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Code Golf

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Harmonic divisor numbers

Consider the \4ドル\$ divisors of \6ドル\$: \1,ドル 2, 3, 6\$. We can calculate the harmonic mean of these numbers as

$$\frac 4 {\frac 1 1 + \frac 1 2 + \frac 1 3 + \frac 1 6} = \frac 4 {\frac {12} 6} = \frac 4 2 = 2$$

However, if we take the \6ドル\$ divisors of \12ドル\$ (\1,ドル 2, 3, 4, 6, 12\$) and calculate their harmonic mean, we get

$$\frac 6 {\frac 1 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 6 + \frac 1 {12}} = \frac 6 {\frac {28} {12}} = \frac {18} {7}$$

Ore numbers or harmonic divisor numbers are positive integers \$n\$ where the harmonic mean of \$n\$'s divisors is an integer, for example \6ドル\$. They are A001599 in the OEIS.

The first few Ore numbers are

1, 6, 28, 140, 270, 496, 672, 1638, 2970, 6200, 8128, 8190, ...

Point of interest: this sequence contains all the perfect numbers (see Wikipedia for a proof).

This is a standard challenge. You may choose which of the following three options to do:

  • Take a positive integer \$n\$ and output the first \$n\$ Ore numbers.
  • Take a positive integer \$n\$ and output the \$n\$th Ore number.
    • You may use 0-indexing (so non-negative input) or 1-indexing, your choice
  • Take no input, and output the never ending list of Ore numbers.

Note that your answer cannot fail due to floating point errors.

This is , so the shortest code in bytes wins.

Answer*

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  • 2
    \$\begingroup\$ 40 bytes \$\endgroup\$ Commented Nov 24, 2021 at 22:49
  • \$\begingroup\$ How do you measure the bytes in Mathematica? \$\endgroup\$ Commented Nov 25, 2021 at 16:04
  • \$\begingroup\$ @EGME Mathematica doesn't have any special encoding, so we just use UTF-8, where both ∣ and ∞ are encoded in 3 bytes. \$\endgroup\$ Commented Nov 25, 2021 at 16:07
  • \$\begingroup\$ So how are you measuring the bytes then, for the whole thing? Do you just take the character string and see how much space it needs? \$\endgroup\$ Commented Nov 25, 2021 at 16:08
  • 1
    \$\begingroup\$ Thanks, I think I got it ... let’s see if I can better this :) \$\endgroup\$ Commented Nov 25, 2021 at 16:20

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