OFFSET
1,2
COMMENTS
The equation n = m*tau(m)/sigma(m) has an integer solution m.
Here tau(n) (A000005) is the number of divisors of n and sigma(n) is the sum of the divisors of n (A000203).
A001600 sorted in order.
The Mersenne exponents (A000043) are in this sequence because the even perfect numbers, 2^(p-1)*(2^p-1) where p is in A000043, are all harmonic numbers (A001599) with harmonic mean of divisors p. - Amiram Eldar, Apr 15 2024
REFERENCES
For further references see A001599.
LINKS
T. Goto and S. Shibata, All numbers whose positive divisors have integral harmonic mean up to 300, Math. Comput. 73 (2004), 475-491.
MATHEMATICA
f[n_] := (n*DivisorSigma[0, n]/DivisorSigma[1, n]); a = Table[ 0, {120}]; Do[ b = f[n]; If[ IntegerQ[b] && b < 121 && a[[b]] == 0, a[[b]] = n], {n, 1, 560000000}]; Select[ Range[120], a[[ # ]] > 0 &] (* Robert G. Wilson v, Feb 14 2004 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
R. K. Guy, Feb 08 2004
EXTENSIONS
More terms from Robert G. Wilson v, Feb 14 2004
STATUS
approved