Am I a 'Redivosite' Number?

Question

_{Redivosite is a portmanteau word invented for the sole purpose of this challenge. It's a mix of Reduction, Division and Composite.}

Definition

Given an integer N > 6:

• If N is prime, N is not a Redivosite Number.
• If N is composite:
  • repeatedly compute N' = N / d + d + 1 until N' is prime, where d is the smallest divisor of N greater than 1
  • N is a Redivosite Number if and only if the final value of N' is a divisor of N

Below are the 100 first Redivosite Numbers (no OEIS entry at the time of posting):

14,42,44,49,66,70,143,153,168,169,176,195,204,260,287,294,322,350,414,462,518,553,572,575,592,629,651,702,726,735,775,806,850,869,889,891,913,950,1014,1023,1027,1071,1118,1173,1177,1197,1221,1235,1254,1260,1302,1364,1403,1430,1441,1554,1598,1610,1615,1628,1650,1673,1683,1687,1690,1703,1710,1736,1771,1840,1957,1974,2046,2067,2139,2196,2231,2254,2257,2288,2310,2318,2353,2392,2409,2432,2480,2522,2544,2635,2640,2650,2652,2684,2717,2758,2760,2784,2822,2835

Examples

• N = 13: 13 is prime, so 13 is not a Redivosite Number
• N = 32: 32 / 2 + 3 = 19; 19 is not a divisor or 32, so 32 is not a Redivosite Number
• N = 260: 260 / 2 + 3 = 133, 133 / 7 + 8 = 27, 27 / 3 + 4 = 13; 13 is a divisor or 260, so 260 is a Redivosite Number

Your task

• Given an integer N, return a truthy value if it's a Redivosite Number or a falsy value otherwise. (You may also output any two distinct values, as long as they're consistent.)
• The input is guaranteed to be larger than 6.
• This is code-golf, so the shortest answer in bytes wins!

Answer 1

Haskell, 91 85 83 80 75 74 bytes

n#m=([n#(div m d+d+1)|d<-[2..m-1],mod m d<1]++[mod n m<1&&m<n])!!0
f x=x#x

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f x=x#x                           -- call # with x for both parameters
n#m               
         |d<-[2..m-1],mod m d<1   -- for all divisors d of m
    [n#(div m d+d+1)           ]  -- make a list of recursive calls to #,
                                  -- but with m set to m/d+d+1
   ++ [mod n m<1&&m<n]            -- append the Redivosite-ness of n (m divides n,
                                  -- but is not equal to n)
                           !!0    -- pick the last element of the list
                                  -- -> if there's no d, i.e. m is prime, the
                                  --    Redivosite value is picked, else the
                                  --    result of the call to # with the smallest d

Edit: -2 bytes thanks to @BMO, -3 bytes thanks to @H.PWiz and -5 -6 bytes thanks to @Ørjan Johansen

Answer 2

Husk, 14 bytes

?¬Ṡ¦ΩṗoΓ~+→Πpṗ

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-3 thanks to H.PWiz.

Answer 3

C (gcc), 94 89 bytes

m,n;o(k){for(m=1;m++<k;)if(k%m<1)return m;}
F(N){for(n=N;m=o(n),m-n;n=n/m-~m);N=m<N>N%n;}

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Explanation

m,n;                  // two global integers
o(k){                 // determine k's smallest divisor
 for(m=1;m++<k;)      // loop through integers 2 to n (inclusive)
  if(k%m<1)return m;} // return found divisor
F(N){                 // determine N's redivosity
 for(n=N;             // save N's initial value
  m=o(n),             // calculate n's smallest divisor (no name clash regarding m)
  m-n;                // loop while o(n)!=n, meaning n is not prime
                      //  (if N was prime, the loop will never be entered)
  n=n/m-~m);          // redivosite procedure, empty loop body
 N=m<N>N%n;}          // m<N evaluates to 0 or 1 depending on N being prime,
                      //  N%n==0 determines whether or not N is divisible by n,
                      //  meaning N could be redivosite => m<N&&N%n==0
                      //  <=> m<N&&N%n<1 <=> m<N&&1>N%n <=> (m<N)>N%n <=> m<N>N%n

Answer 4

Jelly, 14 bytes

ÆḌḊ
Ç.ịS‘µÇ¿eÇ

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How it works

ÆḌḊ         Helper link. Argument: k

ÆḌ          Yield k's proper (including 1, but not k) divisors.
  Ḋ         Dequeue; remove the first element (1).


Ç.ịS‘µÇ¿eÇ  Main link. Argument: n

     µ      Combine the links to the left into a chain.
      Ç¿    While the helper link, called with argument n, returns a truthy result,
            i.e., while n is composite, call the chain to the left and update n.
Ç             Call the helper link.
 .ị           At-index 0.5; return the elements at indices 0 (last) and 1 (first).
              This yields [n/d, d].
   S          Take the sum.
    ‘         Increment.
        Ç   Call the helper link on the original value of n.
       e    Test if the result of the while loop belongs to the proper divisors.

Answer 5

Python 2, 97 91 bytes

r=0;e=d=i=input()
while r-e:e=i;r=[j for j in range(2,i+1)if i%j<1][0];i=i/r-~r
d%e<1<d/e<q

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Outputs via exit code.

Ungolfed:

r = 0                             # r is the lowest divisor of the current number,
                                  # initialized to 0 for the while loop condition.
e = d = i = input()               # d remains unchanged, e is the current number
                                  # and i is the next number.
while r != e:                     # If the number is equal to its lowest divisor,
                                  # it is prime and we need to end the loop.
    e = i                         # current number := next number
    r = [j for j in range(2, i+1) # List all divisors of the number in the range [2; number + 1)
         if i%j < 1][0]           # and take the first (lowest) one.
    i = i/r+r+1                   # Calculate the next number.
                                  # We now arrived at a prime number.
print d%e == 0 and d != e         # Print True if our current number divides the input
                                  # and is distinct from the input.
                                  # If our current number is equal to the input,
                                  # the input is prime.

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Answer 6

05AB1E, 17 16 bytes

[Dp#Òć©sP+>]Ö®p*

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Explanation

[                  # start loop
 Dp#               # break if current value is prime
    Ò              # get prime factors of current value
     ć©            # extract the smallest (d) and store a copy in register
       sP          # take the product of the rest of the factors
         +>        # add the smallest (d) and increment
           ]       # end loop
            Ö      # check if the input is divisible by the resulting prime
             ®p    # check if the last (d) is prime (true for all composite input)
               *   # multiply

Answer 7

Pyth, 20 bytes

<P_QiI.WtPHh+/ZKhPZK

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How it works

iI.WtPHh+/ZKhPZK || Full program.

  .W             || Functional while. It takes two functions as arguments, A and B.
                 || While A(value) is truthy, turn the value into B(value). The 
                 || starting value is the input.
    tPH          || First function, A. Takes a single argument, H.
     PH          || .. The prime factors factors of H.
    t            || .. Tail (remove first element). While truthy (H is composite):
       h+/ZKhPZK || The second function, B. Takes a single argument, Z:
         /Z      || .. Divide Z, by:
           KhP   || .. Its lowest prime factor, and assign that to K.   
       h         || .. Increment. 
        +      K || And add K.
iI               || Check if the result (last value) divides the input.

And the first 4 bytes (<P_Q) just check if the input is not prime.

With help from Emigna, I managed to save 3 bytes.

Answer 8

Python 3, 149 bytes

def f(N):
	n=N;s=[0]*-~N
	for p in range(2,N):
		if s[p]<1:
			for q in range(p*p,N+1,p):s[q]=s[q]or p
	while s[n]:n=n//s[n]-~s[n]
	return s[N]>1>N%n

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Using a sieve approach. Should be fast (O(N log log N) = time complexity of the sieve of Eratosthenes) even with large N (but stores O(N) integers in memory)

Note:

• It can be proven that all intermediate values of n does not exceed N, and for N > 7 p can be in range(2,N) instead of range(2,N+1) for sieving.
• / doesn't work, // must be used, because of list index.
• Storing range into another variable doesn't help, unfortunately.

Explanation:

• -~N == N+1.
• At first, the array s is initialized with N+1 zeroes (Python is 0-indexing, so the indices are 0..N)
• After initialization, s[n] is expected to be 0 if n is a prime, and p for p the minimum prime which divides n if n is a composite. s[0] and s[1] values are not important.

• For each p in range [2 .. N-1]:

  • If s[p] < 1 (that is, s[p] == 0), then p is a prime, and for each q being a multiple of p and s[q] == 0, assign s[q] = p.

• The last 2 lines are straightforward, except that n//s[n]-~s[n] == (n // s[n]) + s[n] + 1.

---

Python 3, 118 bytes

def f(N):
	n=N;s=[0]*-~N
	for p in range(N,1,-1):s[2*p::p]=(N-p)//p*[p]
	while s[n]:n=n//s[n]-~s[n]
	return s[N]>1>N%n

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At the cost of slightly worse performance. (This one runs in O(N log N) time complexity, assume reasonable implementation of Python slices)

---

The equivalent full program is 117 bytes.

Answer 9

Haskell, 110 bytes

f n|mod(product[1..n-1]^2)n>0=1<0|1>0=n`mod`u n<1
u n|d<-[i|i<-[2..],n`mod`i<1]!!0=last$n:[u$n`div`d+d+1|d/=n]

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Not very happy...

Answer 10

Octave, 92 bytes

function r=f(n)k=n;while~isprime(k)l=2:k;d=l(~mod(k,l))(1);k=k/d+d+1;end;r=~mod(n,k)&k<n;end

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Answer 11

APL (Dyalog), 50 bytes

{(0=n|⍵)∧⍵>n←{⍬≢k←o/⍨0=⍵|⍨o←2↓⍳⍵:∇1+d+⍵÷d←⌊/k⋄⍵}⍵}

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Answer 12

Japt, 25 24 bytes

I fear I may have gone in the wrong direction with this but I've run out of time to try a different approach.

Outputs 0 for false or 1 for true.

j ?V©vU :ßU/(U=k g)+°UNg

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Answer 13

Perl 5, 291+1(-a) = 292 bytes

Darn Perl for not having a native prime checker.

use POSIX;&r($_,$_);
sub p{$n=shift;if($n<=1){return;}if($n==2||$n==3){return 1;}if($n%2==0||$n%3==0){return;}for(5..ceil($n/2)){if($n%$_==0){return;}}return 1;}
sub r{$n=shift;$o=shift;if(&p($n)){print $o%$n==0&&$n!=$o?1:0;last;}for(2..ceil($n/2)){if($n%$_==0){&r(($n/$_)+$_+1, $o);last;}}}

Ungolfed version,

use POSIX;
&r($_,$_);
sub p{
    my $n=shift;
    if($n<=1){
        return;
    }
    if($n==2||$n==3){
        return 1;
    }
    if($n%2==0||$n%3==0){
        return;
    }
    for(5..ceil($n/2)){
        if($n%$_==0){
            return;
        }
    }
    return 1;
}
sub r{
    my $n=shift;
    my $o=shift;
    if(&p($n)){
        print $o%$n==0&&$n!=$o ? 1 : 0;
        last;
    }
    for(2..ceil($n/2)){
        if($n%$_==0){
            &r(($n/$_)+$_+1, $o);
            last;
        }
    }
}

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Answer 14

Wolfram Language (Mathematica), 64 bytes

Straightforward implementation using recursive pattern replacement

(replacing "\[Divides]" with the "∣" symbol saves 7 bytes)

(g=!PrimeQ@#&)@#&&(#//.x_/;g@x:>x/(c=Divisors[x][[2]])+c+1)\[Divides]#&

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Answer 15

Clean, 128 117 114 bytes

import StdEnv
@n#v=hd[p\\p<-[2..]|and[gcd p i<2\\i<-[2..p-1]]&&n rem p<1]
|v<n= @(n/v+v+1)=n
?n= @n<n&&n rem(@n)<1

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Answer 16

J, 35 bytes

(~:*0=|~)(1+d+]%d=.0{q:)^:(0&p:)^:_

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The minimum divisor being the first prime factor was stolen from @Dennis's Jelly solution (previously I was using <./@q:).

There should be a better way to do the iteration, but I can't seem to find it. I thought of avoiding doing the primality test (^:(0&p:)) and instead using adverse but it seems like that will be a bit longer since you'll need a _2{ and some changes which might not give a net reduction.

I really feel like there must be a way to avoid having parens around the primality check, too.

Explanation (expanded)

(~: * 0 = |~)(1 + d + ] % d =. 0 { q:) ^: (0&p:) ^:_ Input: N
             (1 + d + ] % d =. 0 { q:) ^: (0&p:) ^:_ Find the final N'
                                       ^:        ^:_  Do while
                                           0&p:       N is not prime
                                   q:                 Get prime factors (in order)
                               0 {                    Take first (smallest divisor)
                          d =.                        Assign this value to d
             1 + d + ] %  d                           Compute (N/d) + 1 + d
(~: * 0 = |~)                                        Is it redivosite?
      0 = |~                                          N = 0 (mod N'), i.e. N'|N
    *                                                 And
 ~:                                                   N =/= N', i.e. N is not prime

Answer 17

APL NARS, 43 chars, 85 bytes

{(⍵≤6)∨0π⍵:0⋄⍵{1=⍴t←π⍵:0=⍵|⍺⋄⍺∇1+↑t+⍵÷↑t}⍵}

(hoping that it converge for all number>6) test:

h←{(⍵≤6)∨0π⍵:0⋄⍵{1=⍴t←π⍵:0=⍵|⍺⋄⍺∇1+↑t+⍵÷↑t}⍵}
v←⍳100     
v,¨h¨v
   1 0  2 0  3 0  4 0  5 0  6 0  7 0  8 0  9 0  10 0  11 0
   12 0  13 0  14 1  15 0  16 0  17 0  18 0  19 0  20 0  
   21 0  22 0  23 0  24 0  25 0  26 0  27 0  28 0  29 0  
   30 0  31 0  32 0  33 0  34 0  35 0  36 0  37 0  38 0  
   39 0  40 0  41 0  42 1  43 0  44 1  45 0  46 0  47 0  
   48 0  49 1  50 0  51 0  52 0  53 0  54 0  55 0  56 0  
   57 0  58 0  59 0  60 0  61 0  62 0  63 0  64 0  65 0  
   66 1  67 0  68 0  69 0  70 1  71 0  72 0  73 0  74 0  
   75 0  76 0  77 0  78 0  79 0  80 0  81 0  82 0  83 0  
   84 0  85 0  86 0  87 0  88 0  89 0  90 0  91 0  92 0  
   93 0  94 0  95 0  96 0  97 0  98 0  99 0  100 0  

The idea of using 2 anonymous functions I get to other Apl solution.

 {(⍵≤60)∨π⍵:0⋄ -- if arg ⍵ is prime or <=6 return 0
  ⍵{1=⍴t←π⍵:0=⍵|⍺⋄ -- if list of factors ⍵ has length 1 (it is prime)
                    -- then return ⍺mod⍵==0
  ⍺∇1+↑t+⍵÷↑t}⍵}   -- else recall this function with args ⍺ and 1+↑t+⍵÷↑t
   

Answer 18

Pyt, 44 bytes

←⁻0`ŕ⁺ĐĐϼ↓Đ3Ș⇹÷+⁺Đṗ¬⇹⁻⇹łŕáĐ0⦋Đ↔ĐŁ⁻⦋⁺|¬⇹ṗ⇹3Ș⊽

See the code below for an explanation - the only differences are (1) that N is decremented before to account for the incrementation at the beginning of the loop, and (2) it uses NOR instead of OR.

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---

I made this before I re-read the question and noticed that it only wanted a true/false.

Pyt, 52 bytes

60`ŕ⁺ĐĐϼ↓Đ3Ș⇹÷+⁺Đṗ¬⇹⁻⇹łŕáĐ0⦋Đ↔ĐŁ⁻⦋⁺|¬⇹Đṗ⇹3Ș∨ł⇹Đƥ⇹ŕ1ł

Prints an infinite list of Redivosite numbers.

Explanation:

6                                                            Push 6
 0                                                           Push unused character
  `                   ł                     ł      ł         Return point for all three loops
   ŕ                                                         Remove top of stack
    ⁺                                                        Increment top of stack (n)
     ĐĐ                                                      Triplicate top of stack (n)
       ϼ↓                                                    Get smallest prime factor of n (returns 1 if n is prime) 
         Đ                                                   Duplicate top of stack
          3Ș⇹                                                Manipulate stack so that the top is (in descending order): [d,(N,N'),d]
             ÷+⁺                                             Calculates N'=(N,N')/d+d+1
                Đṗ¬                                          Is N' not prime?
                   ⇹⁻⇹                                       Decrement N' (so the increment at the beginning doesn't change the value), and keep the boolean on top - end of innermost loop (it loops if top of stack is true)
                       ŕ                                     Remove top of stack
                        á                                    Convert stack to array
                         Đ                                   Duplicate array
                          0⦋Đ                                Get the first element (N)
                             ↔ĐŁ⁻⦋                           Get the last element ((final N')-1)
                                  ⁺                          Increment to get (final N')
                                   |¬                        Does N' not divide N?
                                     ⇹Đṗ                     Is N prime?
                                        ⇹3Ș∨                 Is N prime or does N' not divide N? - end of second loop (loops if top of stack is true)
                                             ⇹Đƥ⇹ŕ           Print N, and reduce stack to [N]
                                                  1          Push garbage (pushes 1 so that the outermost loop does not terminate)

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