OFFSET
1,2
COMMENTS
If k is a harmonic number (A001599) and p is a prime that does not divide k, then k*p is a harmonic number if and only if (p+1)/2 is a divisor of the harmonic mean of the divisors of k, h(k) = k*tau(k)/sigma(k) = k*A000005(k)/A000203(k). The terms of this sequence are harmonic numbers k such that for all the divisors d of h(k), 2*d - 1 is either a nonprime or a prime divisor of k.
The even perfect numbers, 2^(p-1)*(2^p - 1) where p is a Mersenne exponent (A000043), have harmonic mean of divisors p. Therefore, they are in this sequence if p = 2 or if 2*p - 1 is composite (i.e., not in A172461). Of the first 47 Mersenne exponents there are 37 such primes (p = 2, 5, 13, 17, ...), with the corresponding even perfect numbers 6, 496, 33550336, 8589869056, ...
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..246 (terms below 10^14)
Mariano Garcia, On numbers with integral harmonic mean, The American Mathematical Monthly, Vol. 61, No. 2 (1954), pp. 89-96. See page 95.
EXAMPLE
1 is a term since it is a harmonic number, and there is no prime p such that 1*p = p is a harmonic number (if p is a prime, h(p) = 2*p/(p+1) cannot be an integer).
MATHEMATICA
harmNums = Cases[Import["https://oeis.org/A001599/b001599.txt", "Table"], {_, _}][[;; , 2]]; harMean[n_] := n * DivisorSigma[0, n]/DivisorSigma[1, n]; primeCountQ[n_] := Module[{d = Divisors[harMean[n]]}, Select[2*d - 1, PrimeQ[#] && ! Divisible[n, #] &] == {}]; Select[harmNums, primeCountQ]
CROSSREFS
KEYWORD
nonn
AUTHOR
Amiram Eldar, Jun 03 2020
STATUS
approved