Harmonic divisor numbers

Question

Consider the \$4\$ divisors of \$6\$: \$1, 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\$ divisors of \$12\$ (\$1, 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\$. 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 sequence 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 code-golf, so the shortest code in bytes wins.

Answer 1

Factor + lists.lazy math.primes.factors math.unicode, ⁶⁹ 65 bytes

[ 1 lfrom [ divisors [ length 1 ] keep n/v Σ mod 0 = ] lfilter ]

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It's a quotation that returns an infinite lazy list of the harmonic divisor numbers.

Explanation

• 1 lfrom an infinite lazy list of natural numbers
• [ ... ] lfilter select numbers for which the quotation returns true
• divisors get the divisors of a number (e.g. 6 divisors -> { 1 2 3 6 })
• [ length 1 ] keep (e.g. { 1 2 3 6 } [ length 1 ] keep -> 4 1 { 1 2 3 6 })
• n/v divide number by vector (e.g. 1 { 1 2 3 6 } n/v -> { 1 1/2 1/3 1/6 })
• Σ take the sum
• mod 0 = is it a divisor?

Answer 2

Raku, 46 bytes

grep {{@_%%sum 1 X/@_}(grep $_%%*,1..$_)},^∞

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This is a lazy infinite sequence of the harmonic divisor numbers.

Answer 3

R, 55 52 bytes

-3 bytes thanks to @Dominic van Essen.

while(F<-F+1)(1/mean(1/(y=1:F)[!F%%y]))%%1||print(F)

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Prints Ore numbers infinitely.

Answer 4

Jelly, 9 bytes

ÆDpWSḍ/µ#

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This is horribly scuffed because I couldn't figure out how to get it working with precision. I had the same idea as ovs it turns out, but ÆDÆmḍµ# fails due to precision issues.

I honestly hate how this is written.

ÆDpWSḍ/µ#    Main Link
       µ#    nfind; return first n values satisfying:
ÆD           divisors of n
  p          cartesian product with
   W         [n] (returns [[a, n], [b, n], ...])
    S        sum (returns [divisor sum, divisor count * n])
     ḍ/      reduce by divisibility check

Answer 5

Pari/GP, 42 bytes

for(n=1,oo,numdiv(n)*n%sigma(n)||print(n))

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Answer 6

Wolfram Language (Mathematica), 42 40 bytes

-2 bytes thanks to att!

Do[Mean@Divisors@n∣n&&Print@n,{n,∞}]

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Answer 7

Charcoal, 37 bytes

Ｎθ≔⁰ηＷθ«≦⊕η≔Φ⊕η∧κ¬﹪ηκζ¿¬﹪×ηＬζΣζ≦⊖θ»Ｉη

Try it online! Link is to verbose version of code. Outputs the 1-indexed nᵗʰ Ore number. Explanation:

Ｎθ

Input n.

≔⁰η

Start looking for Ore numbers greater than zero.

Ｗθ«

Repeat until the nᵗʰ number has been found.

≦⊕η

Try the next integer.

≔Φ⊕η∧κ¬﹪ηκζ

Get its factors.

¿¬﹪×ηＬζΣζ

If the harmonic mean is an integer, then...

≦⊖θ

Decrement the count of remaining Ore numbers to find.

»Ｉη

Print the found Ore number.

Answer 8

Ruby, 71 56 bytes

1.step{|n|k=0;(1..n).count{|x|n%x<1&&k+=1r/x}%k>0||p(n)}

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• Saved 15 thanks to @G B lots of golfs

Outputs the sequence indefinitely.

Answer 9

Jelly, 9 bytes

1Æd×ọÆsƲ#

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More boring than the other answer:

             Implicit input: an integer z.
1      Ʋ#    Count up from 1, finding z numbers for which...
 Æd×           divisor_count(n) × n
    ọ          is divisible by
     Æs        divisor_sum(n).

Answer 10

Core Maude, 248 bytes

mod H is pr LIST{Rat}. ops o h : Rat ~> Rat . var A B C D : Rat . eq o(A)= o(2
A). eq o(s A 0)= A . eq o(A s B)= o(s A(B + ceiling(frac(h(A A 0 0))))). eq h(A
s B C D)= h(A B(0 ^(A rem s B)/ s B + C)(0 ^(A rem s B)+ D)). eq h(A 0 C D)=
D / C . endm

The result is obtained by reducing the o function with the zero-indexed input \$n\$.

Example Session

Maude> red o(0) .  --- 1
result NzNat: 1
Maude> red o(1) .  --- 6
result NzNat: 6
Maude> red o(2) .  --- 28
result NzNat: 28
Maude> red o(3) .  --- 140
result NzNat: 140
Maude> red o(4) .  --- 270
result NzNat: 270
Maude> red o(5) .  --- 496
result NzNat: 496
Maude> red o(6) .  --- 672
result NzNat: 672
Maude> red o(7) .  --- 1638
result NzNat: 1638
Maude> red o(8) .  --- 2970
result NzNat: 2970
Maude> red o(9) .  --- 6200
result NzNat: 6200
Maude> red o(10) .  --- 8128
result NzNat: 8128
Maude> red o(11) .  --- 8190
result NzNat: 8190

Ungolfed

mod H is
    pr LIST{Rat} .

    ops o h : Rat ~> Rat .

    var A B C D : Rat .

    eq o(A) = o(2 A) .
    eq o(s A 0) = A .
    eq o(A s B) = o(s A (B + ceiling(frac(h(A A 0 0))))) .

    eq h(A s B C D) = h(A B (0 ^ (A rem s B) / s B + C) (0 ^ (A rem s B) + D)) .
    eq h(A 0 C D) = D / C .
endm

Maude has built-in support for rational arithmetic, so we just compute the harmonic mean of the divisors with h. Then, ceiling(frac(h(...))) will be 0 if h(...) is a natural number or 1 otherwise. Also, note that in Maude 0 ^ 0 == 1 and 0 ^ X = 0 for X =/= 1.

Answer 11

Stax, 11 bytes

¡♪♫ö╪ü♣↕¥Vv

Run and debug it

Runs an infinite loop with no input.

Answer 12

Zephyr, 151 bytes

set n to 1
while 1=1
set s to 0
set c to 0
for d from 1to n
if(n mod d)=0
set s to(/d)+s
inc c
end if
next
if((c/s)mod 1)=0
print n
end if
inc n
repeat

Try it online! Uses the output-infinitely strategy; you'll need to kill the program before 60 seconds in order to see any output.

Ungolfed

# Start from 1
set num to 1
# Loop forever
while true
  # Calculate the sum of the reciprocals of the divisors
  # and also the total number of divisors
  set reciprocalSum to 0
  set divisorCount to 0
  for divisor from 1 to num
    if (num mod divisor) = 0
      set reciprocalSum to reciprocalSum + (/ divisor)
      inc divisorCount
    end if
  next
  # Print the number if the divisor count divided by the
  # divisor-reciprocal sum is an integer
  if ((divisorCount / reciprocalSum) mod 1) = 0
    print num
  end if
  # Go to the next number
  inc num
repeat

Answer 13

Pyt, 22 bytes

1`ĐðĐĐΠ⇹/Ʃ⇹Π|?ŕĐƥ:ŕ;⁺ł

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Prints Ore numbers forever.

1                              Push 1
 `                   ł         Do... while top of stack is truthy
  Đ                            Đuplicate
   ð                           Get list of ðivisors
    ĐĐ                         Đuplicate twice
      Π                        Get product
       ⇹                       Swap top two elements on stack
        /                      Divide
         Ʃ                     Ʃum
          ⇹                    Swap top two elements
           Π|?                 Does the sum divide the product cleanly?
              ŕĐƥ              If so, ŕemove the boolean, Đuplicate, and ƥrint
                 :ŕ            Otherwise, ŕemove the boolean
                   ;⁺          Either way, increment

Answer 14

MathGolf, 13 bytes

î∙─‼Σ£î*\÷╛p∟

Outputs indefinitely.

Try it online. (You do have to manually cancel it during runtime to see output apparently, before it times out after 60 seconds..)

Explanation:

            ∟  # Do-while true without popping:
î              #  Push the 1-based loop-index
 ∙             #  Triplicate it
  ─            #  Pop the top, and get a list of its divisors
   ‼           #  Apply the following two commands separately:
    Σ          #   Sum the divisors-list
     £         #   Get the length of the divisors-list
      î*       #  Multiply the length by the 1-based loop-index
        \      #  Swap the top two values on the stack
         ÷     #  Check that the length*î is divisible by the sum
          ╛    #  If this is truthy:
           p   #   Pop the remaining copy of the index, and print it

Answer 15

Python 3, 79 bytes

n=0
while 1:n+=1;a=[i for i in range(1,n+1)if n%i<1];n*len(a)%sum(a)or print(n)

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Outputs indefinitely.

Answer 16

JavaScript (V8), 63 bytes

Prints the sequence forever.

{for(n=0;;s%t||print(n))for(k=++n,t=s=0;k;)n%k--||(s+=n,t-=~k)}

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Answer 17

Husk, 9 bytes

fo§¦ṁ\LḊN

Try it online! (header outputs the first few elements to avoid timing-out)

         N  # from the infinite list of integers
fo          # output those for which
     ṁ\     # the sum of the reciprocals of their divisors
  §¦        # exactly divides
       LḊ   # the length (number) of their divisors

Answer 18

Scala, 111 bytes

Stream.iterate(1:BigInt)(_+1)filter{n=>val d=n to(1,-1)filter(n%_<1)
val p=d.product
p*d.size%d.map(p./).sum<1}

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Returns an infinite Stream.

Scala, 92 bytes

Stream from 1 filter{n=>val d=1 to n filter(n%_<1)
val p=d.product
p*d.size%(0/:d)(_+p/_)<1}

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This one uses normal Ints, evading some of the boilerplate above, but it only generates the first three elements correctly due to integer overflows.

Answer 19

Python 3, 94 bytes

v=0
while 1:p=q=1;v+=1;-~len([(p:=p*d+q,q:=q*d)for d in range(2,v+1)if v%d<1])*q%p or print(v)

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Outputs indefinitely. This is otherwise like (my edit to) @Jitse's answer, but computes the sum of the reciprocals p/q exactly as a pair of (big)ints (p,q).

Answer 20

Vyxal 2.6.1, 5 bytes

≬KṁḊȯ

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This doesn't work in 2.4 because 2.6 uses sympy rationals to store non-integers. Essentially a port of hyper's hypothetical Jelly answer.

Explained

≬KṁḊȯ
≬     # The next three elements as a function, taking single argument n:
 K    #    divisors of n
  ṁ   #    the average of that
   Ḋ  #    does that divide n?
    ȯ # First input numbers that satisfy the above function.

Answer 21

Nibbles, 8.5 bytes

|,~~%*,;|,$~%@$@+

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