How hyperperfect am I?

Question

A \$k\$-hyperperfect number is a natural number \$n \ge 1\$ such that

$$n = 1 + k(\sigma(n) − n − 1)$$

where \$\sigma(n)\$ is the sum of the divisors of \$n\$. Note that \$\sigma(n) - n\$ is the proper divisor sum of \$n\$. The sequence of \$k\$-hyperperfect numbers begins

$$6, 21, 28, 301, 325, 496, 697, \dots$$

This is A034897 on the OEIS.

For example:

$$\begin{align} \sigma(21) & = 1 + 3 + 7 + 21 = 32 \\ 21 & = 1 + 2(32 - 21 - 1) \\ & = 1 + 2 \times 10 \\ & = 21 \end{align}$$

Therefore, \$21\$ is a \$2\$-hyperperfect number.

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You are to take a \$k\$-hyperperfect number \$n\$ as input and output the value of \$k\$. You can assume you will never have to handle numbers greater than your language's limit at any point (i.e. \$k\sigma(n)\$ will always be within the bounds of your language), and you may input and output in any convenient format.

The input is guaranteed to be a \$k\$-hyperperfect number, you don't have to handle inputs such as \$2, 87, 104\$ etc. that aren't \$k\$-hyperperfect.

This is code-golf, so the shortest code in bytes wins.

Test cases

These are the outputs for all the listed values in the OEIS for this sequence, and are the exhaustive list of inputs for \$n < 1055834\$

      n       k
      6       1
     21       2
     28       1
    301       6
    325       3
    496       1
    697      12
   1333      18
   1909      18
   2041      12
   2133       2
   3901      30
   8128       1
  10693      11
  16513       6
  19521       2
  24601      60
  26977      48
  51301      19
  96361     132
 130153     132
 159841      10
 163201     192
 176661       2
 214273      31
 250321     168
 275833     108
 296341      66
 306181      35
 389593     252
 486877      78
 495529     132
 542413     342
 808861     366
1005421     390
1005649     168
1055833     348

Answer 1

Jelly, 5 bytes

:Æṣ’$

:       integer divide
Æṣ’$    the decremented divisor sum

This uses \$\lfloor\frac{n}{\sigma(n)-n-1}\rfloor\$ instead of \$\frac{n-1}{\sigma(n)-n-1}\$, but it still works because \$\frac{1}{\sigma(n)-n-1}\$ can never be greater than or equal to 1.

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Answer 2

J, ^{24 23} 20 bytes

<.@%1<:@#.[:I.0=i.|]

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^{-3 thanks to rak's rounding down trick}

Let n be the input.

• <.@% round down n divided by
• 1<:@#. 1 minus the sum of
• [:I. the indexes where
• 0= 0 is equal to the remainder when
• i. the list 0..n-1
• | is divided elementwise into
• ] n.

Answer 3

JavaScript (ES6), 36 bytes

n=>~-n/(g=k=>~k--&&!(n%k)*k+g(k))(n)

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How?

\$\sigma(n) − n − 1\$ is the sum of the divisors of \$n\$ in \$[2..n-1]\$. This is also the sum of the divisors of \$n\$ in \$[-1..n-1]\$ because \$1\$ and \$-1\$ cancel each other out. The helper function \$g\$ computes the latter, so that we can use the slightly more golf-friendly halt condition ~k--.

Note: From a mathematical perspective, we really should remove \$0\$ from the list of possible divisors. But it is quietly ignored in this code as we get !NaN*0, which is \$0\$.

Answer 4

Wolfram Language (Mathematica), 25 bytes

--+#/(Tr@Divisors@#-#-1)&

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-1 byte from @att

Answer 5

Haskell, 38 bytes

k n=div n$sum[d|d<-[2..n-1],mod n d<1]

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Answer 6

Japt, 12 6 bytes

zUâ ¤x

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• Uses Jelly answer formula

Answer 7

V (vim), ^{81 79} 70 bytes

"aD@ai0
<esc>V{g<c-a>{dd}dkqqC<c-r>=!(<c-r>a%<c-r>")*<c-r>"
<esc>k@qq@q:%s/\n/+
$x0C<c-r>=<c-r>a/(<c-r>")

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Uses the observation in rak1507's Jelly answer.

-2 bytes from kops.

-9 more bytes from kops, removing the entire third line!

Answer 8

05AB1E, 5 bytes

Ties rak1507's Jelly answer for #1.

ѨO<÷

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ѨO<÷  # full program
       # implicit input...
    ÷  # divided by...
  O    # sum of...
Ñ      # divisors of...
       # implicit input...
 ¨     # excluding the last...
    ÷  # rounded down
       # implicit output

Answer 9

R, 35 bytes

function(x,y=x-1)y/sum(2:y*!x%%2:y)

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Inspired by @Dominic's answer to linked challenge.

Uses the fact that \$\sigma(n)-n-1\$ for n>2 is just sum of divisors that lie between 2 and n-1. Then, we simply divide n-1 by the obtained value to get k (as n is guaranteed to be hyperperfect).

R, 34 bytes

function(x)x%/%sum(2:x*!x%%2:x,-x)

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Using @rak's formula.

Answer 10

C (gcc), 45 44 bytes

i,t;k(x){for(t=i=1;++t<x;)i-=x%t?0:t;x/=-i;}

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• uses Jelly's formula

• saved 1 by sum negatively and starting from 1

Explanation

   i and t used to get divisors sum
   for(t=i=1;++t<x;) - we start our loop with t=1 and end before x to exclude them from dividers 
   i-=x%t?0:t;       - we iterate all values and add to i negatively if modulo t is 0
   x/=-i;           - finally we return trough eax trick x divided by -sum which started from 1 so we have the +1 term added to k(sum)

Answer 11

Retina 0.8.2, 54 bytes

.+
$*
1(?=(1(?<=(?=(?(\3+$)(\2?\3)))(1+)))+1)(\2)+
$#4

Try it online! Link includes faster test cases. Explanation:

.+
$*

Convert the input to unary.

1(?=(1(?<=(?=(?(\3+$)(\2?\3)))(1+)))+1)(\2)+
$#4

Based on @Deadcode's answer to Am I not good enough for you? this calculates the sum of divisors from 2 to n-1, but golfed down (removed a ^ and two $s) by assuming that the input is a hyperperfect number. After the sum of divisors is calculated the result is then divided into n-1.

Answer 12

Python 3, 46 45 bytes

1 byte off thanks to @dingledooper! Also, this uses the floor trick from @rak1507's answer.

lambda n:n//sum(k*(n%k<1)for k in range(2,n))

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Answer 13

K (oK), 18 bytes

{_x%+/1_&~(!x)!'x}

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A port of @Jonah's J answer - don't forget to upvote him!

Answer 14

Vyxal, 4 bytes

∆K‹ḭ

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Porting the jelly answer ftw

Explained

∆K‹ḭ
∆K   # sum of proper divisors of input
  ‹  # ↑ - 1
   ḭ # input // ↑ (integer division)

Answer 15

Excel, 52 bytes

=LET(n,A2-1,q,SEQUENCE(n),n/SUM((MOD(A2,q)=0)*q,-1))

Works except for the last test case. SEQUENCE is limited to 2^20. The following works up to 2^40 and is 79 bytes.

=LET(q,SEQUENCE(A2^0.5),a,(MOD(A2,q)=0)*q,(A2-1)/(SUM(a,IFERROR(A2/a,0))-A2-1))

Answer 16

Pari/GP, 23 bytes

n\sum(i=2,n-1,!(n%i)*i)