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.

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

Offset: 1

Author

Arkadiusz Wesolowski, Sep 13 2013

Keywords

• nonn

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.

Examples

			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
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Florian Luca and Francesco Pappalardi, Composite positive integers with an average prime factor, Acta Arithmetica, Vol. 129, No. 2 (2007), pp. 197-201.

Crossrefs

Subsequence of A120944.

Cf. A185642.

Programs

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 *)

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, ", "))));

Formula

omega(a(n)) > 2. - David A. Corneth, May 01 2017