Harmonic divisor numbers

Question 1

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 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.

Question 2

Brownie points for beating/matching my 9 byte Jelly answer

Question 3

Can our answers fail due to integer overflow errors? (I assume not, but just checking)

Question 4

@user I don't really care if they fail because the numbers involved exceeded a practical size, so long as it would work in theory, and that the failure isn't because of any floating point errors

Question 5

Factor + `lists.lazy math.primes.factors math.unicode`, ^{(削除) 69 (削除ここまで)} 65 bytes

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

Try it online!

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?

Question 6

Raku, 46 bytes

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

Try it online!

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

Question 7

R, (削除) 55 (削除ここまで) 52 bytes

-3 bytes thanks to @Dominic van Essen.

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

Try it online!

Prints Ore numbers infinitely.

Question 8

I was slower than you, but got 54 bytes; the approach is pretty similar, though...

Question 9

@Dominic, nice, combination of these approaches can give us 52

Question 10

Hmm... I'm not sure that the step of taking the mean (or summing) reciprocals (mean(1/(y=1:F)[!F%%y]), or sum(1/k[d]) in the first version) will always satisfy "your answer cannot fail due to floating point errors".

Question 11

I think the FP error issue can be kind of avoided by summing the product-of-divisors divided by each divisor (so the division always yields an integer) in 70 bytes, but it's not really much of an improvement, because it goes out of R's integer range before encountering any 'Ore numbers' that would cause a floating point error...

Question 12

@Dominic, that's a tough one. I assumed that since (within reasonable limits) we don't encounter floating point errors, then current solution is fine. It's also not clear to me from this meta discussion, as this solution doesn't currently fail due to floats, but may - over some unreasonable bound.

Question 13

Jelly, 9 bytes

ÆDpWSḍ/μ#

Try It Online!

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

Question 14

Pari/GP, 42 bytes

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

Try it online!

Question 15

Wolfram Language (Mathematica), (削除) 42 (削除ここまで) 40 bytes

-2 bytes thanks to att!

Do[Mean@Divisors@n∣n&&Print@n,{n,∞}]

Try it online!

Question 16

40 bytes

Question 17

How do you measure the bytes in Mathematica?

Question 18

@EGME Mathematica doesn't have any special encoding, so we just use UTF-8, where both ∣ and ∞ are encoded in 3 bytes.

Question 19

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?

Question 20

Thanks, I think I got it ... let’s see if I can better this :)

Question 21

Pyt, 22 bytes

1`ĐðĐĐΠ⇹/Ʃ⇹Π|?ŕĐƥ:ŕ;+ł

Try it online!

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

Question 22

Charcoal, 37 bytes

×ばつ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.

×ばつ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.

Question 23

Ruby, (削除) 71 (削除ここまで) 56 bytes

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

Try it online!

Saved 15 thanks to @G B lots of golfs

Outputs the sequence indefinitely.

Question 24

Shorter

Question 25

Even shorter

Question 26

Jelly, 9 bytes

1Æd×ばつọÆsƲ#

Try it online!

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.

Question 29

Stax, 11 bytes

¡♪♫ö╪ü♣↕\Vv

Run and debug it

Runs an infinite loop with no input.

Question 30

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

Question 31

Nibbles, 8.5 bytes

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

Attempt This Online!

Question 32

TI-BASIC (TI-84 Plus CE Python), (削除) 74 67 (削除ここまで) 65 bytes

While 1
Ans+1
If not(fPart(round(AnsΣ(int(Ans/K)-int((Ans-1)/K),K,1,1+Ans)/Σ(Knot(fPart(Ans/K)),K,1,1+Ans),9
Disp Ans
End

make sure Ans is set to 0 before running.

Question 33

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

Question 34

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)

Try it online!

Outputs indefinitely.

Question 35

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)}

Try it online!

Question 36

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

Question 37

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}

Try it online!

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}

Try it online!

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

Question 38

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)

Try it online!

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).

Question 39

For what it's worth, your edit was rejected, as Jitse's answer works at the moment, and your edit would have made it reliant on floating points, making it prone to failing due to floating point errors. This answer is perfectly fine though

Question 40

Vyxal 2.6.1, 5 bytes

≬KṁḊȯ

Try it Online!

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.

Question 41

Why not check to see if the number of divisors is divisible by the sum of the reciprocals, rather than checking if the division is an integer?

Question 42

Because it doesn't work for cases like 28 @cairdcoinheringaahing

Question 43

@cairdcoinheringaahing also, floating point inaccuracies

Question 44

Having said that, that isn't an issue in the 2.6 pre-releases lol

Question 45

@Radek you need to download the v2.6.0rc2 build of vyxal and use the REPL

Question 46

YASEPL, 81 bytes

=n!+=k$=f=o`1=g$n-/k(=l$n/k(-g!f+l=u$n/k%1+[2!o+k`2!k+}4,n!o+.9(!f/o*n%1+[3>n`3?3

this prints them infinitely

explanation

=n main increment
 (after this is where the loop starts)
 !+ add 1 to n
 =k$ K variable for summation
 =f F is total amount of divisors of N
 =o O is sum of divisors of N
 `1 start sum
 ( calculating number of divisors )
 =g$n-/k(=l$n/k(-g l = floor(n/k) - floor(n-1/k)
 !f+l add to sum
 ( calculating sum of divisors )
 =u$n/k%1+[2 if (n/k) = 0...
 !o+k add K to O
 `2 end if
 !k+ increment K
 }4,n if K <= N, go back to `1
 !o+.9( ceil O
 !f/o*n%1+[3 if (f/o)*n = 0...
 >n print n
 `3 endif
?3 go back to the third char of the program (where !+ is)

this assumes that \$ n \$ is a harmonic divisor number if the following is true:

$$ 0=(\frac{\sum_{k=1}^{n}(\left\lfloor\frac{n}{k}\right\rfloor-\left\lfloor\frac{n-1}{k} \right\rfloor)}{\left\lceil\sum_{i=1}^{n}(0^{\frac{n}{i}\mod1}i)\right\rceil}\times n)\mod1 $$ I'm not the best with writing math equations so take it with a grain of salt.

this thing gets really slow.

Question 47

APL(NARS), 80 chars

r←F w;x;m;k×ばつ⍳∼1=÷1∨(≢m)÷+/÷m←k/⍨0=x∣⍨k←⍳x
r,←x⋄w-←1⋄→2

// +/ 12 9 8 38 13=80

F 1-indexed, and it is slow it seems has at last O(n^5) if goes well. F use rationals.
÷1∨k finds the denominator of k rational.

Test:

 F 5
1 6 28 140 270 
 F 10
1 6 28 140 270 496 672 1638 2970 6200 
 F 1
1

Question 48

jBasher2, 592 bytes

create n with type number
while 1 == 1
add n by 1
set that to n
create k with type number
set 1 to k
create f with type number
create o with type number
while k <= n
create g with type number
subtract n by 1
divide that by k
parse that as int
set that to g
divide n by k
parse that as int
subtract that by g
add f by that
set that to f
divide n by k
modulo that by 1
if that == 0
add k by o
set that to o
endif
add 1 by k
set that to k
endwhile
divide 1 by 10
add o by that
parse that as int
set that to o
divide f by o
multiply that by n
modulo that by 1
if that == 0
output n
endif
endwhile

awesome translation of my YASEPL answer

chunes 27.8k3 gold badges32 silver badges55 bronze badges · Answer 1 · 2021-11-23 02:24:00Z

Factor + `lists.lazy math.primes.factors math.unicode`, ^{(削除) 69 (削除ここまで)} 65 bytes

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

Try it online!

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?

Sean 8,16612 silver badges16 bronze badges · Answer 2 · 2021-11-23 07:53:27Z

4

\$\begingroup\$

Raku, 46 bytes

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

Try it online!

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

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answered Nov 23, 2021 at 7:53

Sean's user avatar

Sean

8,16612 silver badges16 bronze badges

\$\endgroup\$

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pajonk 19.4k5 gold badges29 silver badges78 bronze badges · Answer 3 · 2021-11-23 07:06:57Z

4

\$\begingroup\$

R, (削除) 55 (削除ここまで) 52 bytes

-3 bytes thanks to @Dominic van Essen.

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

Try it online!

Prints Ore numbers infinitely.

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edited Nov 23, 2021 at 9:10

answered Nov 23, 2021 at 7:06

pajonk's user avatar

pajonk

19.4k5 gold badges29 silver badges78 bronze badges

\$\endgroup\$

5

\$\begingroup\$ I was slower than you, but got 54 bytes; the approach is pretty similar, though... \$\endgroup\$

Dominic van Essen
– Dominic van Essen

2021年11月23日 09:02:19 +00:00
Commented Nov 23, 2021 at 9:02
\$\begingroup\$ @Dominic, nice, combination of these approaches can give us 52 \$\endgroup\$

pajonk
– pajonk

2021年11月23日 09:04:27 +00:00
Commented Nov 23, 2021 at 9:04
\$\begingroup\$ Hmm... I'm not sure that the step of taking the mean (or summing) reciprocals (mean(1/(y=1:F)[!F%%y]), or sum(1/k[d]) in the first version) will always satisfy "your answer cannot fail due to floating point errors". \$\endgroup\$

Dominic van Essen
– Dominic van Essen

2021年11月23日 10:47:35 +00:00
Commented Nov 23, 2021 at 10:47
\$\begingroup\$ I think the FP error issue can be kind of avoided by summing the product-of-divisors divided by each divisor (so the division always yields an integer) in 70 bytes, but it's not really much of an improvement, because it goes out of R's integer range before encountering any 'Ore numbers' that would cause a floating point error... \$\endgroup\$

Dominic van Essen
– Dominic van Essen

2021年11月23日 10:48:52 +00:00
Commented Nov 23, 2021 at 10:48
\$\begingroup\$ @Dominic, that's a tough one. I assumed that since (within reasonable limits) we don't encounter floating point errors, then current solution is fine. It's also not clear to me from this meta discussion, as this solution doesn't currently fail due to floats, but may - over some unreasonable bound. \$\endgroup\$

pajonk
– pajonk

2021年11月23日 11:47:07 +00:00
Commented Nov 23, 2021 at 11:47

Add a comment |

hyperneutrino ♦ 42.8k5 gold badges72 silver badges227 bronze badges · Answer 4 · 2021-11-23 01:24:30Z

Jelly, 9 bytes

ÆDpWSḍ/μ#

Try It Online!

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

alephalpha 51.9k7 gold badges75 silver badges196 bronze badges · Answer 5 · 2021-11-23 01:25:24Z

3

\$\begingroup\$

Pari/GP, 42 bytes

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

Try it online!

Share

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answered Nov 23, 2021 at 1:25

alephalpha's user avatar

alephalpha

51.9k7 gold badges75 silver badges196 bronze badges

\$\endgroup\$

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ovs 61.2k3 gold badges49 silver badges164 bronze badges · Answer 6 · 2021-11-23 09:35:11Z

3

\$\begingroup\$

Wolfram Language (Mathematica), (削除) 42 (削除ここまで) 40 bytes

-2 bytes thanks to att!

Do[Mean@Divisors@n∣n&&Print@n,{n,∞}]

Try it online!

Share

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edited Nov 25, 2021 at 16:06

answered Nov 23, 2021 at 9:35

ovs's user avatar

ovs

61.2k3 gold badges49 silver badges164 bronze badges

\$\endgroup\$

15

2

\$\begingroup\$ 40 bytes \$\endgroup\$

att
– att

2021年11月24日 22:49:34 +00:00
Commented Nov 24, 2021 at 22:49
\$\begingroup\$ How do you measure the bytes in Mathematica? \$\endgroup\$

EGME
– EGME

2021年11月25日 16:04:39 +00:00
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\$

ovs
– ovs

2021年11月25日 16:07:01 +00:00
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\$

EGME
– EGME

2021年11月25日 16:08:26 +00:00
Commented Nov 25, 2021 at 16:08
1

\$\begingroup\$ Thanks, I think I got it ... let’s see if I can better this :) \$\endgroup\$

EGME
– EGME

2021年11月25日 16:20:33 +00:00
Commented Nov 25, 2021 at 16:20

| Show 10 more comments

Kip the Malamute 1,8033 silver badges13 bronze badges · Answer 7 · 2023-02-25 17:26:00Z

Pyt, 22 bytes

1`ĐðĐĐΠ⇹/Ʃ⇹Π|?ŕĐƥ:ŕ;+ł

Try it online!

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

Neil 184k12 gold badges76 silver badges290 bronze badges · Answer 8 · 2021-11-23 00:37:17Z

Charcoal, 37 bytes

×ばつ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.

×ばつ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.

AZTECCO 11k1 gold badge17 silver badges60 bronze badges · Answer 9 · 2021-11-23 08:12:06Z

2

\$\begingroup\$

Ruby, (削除) 71 (削除ここまで) 56 bytes

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

Try it online!

Saved 15 thanks to @G B lots of golfs

Outputs the sequence indefinitely.

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edited Nov 23, 2021 at 16:26

answered Nov 23, 2021 at 8:12

AZTECCO's user avatar

AZTECCO

11k1 gold badge17 silver badges60 bronze badges

\$\endgroup\$

2

\$\begingroup\$ Shorter \$\endgroup\$

G B
– G B

2021年11月23日 12:03:51 +00:00
Commented Nov 23, 2021 at 12:03
\$\begingroup\$ Even shorter \$\endgroup\$

G B
– G B

2021年11月23日 12:06:43 +00:00
Commented Nov 23, 2021 at 12:06

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lynn 69.7k11 gold badges137 silver badges285 bronze badges · Answer 10 · 2021-11-23 18:58:42Z

2

\$\begingroup\$

Jelly, 9 bytes

1Æd×ばつọÆsƲ#

Try it online!

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.

recursive 10.5k21 silver badges36 bronze badges · Answer 12 · 2021-11-25 01:01:17Z

2

\$\begingroup\$

Stax, 11 bytes

¡♪♫ö╪ü♣↕\Vv

Run and debug it

Runs an infinite loop with no input.

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edited Nov 25, 2021 at 1:09

answered Nov 25, 2021 at 1:01

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recursive

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\$\endgroup\$

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DLosc 40.7k6 gold badges87 silver badges142 bronze badges · Answer 13 · 2021-11-25 19:54:35Z

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

Adamátor 8,2932 gold badges15 silver badges28 bronze badges · Answer 14 · 2023-02-26 13:00:28Z

2

\$\begingroup\$

Nibbles, 8.5 bytes

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

Attempt This Online!

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answered Feb 26, 2023 at 13:00

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Adamátor

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\$\endgroup\$

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madeforlosers 2,0181 silver badge11 bronze badges · Answer 15 · 2025-09-12 13:56:15Z

TI-BASIC (TI-84 Plus CE Python), (削除) 74 67 (削除ここまで) 65 bytes

While 1
Ans+1
If not(fPart(round(AnsΣ(int(Ans/K)-int((Ans-1)/K),K,1,1+Ans)/Σ(Knot(fPart(Ans/K)),K,1,1+Ans),9
Disp Ans
End

make sure Ans is set to 0 before running.

Kevin Cruijssen 136k14 gold badges154 silver badges394 bronze badges · Answer 16 · 2021-11-23 08:00:27Z

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

Jitse 8,1243 gold badges21 silver badges45 bronze badges · Answer 17 · 2021-11-23 08:59:42Z

1

\$\begingroup\$

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)

Try it online!

Outputs indefinitely.

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edited Nov 23, 2021 at 9:12

answered Nov 23, 2021 at 8:59

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Arnauld 206k21 gold badges187 silver badges670 bronze badges · Answer 18 · 2021-11-23 01:55:01Z

1

\$\begingroup\$

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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edited Nov 23, 2021 at 16:26

answered Nov 23, 2021 at 1:55

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Arnauld

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\$\endgroup\$

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Dominic van Essen 37.2k2 gold badges24 silver badges61 bronze badges · Answer 19 · 2021-11-24 23:10:36Z

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

user 4572 gold badges21 silver badges71 bronze badges · Answer 20 · 2021-11-26 20:31:02Z

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}

Try it online!

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}

Try it online!

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

empii 311 bronze badge · Answer 21 · 2021-12-04 21:38:23Z

1

\$\begingroup\$

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)

Try it online!

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).

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answered Dec 4, 2021 at 21:38

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\$\endgroup\$

1

\$\begingroup\$ For what it's worth, your edit was rejected, as Jitse's answer works at the moment, and your edit would have made it reliant on floating points, making it prone to failing due to floating point errors. This answer is perfectly fine though \$\endgroup\$

caird coinheringaahing
– caird coinheringaahing ♦

2021年12月04日 21:51:24 +00:00
Commented Dec 4, 2021 at 21:51

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lyxal ♦ 35.6k2 gold badges69 silver badges148 bronze badges · Answer 22 · 2021-11-23 00:26:59Z

1

\$\begingroup\$

Vyxal 2.6.1, 5 bytes

≬KṁḊȯ

Try it Online!

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.

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edited Dec 6, 2021 at 3:47

answered Nov 23, 2021 at 0:26

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lyxal ♦

35.6k2 gold badges69 silver badges148 bronze badges

\$\endgroup\$

6

\$\begingroup\$ Why not check to see if the number of divisors is divisible by the sum of the reciprocals, rather than checking if the division is an integer? \$\endgroup\$

caird coinheringaahing
– caird coinheringaahing ♦

2021年11月23日 01:33:22 +00:00
Commented Nov 23, 2021 at 1:33
\$\begingroup\$ Because it doesn't work for cases like 28 @cairdcoinheringaahing \$\endgroup\$

lyxal
– lyxal ♦

2021年11月23日 01:43:26 +00:00
Commented Nov 23, 2021 at 1:43
\$\begingroup\$ @cairdcoinheringaahing also, floating point inaccuracies \$\endgroup\$

lyxal
– lyxal ♦

2021年11月23日 01:46:19 +00:00
Commented Nov 23, 2021 at 1:46
\$\begingroup\$ Having said that, that isn't an issue in the 2.6 pre-releases lol \$\endgroup\$

lyxal
– lyxal ♦

2021年11月23日 01:49:48 +00:00
Commented Nov 23, 2021 at 1:49
1

\$\begingroup\$ @Radek you need to download the v2.6.0rc2 build of vyxal and use the REPL \$\endgroup\$

lyxal
– lyxal ♦

2021年12月05日 04:29:38 +00:00
Commented Dec 5, 2021 at 4:29

| Show 1 more comment

madeforlosers 2,0181 silver badge11 bronze badges · Answer 23 · 2025-09-04 16:39:17Z

YASEPL, 81 bytes

=n!+=k$=f=o`1=g$n-/k(=l$n/k(-g!f+l=u$n/k%1+[2!o+k`2!k+}4,n!o+.9(!f/o*n%1+[3>n`3?3

this prints them infinitely

explanation

=n main increment
 (after this is where the loop starts)
 !+ add 1 to n
 =k$ K variable for summation
 =f F is total amount of divisors of N
 =o O is sum of divisors of N
 `1 start sum
 ( calculating number of divisors )
 =g$n-/k(=l$n/k(-g l = floor(n/k) - floor(n-1/k)
 !f+l add to sum
 ( calculating sum of divisors )
 =u$n/k%1+[2 if (n/k) = 0...
 !o+k add K to O
 `2 end if
 !k+ increment K
 }4,n if K <= N, go back to `1
 !o+.9( ceil O
 !f/o*n%1+[3 if (f/o)*n = 0...
 >n print n
 `3 endif
?3 go back to the third char of the program (where !+ is)

this assumes that \$ n \$ is a harmonic divisor number if the following is true:

$$ 0=(\frac{\sum_{k=1}^{n}(\left\lfloor\frac{n}{k}\right\rfloor-\left\lfloor\frac{n-1}{k} \right\rfloor)}{\left\lceil\sum_{i=1}^{n}(0^{\frac{n}{i}\mod1}i)\right\rceil}\times n)\mod1 $$ I'm not the best with writing math equations so take it with a grain of salt.

this thing gets really slow.

Rosario 1,7045 silver badges9 bronze badges · Answer 24 · 2025-09-05 16:07:45Z

APL(NARS), 80 chars

r←F w;x;m;k×ばつ⍳∼1=÷1∨(≢m)÷+/÷m←k/⍨0=x∣⍨k←⍳x
r,←x⋄w-←1⋄→2

// +/ 12 9 8 38 13=80

F 1-indexed, and it is slow it seems has at last O(n^5) if goes well. F use rationals.
÷1∨k finds the denominator of k rational.

Test:

 F 5
1 6 28 140 270 
 F 10
1 6 28 140 270 496 672 1638 2970 6200 
 F 1
1

madeforlosers 2,0181 silver badge11 bronze badges · Answer 25 · 2025-09-09 15:49:11Z

jBasher2, 592 bytes

create n with type number
while 1 == 1
add n by 1
set that to n
create k with type number
set 1 to k
create f with type number
create o with type number
while k <= n
create g with type number
subtract n by 1
divide that by k
parse that as int
set that to g
divide n by k
parse that as int
subtract that by g
add f by that
set that to f
divide n by k
modulo that by 1
if that == 0
add k by o
set that to o
endif
add 1 by k
set that to k
endwhile
divide 1 by 10
add o by that
parse that as int
set that to o
divide f by o
multiply that by n
modulo that by 1
if that == 0
output n
endif
endwhile

awesome translation of my YASEPL answer

Stack Exchange Network

Harmonic divisor numbers

25 Answers 25

Factor + `lists.lazy math.primes.factors math.unicode`, ^{(削除) 69 (削除ここまで)} 65 bytes

Explanation

Raku, 46 bytes

R, (削除) 55 (削除ここまで) 52 bytes

Jelly, 9 bytes

Pari/GP, 42 bytes

Wolfram Language (Mathematica), (削除) 42 (削除ここまで) 40 bytes

Pyt, 22 bytes

Charcoal, 37 bytes

Ruby, (削除) 71 (削除ここまで) 56 bytes

Jelly, 9 bytes

Core Maude, 248 bytes

Example Session

Ungolfed

Stax, 11 bytes

Zephyr, 151 bytes

Ungolfed

Nibbles, 8.5 bytes

TI-BASIC (TI-84 Plus CE Python), (削除) 74 67 (削除ここまで) 65 bytes

MathGolf, 13 bytes

Python 3, 79 bytes

JavaScript (V8), 63 bytes

Husk, 9 bytes

Scala, 111 bytes

Scala, 92 bytes

Python 3, 94 bytes

Vyxal 2.6.1, 5 bytes

Explained

YASEPL, 81 bytes

APL(NARS), 80 chars

jBasher2, 592 bytes

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Harmonic divisor numbers

25 Answers 25

Factor + lists.lazy math.primes.factors math.unicode, (削除) 69 (削除ここまで) 65 bytes

Explanation

Raku, 46 bytes

R, (削除) 55 (削除ここまで) 52 bytes

Jelly, 9 bytes

Pari/GP, 42 bytes

Wolfram Language (Mathematica), (削除) 42 (削除ここまで) 40 bytes

Pyt, 22 bytes

Charcoal, 37 bytes

Ruby, (削除) 71 (削除ここまで) 56 bytes

Jelly, 9 bytes

Core Maude, 248 bytes

Example Session

Ungolfed

Stax, 11 bytes

Zephyr, 151 bytes

Ungolfed

Nibbles, 8.5 bytes

TI-BASIC (TI-84 Plus CE Python), (削除) 74 67 (削除ここまで) 65 bytes

MathGolf, 13 bytes

Python 3, 79 bytes

JavaScript (V8), 63 bytes

Husk, 9 bytes

Scala, 111 bytes

Scala, 92 bytes

Python 3, 94 bytes

Vyxal 2.6.1, 5 bytes

Explained

YASEPL, 81 bytes

APL(NARS), 80 chars

jBasher2, 592 bytes

Your Answer

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Hot Network Questions

Factor + `lists.lazy math.primes.factors math.unicode`, ^{(削除) 69 (削除ここまで)} 65 bytes