Can the number reach 1 by repeatedly subtracting the largest prime less than it?

Question

Challenge:

Given a number, take the largest prime strictly less than it, subtract it from this number, do this again to this new number with the biggest prime less than it, and continue doing this until it's less than 3. If it reaches 1, your program should output a truthy value, else, the program should output a falsey value.

Examples:

All of these should give a truthy value:

3
4
6
8
10
11
12
14
16
17
18
20
22
23
24
26
27
29
30
32
34
35
37
38
40
41
42
44
46
47
48
50

All of these should give falsey values:

5
7
9
13
15
19
21
25
28
31
33
36
39
43
45
49

Rules:

• You can either write a program or function.
• You can assume that the input is bigger than 2.
• Standard loopholes apply
• This is code-golf so the shortest answer wins!

Answer 1

Retina, 31 bytes

.+
$*
+`1(?!(11+)\1+$)11+
1
^1$

Prints 0 (falsy) or 1 (truthy).

Try it online! (The first line enables a linefeed-separated test suite.)

Explanation

.+
$*

Convert input to unary by turning input N into N copies of 1.

+`1(?!(11+)\1+$)11+
1

Repeatedly remove the largest prime less than the input. This is based on the standard primality test with regex.

^1$

Check whether the result is a single 1.

Answer 2

Jelly, 9 8 bytes

’ÆRṪạµ¡Ḃ

Try it online! or verify all test cases.

How it works

’ÆRṪạµ¡Ḃ  Main link. Argument: n

     µ    Combine all atoms to the left into a chain.
’           Decrement; yield n - 1.
 ÆR         Prime range; yield all primes in [2, ..., n -1].
   Ṫ        Tail; yield p, the last prime in the range.
            If the range is empty, this yields p = 0.
    ạ       Compute the absolute difference of p and n.
      ¡   Call the chain to the left n times.
          This suffices since each iteration decreases n, until one of the fixed
          points (1 or 2) is reached.
       Ḃ  Bit; return the parity of the fixed point.

Answer 3

Regex (ECMAScript or better), 33 32 bytes

^((?=.+?(?=(xx+?)\2*$)\2$)\2)*x$

Takes its input in unary, as a string of x characters whose length represents the number.

Try it online! - ECMAScript
Try it online! - Java
Try it online! - Perl
Try it online! - PCRE
Try it online! - Boost
Try it online! - Python
Try it online! - Ruby
Try it online! - .NET

I finally found a use for the positive lookahead primality test I came up with on 2018-12-07! This is 2 bytes longer than the "standard" negative lookahead primality test, but it allows the full regex to be 1 byte shorter:

^                           # tail = input number
(                           # Loop the following:
    (?=                     # Atomic lookahead - finds the first match, and once
                            # finished, its result won't be changed by backtracking
        .+?                 # tail = largest number that is less than the current tail,
                            #        for which the following matches:
        (?=(xx+?)\2*$)\2$   # Assert tail is prime; \2 = tail
    )
    \2                      # tail -= \2
)*                          # Iterate the above loop zero or more times
x$                          # Assert tail==1

A neat thing about this regex is that it can be logically negated for all inputs other than zero simply by adding one byte, because after the repeated subtraction, all \$n>0\$ end up at either \$1\$ or \$2\$:

^((?=.+?(?=(xx+?)\2*$)\2$)\2)*xx$

Try it online! - ECMAScript
Try it online! - Java
Try it online! - Perl
Try it online! - PCRE
Try it online! - Boost
Try it online! - Python
Try it online! - Ruby
Try it online! - .NET

---

Here is the regex implemented in some languages that have golf-efficient regex calls, where it either beats the other submitted solution(s) or is the only one in that language:

\$\large\textit{Anonymous functions}\$

Julia v0.4+, 58 bytes

n->split("x"^n,r"^((?=.+?(?=(xx+?)\2*$)\2$)\2)*x$")[1]==""

Try it online!

The above returns truthy for an input of 0. If returning falsey for 0 were needed, it'd be 60 bytes:

n->match(r"^((?=.+?(?=(xx+?)\2*$)\2$)\2)*x$","x"^n)!=nothing

Try it online!

or 59 bytes in Julia v0.7 and earlier with !=Void(): Try it online!

Of course as far as Julia v0.4 goes, Glen O's answer outgolfs this at 32 bytes, but for v0.5 and later, this and the below may be optimal golf.

Julia v0.7+, 54 bytes

n->occursin(r"^((?=.+?(?=(xx+?)\2*$)\2$)\2)*x$",'x'^n)

Try it online!

Julia v1.2+, 53 bytes

n->endswith('x'^n,r"^((?=.+?(?=(xx+?)\2*$)\2$)\2)*x")

Attempt This Online!

PowerShell, 52 51 bytes

'x'*"$args"-match'^((?=.+?(?=(xx+?)\2*$)\2$)\2)*x$'

Try it online!

-1 bytes thanks to Julian

Python, 70 69 bytes

lambda n:re.match(r'((?=.+?(?=(xx+?)\2*$)\2$)\2)*x$','x'*n);import re

Try it online!

Python, 73 bytes

lambda n:__import__('re').match(r'((?=.+?(?=(xx+?)\2*$)\2$)\2)*x$','x'*n)

Try it online!

(If it must be a pure lambda.)

JavaScript (ES6), 58 54 bytes

n=>/^((?=.+?(?=(..+?)\2*$)\2$)\2)*.$/.test(Array(n+1))

Try it online!

-4 bytes thanks to a technique used by RK. and CubeyTheCube

Perl, 49 47 bytes

-2 bytes thanks to dingledooper

sub{1x pop~~/^((?=.+?(?=(..+?)\2*$)\2$)\2)*.$/}

Try it online!

This beats a port of Ton Hospel's 41 byte answer (57 bytes):

sub{my$x=1x pop;$x=$`while$x=~/\B(?!(11+)\1+$|$)|11$/;$x}

Try it online!

Even if it's allowed to modified the global variable $_ (51 bytes):

sub{$_=1x pop;$_=$`while/\B(?!(11+)\1+$|$)|11$/;$_}

Try it online!

Ruby, 48 45 bytes

->n{?x*n=~/^((?=.+?(?=(xx+?)\2*$)\2$)\2)*x$/}

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-2 bytes by switching the truthy value from "string of x characters whose length is the input number" to "the integer value 0", while keeping the same falsey value of nil
-1 byte by using ?x instead of "x"

PHP, 89 75 71 bytes

fn($n)=>preg_match('/^((?=.+?(?=(..+?)\2*$)\2$)\2)*.$/',str_pad('',$n))

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-4 bytes by switching from x to as the repeated character

R, 73 72 68 63 bytes

\(n)grepl('^((?=.+?(?=(..+?)\\2*$)\\2$)\\2)*.$',strrep(1,n),,1)

Attempt This Online! / Try it online!

\(n)grepl('^((?=.+?(?!(..+)\\2+$)(..+))\\3)*.$',strrep(1,n),,1)

Attempt This Online! / Try it online!

-1 byte thanks to Giuseppe
-4 bytes by using grepl() instead of sum(grep()) or any(grep())
-5 bytes by using a new anonymous function syntax introduced in R v4.1.0

Since TIO uses R v3.5.2, the TIO links have the old 68 byte versions of the function.

Java 8, 92 89 88 71 bytes

n->new String(new char[n]).matches("((?=.+?(?=(..+?)\\2*$)\\2$)\\2)*.")

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n->new String(new char[n]).matches("((?=.+?(?!(..+)\\2+$)(..+))\\3)*.")

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-17 bytes by just matching the regex straight against a string of NUL characters instead of replacing the NULs with x

Java 11, 61 bytes

n->"x".repeat(n).matches("((?=.+?(?=(xx+?)\\2*$)\\2$)\\2)*x")

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-10 bytes relative to Java <11, thanks to Kevin Cruijssen

\$\large\textit{Full programs}\$

Python 2, 69 bytes

import re
re.match(r'((?=.+?(?=(xx+?)\2*$)\2$)\2)*x$','x'*input()).re

Outputs via its exit code, value 0 for truthy (by exiting normally), and 1 for falsey (by triggering an AttributeError).

Try it online!

Python 2, 71 70 bytes

import re
re.match(r'((?=.+?(?=(xx+?)\2*$)\2$)\2)*xx$','x'*input()).re

Outputs via its exit code, value 1 for truthy and 0 for falsey, using the truth-inverted regex.

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Alternative method using the uninverted regex, outputting exit code 1 for truthy (by triggering a NameError), and 0 for falsey (by exiting normally), at 71 bytes:

import re
if re.match(r'((?=.+?(?=(xx+?)\2*$)\2$)\2)*x$','x'*input()):z

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Python 2, 79 74 bytes

import re
print re.match(r'((?=.+?(?=(xx+?)\2*$)\2$)\2)*x$','x'*input())>0

Outputs True or False via stdout.

Try it online!

Python 3, 84 83 bytes

import re
print(not re.match(r'((?=.+?(?=(xx+?)\2*$)\2$)\2)*xx$','x'*int(input())))

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The above uses the truth-inverted regex. Without it is 84 bytes:

import re
print(bool(re.match(r'((?=.+?(?=(xx+?)\2*$)\2$)\2)*x$','x'*int(input()))))

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PHP -F, 70 bytes

<?=preg_match('/^((?=.+?(?=(..+?)\2*$)\2$)\2)*.$/',str_pad('',$argn));

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Perl -p, 45 43 bytes

-2 bytes thanks to dingledooper

$_=1x$_~~/^((?=.+?(?=(..+?)\2*$)\2$)\2)*.$/

Try it online!

Beaten by Ton Hospel's 41 byte answer which mixes regex and code: Try it online!

Perl, 46 44 bytes

say 1x<>~~/^((?=.+?(?=(..+?)\2*$)\2$)\2)*.$/

Try it online!

Now beats a port of Ton Hospel's answer (45 bytes):

$_=1x<>;$_=$`while/\B(?!(11+)\1+$|$)|11$/;say

Try it online!

Answer 4

Pyth, 18 15 14 bytes

Thanks to @Maltysen for -1 byte

#=-QefP_TUQ)q1

A program that takes input on STDIN and prints True or False as appropriate.

Try it online

How it works

#=-QefP_TUQ)q1  Program. Input: Q
#          )    Loop until error statement (which occurs when Q<3):
         UQ      Yield [0, 1, 2, 3, ..., Q-1]
     fP_T        Filter that by primality
    e            Yield the last element of that
 =-Q             Q = Q - that
            q1  Q is 1 (implicit variable fill)
                Implicitly print

Old version with reduce, 18 bytes

qu-G*<HGH_fP_TSQQ1

Try it online

How it works

qu-G*<HGH_fP_TSQQ1  Program. Input: Q
              SQ    Yield [1, 2, 3, ..., Q]
          fP_T      Filter that by primality
         _          Reverse it
 u                  Reduce it:
                Q    with base case Q and
                     function G, H -> 
     <HG              H<G
    *   H             *H (yields H if H<G, else 0)
  -G                  Subtract that from G
q                1  The result of that is 1
                    Implicitly print

Answer 5

JavaScript (ES6), 64 63 bytes

Saved 1 byte thanks to @Neil

g=(x,n=x-1)=>n<2?x:x%n?g(x,n-1):g(x-1)
f=x=>x<3?x%2:f(x-g(x-1))

I wrote this in 2 minutes... and it worked perfectly the first time. First user to find the inevitable bug wins....

Try it out

g=(x,n=x-1)=>n<2?x:x%n?g(x,n-1):g(x-1)
f=x=>x<3?x%2:f(x-g(x-1))

<input type="number" value=3 min=3 onchange="A.innerHTML=f(this.value)"><br>
<p id=A>1</p>

How it works

First we define g(x) as the function that finds the first prime number p <= x. This is done using the following process:

1. Start with n = x-1.
2. If n < 2, x is prime; return x.
3. If x is divisible by n, decrement x and go to step 1.
4. Otherwise, decrement n and go to step 2.

The solution to this challenge, f(x), is now fairly straightforward:

1. If x < 3, return x = 1.
2. Otherwise, subtract g(x-1) and try again.

Answer 6

Julia (0.4), 32 bytes

(Update (as of 1.4): This is rather out-of-date, now - primes is no longer in Base, and ?: needs spaces around the ? and the :)

While it's not going to be the shortest solution among the languages, this might be the shortest of the human-readable ones...

!n=n>2?!(n-primes(n-1)[end]):n<2

Try it online!

Or, to put it in slightly clearer terms

function !(n)
  if n>2
    m=primes(n-1)[end]   # Gets largest prime less than n
    return !(n-m)        # Recurses
  else
    return n<2           # Gives true if n is 1 and false if n is 2
  end
end

Called with, for example, !37.

Answer 7

Pyke, 15 11 bytes

WDU#_P)e-Dt

Try it here!

            - stack = input
W           - while continue:
  U#_P)     -     filter(is_prime, range(stack))
       e    -    ^[-1]
 D      -   -   stack-^
         Dt -  continue = ^ != 1

Returns 1 if true and raises an exception if false

Answer 8

Husk, ¹¹ 9 bytes

εωλ-→fṗŀ¹

Try it online!

-2 bytes from Leo.

None of the testcases reach zero because the max number we can subtract from n can only be n-1. Hence, we can check insignificance instead of equality with 1.

Explanation

εωλ-→fṗŀ¹
 ωλ       iterate on the input till fixpoint
       ŀ¹  range 0..n-1
     fṗ    filter out nonprimes
    →      last element
   -       subtract from n
ε         is fixed point insignificant? (<= 1?) 

Answer 9

Python3, 102 92 90 89 88 bytes

f=lambda n:n<2if n<3else f(n-[x for x in range(2,n)if all(x%y for y in range(2,x))][-1])

Golfing suggestions welcome! I see that gmpy contains a function next_prime, but I can't test it yet :(

-2 bytes, thanks to @JonathanAllan!

-1 byte, thanks to @Aaron!

Testcases

f=lambda n:n<2if n<3else f(n-[x for x in range(2,n)if all(x%y for y in range(2,x))][-1])

s="3 4 6 8 10 11 12 14 16 17 18 20 22"
h="5 7 9 13 15 19 21 25 28 31 33 36 39"

for j in s.split(" "):print(f(int(j)))
for j in h.split(" "):print(f(int(j)))

Output is 13 truthy values and 13 falsey values. s contains the truthy cases and h the falseys.

Answer 10

Haskell, ⁶⁵ 63 bytes

f n|n>2=f$n-last[p|p<-[2..n-1],all((>0).mod p)[2..p-1]]|1>0=n<2

Try it online!

Answer 11

Mathematica, 32 bytes

2>(#//.x_/;x>2:>x+NextPrime@-x)&

This is an unnamed function which takes an integer and returns a boolean.

Explanation

There's a lot of syntax and funny reading order here, so...

   #                               This is simply the argument of the function.
    //.                            This is the 'ReplaceRepeated' operator, which applies
                                   a substitution until the its left-hand argument stops
                                   changing.
       x_/;x>2                     The substitution pattern. Matches any expression x as
                                   long as that expression is greater than 2.
              :>                   Replace that with...
                  NextPrime@-x     Mathematica has a NextPrime built-in but no
                                   PreviousPrime built-in. Conveniently, NextPrime
                                   works with negative inputs and then gives you the 
                                   next "negative prime" which is basically a
                                   PreviousPrime function (just with an added minus sign).
                x+                 This gets added to x, which subtracts the previous
                                   prime from it.
2>(                           )    Finally, we check whether the result is less than 2.

Answer 12

Python, with sympy, 60 bytes

import sympy
f=lambda n:n>2and f(n-sympy.prevprime(n))or n<2

---

My previous method was 83 bytes without sympy using recursion, but I took truthy/falsey to mean distinguishable and consistent, but I have been informed that's an incorrect interpretation. I can't seem to salvage it due to the tail, but I'll leave it here in case someone knows how to do so:

f=lambda n,p=0:n>2and(any(p%x==0for x in range(2,p))and f(n,p-1)or f(n-p,n+~p))or n

Answer 13

APL (Dyalog Unicode), 27 bytes

{⍵<3:⍵=1⋄∇⍵-⊃⌽(⊢~∘.×⍨)2↓⍳⍵}
 ⍵<3 ⍝ Termination condition
   :⍵=1 ⍝ return value
     ⋄ ⍝ Separator
       ∇ ⍝ Recursive call
         ⍵-⊃⌽ ⍝ Subtract largest prime
           (⊢~∘.×⍨) ⍝ Calculate primes less than argument
             2↓⍳⍵ ⍝ Range 2–(N-1)

Try it online!

Answer 14

Brachylog, 11 bytes

1|-↙Xṗ∧X≜!↰

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1              The input is 1,
 |             or:
       X≜      X is the integer with the least absolute value such that
  -↙X          it is positive, and if it is subtracted from the input
     ṗ∧        the result is prime.
         !     Consider no other value of X.
          ↰    Call this predicate again with X as the input.

Answer 15

Vyxal, 7 bytes

‡∆ṗεẊ1=

Try it Online!

Very nice and neat answer (flagless too!)

Explained

‡∆ṗεẊ1=
‡∆ṗε    # lambda x: abs(x - prev_prime(x))
    Ẋ   # repeat that on the input until it doesn't change
     1= # does that equal 1?

Answer 16

MATL, 13 bytes

`tqZq0)-t2>}o

Try it online! Or verify all test cases at once.

Explanation

`        % Do...while
  t      %   Duplicate. Takes input implicitly in the first iteration
  qZq    %   All primes less than that
  0)     %   Get last one
  -      %   Subtract (this result will be used in the next iteration, if any)
  t      %   Duplicate
  2>     %   Does it exceed 2? If so: next iteration. Else: execute the "finally" 
         %   block and exit do...while loop
}        % Finally
  o      %   Parity. Transforms 2 into 0 and 1 into 1
         % End do...while implicitly
         % Display implicitly

Answer 17

CJam, 21 16 bytes

Thanks to Dennis for saving 4 bytes.

ri{_1|{mp},W=-}h

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Explanation

ri       e# Read input and convert to integer N.
{        e# Run this block as long as N is positive (or until the program aborts
         e# with an error)...
  _1|    e#   Duplicate and OR 1. This rounds up to an odd number. For N > 2, this
         e#   will never affect the greatest prime less than N.
  {mp},  e#   Get all primes from 0 to (N|1)-1.
         e#   For N > 2, this will contain all primes less than N.
         e#   For N = 2, this will contain only 2.
         e#   For N = 1, this will be empty.
  W=     e#   Select the last element (largest prime up to (N|1)-1).
         e#   For N = 1, this will result in an error and terminate the program, which
         e#   still prints the stack contents though (which are 1, the desired output).
  -      e#   Subtract from N. Note that this gives us 0 for N = 2, which terminates the 
         e#   loop.
}h

Answer 18

Perl, 42 bytes

Includes +1 for -p

Run with input on STDIN

reach1.pl:

#!/usr/bin/perl -p
$_=1x$_;$_=$`while/\B(?!(11+)\1+$|$)|11$/

Uses the classic primality regex

Answer 19

.NET Regex, 38 bytes

Just to show that it can be checked in a single regex.

^(?>(?<=(.*))..+(?<!^\1\2+(.+.)|$))+.$

Input is assumed to be in unary.

Explanation

It simply implements the requirement to the word, repeatedly removing the biggest prime and check whether there remainder is 1.

• (?>(?<=(.*))..+(?<!^\1\2+(.+.)|$))+: Non-backtracking group makes sure the biggest prime we found is not overriden, and + simply repeat the process of matching the biggest prime.

  • (?<=(.*))..+(?<!^\1\2+(.+.)|$): Match the biggest prime less than the remaining number

    • (?<=(.*)): Record how much we have subtracted to establish an "anchor" point for assertion.

    • ..+: Look for the biggest number...

    • (?<!^\1\2+(.+.)|$): ... which is prime and less than the remaining number.
      • (?<!^\1\2+(.+.)): The usual prime test routine, with ^\1 tacked in front to make sure we are checking the amount matched by ..+
      • (?!<$): Assert less than the remaining number

Answer 20

Python 2.7: 88 87 Bytes

r=lambda n:n>2and r(n-[a for a in range(2,n)if all(a%b for b in range(2,a))][-1])or n<2

Thx @TuukkaX for -1 more byte!

Answer 21

Factor + math.primes, 51 bytes

[ [ dup 1 - primes-upto last - dup 2 > ] loop 1 = ]

Try it online!

Answer 22

Vitsy, 28 26 bytes

This can definitely be shortened.

<]xN0)l1)-1[)/3D-];(pD-1[D

‮

<                    Traverse the code in this direction, rotating on the line.
                     For the sake of reading the code easier, I'm reversing the
                     code on this line. This will be the order executed.

 D[1-Dp(;]-D3/)[1-)1l)0Nx]
 D                         Duplicate the top member of the stack.
  [      ]                 Do the stuff in brackets until break is called.
   1-                      Subtract 1 from the top item of the stack.
     D                     Duplicate the top member of the stack.
      p(                   If the top member is a prime...
        ;                  break;
          -                Pop a, b, push a - b.
           D3/)[         ] If this value is less than 3, do the bracketed code.
                1-         Subtract the top item of the stack by 1.
                  )        If the top item is zero...
                   1       Push 1.
                    l)     If the length of the stack is zero...
                      0    Push 0.
                       N   Output the top member of the stack.
                        x  System.exit(0);

Try it online!

Answer 23

Perl 6,  54 53 52  51 bytes

{($_,{$_-($_-1...2).first: *.is-prime}...3>*)[*-1]==1}
{($_,{$_-($_-1...2).first: *.is-prime}...3>*).any==1}
{any($_,{$_-($_-1...2).first: *.is-prime}...3>*)==1}

{any($_,{$_-(^$_).grep(*.is-prime)[*-1]}...3>*)==1}

Explanation:

# bare block lambda with implicit parameter ｢$_｣
# used to generate all of the rest of the elements of the sequence
{
  # create an any Junction of the following list
  any(
    $_, # initialize sequence with the inner block's argument

    # bare block lambda with implicit parameter ｢$_｣
    {
      # take this inner block's argument and subtract
      $_ -

      ( ^$_ )            # Range up-to and excluding ｢$_｣
      .grep(*.is-prime)\ # find the primes
      [ * - 1 ]          # return the last value
    }

    ...   # keep doing that until

    3 > * # the result is less than 3

  # test that Junction against ｢1｣
  # ( returns an ｢any｣ Junction like ｢any(False, False, True)｣ )
  ) == 1
}

Example:

# show what is returned and if it is truthy
sub show ($_) {
  # ｢.&{…}｣ uses the block as a method and implicitly against ｢$_｣
  my $value = .&{any($_,{$_-(^$_).grep(*.is-prime)[*-1]}...3>*)==1}
  say join "\t", $_, ?$value, $value.gist;
}

show 3;  # 3    True    any(False, True)
show 4;  # 4    True    any(False, True)
show 5;  # 5    False   any(False, False)
show 10; # 10   True    any(False, False, True)
show 28; # 28   False   any(False, False, False)
show 49; # 49   False   any(False, False)
show 50; # 50   True    any(False, False, True)

Answer 24

Irregular, 63 bytes

p~?1_$-1p:;
n=i(0)?1_$-1p:;
_~
N=n
1(?!(11+)\1+$)11+~1
^11$~0
N

I created this language two days ago, and the basic premises are that there are no built in loops, the only features are basic arithmetic and decision making, and program evaluation is based on regular expressions.

Explanation

p~?1_$-1p:;
n=i(0)?1_$-1p:;
_~
N=n

This part converts the input into unary. It repeatedly subtracts 1 from the input until it equals 0, prepending 1_ each time. It then removes all of the _s. If I hadn't forgotten a break in my code it could be written as so:

p~?1_$-1p:;
_~
n=i(0)?1_$-1p:;

The next part repeatedly removes the largest prime from the input until it is equal to 1 or 11, with 11 being replaced with 0.

1(?!(11+)\1+$)11+~1
^11$~0
N

I used the regex from Martin Ender's answer.

Answer 25

PowerShell v2+, 81 bytes

param($n)while($n-gt2){$n-=(($n-1)..2|?{'1'*$_-match'^(?!(..+)\1+$)..'})[0]}!--$n

Takes input $n. Enters a while loop so long as $n is still 3 or greater. Each iteration, subtracts a number from $n. The number is the results of the regex primality test applied against a range ($n-1)..2 via the Where-Object (?) operator, then the first [0] of the results (since the range is decreasing, this results in the largest one being selected). After concluding the loop, $n is either going to be 1 or 2, by definition, so we pre-decrement $n (turning it into either 0 or 1), and take the Boolean-not ! thereof. That's left on the pipeline and output is implicit.

Examples

PS C:\Tools\Scripts\golfing> 3..20|%{"$_ --> "+(.\can-the-number-reach-one.ps1 $_)}
3 --> True
4 --> True
5 --> False
6 --> True
7 --> False
8 --> True
9 --> False
10 --> True
11 --> True
12 --> True
13 --> False
14 --> True
15 --> False
16 --> True
17 --> True
18 --> True
19 --> False
20 --> True

Answer 26

Floroid, 45 30 29 bytes

f=Bb:b<2Fb<3Gf(b-en(b-1)[-1])

Answer 27

Clojure, 125 bytes

#(loop[x %](if(> x 2)(recur(- x(loop[y(dec x)](if(some zero?(vec(for[z(range 2 y)](mod y z))))(recur(dec y))y))))(quot 1 x)))

Yikes, that is one long piece of code. The most verbose language strikes again!

Ungolfed:

(defn subprime [n]
  (loop [x n]
    (if (> x 2)
      (recur
        (- x
          (loop [y (dec x)]
            (if (some zero? (vec (for [z (range 2 y)] (mod y z))))
              (recur (dec y)) y))))
      (quot 1 x))))

Answer 28

05AB1E, 7 bytes

ΔD<ÅPà-

Outputs 2 as falsey (only 1 is truthy in 05AB1E).

Try it online or verify all test cases.

Explanation:

Unfortunately the previous prime builtin ÅM goes into an infinite loop for \$\leq2\$, otherwise it could have been used to save a few bytes.

Δ        # Loop until the result no longer changes:
 D       #  Duplicate the current value
         #  (which is the implicit input-integer in the first iteration)
  <ÅP    #  Pop it, and push a list of all primes < this value:
  <      #   Decrease the value by 1
   ÅP    #   Pop and push a list of all primes <= this value-1
     à   #  Pop and push the maximum prime of this list
         #  (or an empty string if the list is empty)
      -  #  Subtract it from the value (where subtracting the empty string is a no-op)
         # (after which the found result is output implicitly)

Answer 29

Haskell, 79 bytes

Not really short but pointfree :)

(<2).until(<3)(until(flip(`until`(+1))2.(.)(<1).mod>>=(==))pred.pred>>=flip(-))

Answer 30

Matlab, 51 bytes

v=@(x)x-max(primes(x-1));while(x>=3)x=v(x);end;x==1

This is VERY similar to the JS6 solution by ETHProductions, but needs the variable to be in the workspace.