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%I #52 Feb 06 2021 10:22:39
%S 1645,88473,63626653506
%N Composite numbers for which the harmonic mean of proper divisors is an integer.
%C Of course, for all prime numbers the harmonic mean of proper divisors is an integer.
%C a(4) >= 2*10^11. - _Hiroaki Yamanouchi_, Nov 20 2014
%C Conjecture: all terms are of the form m*(sigma(m)-1) where sigma(m)-1 is prime. - _Chai Wah Wu_, Dec 15 2020
%C a(4) <= 22351741783447265625. - _Daniel Suteu_, Dec 16 2020
%C From _Chai Wah Wu_, Feb 04 2021: (Start)
%C Other terms of the sequence of the form m*(sigma(m)-1) correspond to the following values of m:
%C 3 * 5^143
%C 3 * 5^623
%C 3 * 5^1423
%C 5 * 7^127
%C 5 * 7^6595
%C 101 * 103^25
%C (End)
%C 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
%e 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.
%t Select[Range[2,100000],(IntegerQ[HarmonicMean[Most[Divisors[#]]]] && Not[PrimeQ[#]])&] (* _Daniel Lignon_, Nov 17 2014 *)
%o (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
%o (PARI) isok(n) = n > 1 && !isprime(n) && (n*(numdiv(n)-1)) % (sigma(n)-1) == 0; \\ _Daniel Suteu_, Feb 05 2021
%Y Cf. A001599 for harmonic mean of all divisors and A247078 for harmonic mean of nontrivial divisors.
%K nonn,more,bref
%O 1,1
%A _Daniel Lignon_, Nov 17 2014
%E a(3) from _Hiroaki Yamanouchi_, Nov 20 2014
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