370953 sequences. (Running on oeis4.)