Œv¹y5×åO

Π        # Compute all substrings from the input.
 v        # For each substring.
   y5×    # Repeat the substring 5 times (42 × 5 = 4242424242).
  ¹   å   # Check if it's in the input string.
       O  # Sum up the result. Non-boring numbers should give 0.

(.+)1{4}

L->java.util.regex.Pattern.compile("(.+)\1{4}").matcher(L).find();

import re;lambda u:re.search(r'(.+)1{4}',u)

process.stdin.on('data',t=>console.log(/(.+)1{4}/.test(t)))

alert(/(.+)1{4}/.test(prompt()))

n=>/(.+)1{4}/.test(n)

/(.+)1{4}/.test

$_=/(.+)1{4}/

s=>new Regex(@"(.+)1{4}").IsMatch(s);

[]

f}*5TQ.:   input as a string stored in Q
f}*5TQ.:Q  implicit arguments
        Q  input
      .:   all substrings of.
f   T      filter for this condition, keep those returning true:
  *5           repeat five times
 }   Q         in Q? (True/False)

b=a__;StringContainsQ[b~~b~~b~~b~~b]

StringContainsQ@RegularExpression@"(.+)1{4}"

<?=preg_match('/(.+)1{4}/U',$n);

<?for($e=$n+1;--$e;)for($f=$e;$f--;)for($a=str_split(substr($n,$f),$e),$k=$c='';strlen($v=array_pop($a));)$c-$v?$k=0&$c=$v:($k++<3?:die(1));

import Data.List
f x=or$tail[isInfixOf(concat$replicate 5 b)x|b<-subsequences x]

f "11111211"
f "11111"
f "12345"
f "1112111"
f "4242424242"
f "-11111"
f "3452514263534543543543543543876514264527473275"

s=int2str(i);all(s==s(1))

all(s==s(1))

DECLARE @t varchar(99)='11111'

,@z bit=0,@a INT=1,@ INT=1WHILE @a<LEN(@t)SELECT @z=IIF(@t LIKE'%'+replicate(SUBSTRING(@t,@a,@),5)+'%',1,@z),@=IIF(@=20,1,@+1),@a+=IIF(@=1,1,0)PRINT @z

DECLARE @t varchar(99)='11111'

,@z bit=0,
@a INT=1,
@ INT=1
WHILE @a<LEN(@t)
  SELECT
    @z=IIF(@t LIKE'%'+replicate(SUBSTRING(@t,@a,@),5)+'%',1,@z),
    @=IIF(@=20,1,@+1),
    @a+=IIF(@=1,1,0)

PRINT @z

$args[0]-match"(.+)1{4}"

#(re-find #"(.+)1{4}"%)

b=s=>{for(let i=0;i<s.length-4;i++){for(let n=1;n<=Math.floor((s.length-i)/5);n++){if([1,2,3,4].every(k=>s.slice(i,i+n)==s.slice(i+n*k,i+n*(k+1))))return 1}}return 0}

// test any start pos, and length
var b = s => {
    for (let i = 0; i <= s.length - 5; i++) {
        for (let n = 1; n <= Math.floor((s.length - i) / 5); n++) {
            // test if s is boring at position i, with length n
            if ([1, 2, 3, 4].every(k => s.slice(i, i + n) === s.slice(i + n * k, i + n * (k + 1)))) {
                return 1;
            }
        }
    }
    return 0;
};

console.log(b('11111'));
console.log(b('12345'));
console.log(b('1112111'));
console.log(b('-11111'));
console.log(b('3452514263534543543543543543876514264527473275'));