OFFSET
1,1
COMMENTS
Of course, for all prime numbers the harmonic mean of proper divisors is an integer.
a(4) >= 2*10^11. - Hiroaki Yamanouchi, Nov 20 2014
Conjecture: all terms are of the form m*(sigma(m)-1) where sigma(m)-1 is prime. - Chai Wah Wu, Dec 15 2020
a(4) <= 22351741783447265625. - Daniel Suteu, Dec 16 2020
From Chai Wah Wu, Feb 04 2021: (Start)
Other terms of the sequence of the form m*(sigma(m)-1) correspond to the following values of m:
3 * 5^143
3 * 5^623
3 * 5^1423
5 * 7^127
5 * 7^6595
101 * 103^25
(End)
Equivalently, composite numbers k such that sigma(k)-1 divides k*(tau(k)-1), where sigma = A000203 and tau = A000005. - Daniel Suteu, Feb 05 2021
EXAMPLE
The proper divisors of 1645 are [1,5,7,35,47,235,329] and their harmonic mean is 7/(1/1 + 1/5 + 1/7 + 1/35 + 1/47 + 1/235 + 1/329) = 5.
MATHEMATICA
Select[Range[2, 100000], (IntegerQ[HarmonicMean[Most[Divisors[#]]]] && Not[PrimeQ[#]])&] (* Daniel Lignon, Nov 17 2014 *)
PROG
(PARI) lista(nn) = forcomposite (n=2, nn, my(d=divisors(n)); if (denominator((#d-1)/sum(i=1, #d-1, 1/d[i])) == 1, print1(n, ", "))); \\ Michel Marcus, Nov 17 2014
(PARI) isok(n) = n > 1 && !isprime(n) && (n*(numdiv(n)-1)) % (sigma(n)-1) == 0; \\ Daniel Suteu, Feb 05 2021
CROSSREFS
KEYWORD
nonn,more,bref
AUTHOR
Daniel Lignon, Nov 17 2014
EXTENSIONS
a(3) from Hiroaki Yamanouchi, Nov 20 2014
STATUS
approved