Python3, 106 bytes

def f(s):s=s.replace(':','');return any((s[:i]*k)[:len(s)]==s for i in range(1,4)for k in range(1,len(s)))

Try it online!

J, 113 99 90

Probably still quite golfable.

f=:t#:[:(>:^:([:(3&g+:2&g=.[:*/]=(]$)${.)@(":@{.,3}.7":100#.1,])t#:])^:_)1+(t=.24,2$60)#.]

Takes a vector (h m s) as input, and returns a vector in the same format as output.

Examples:

   f 0 0 0
0 1 0
   f 4 14 14
4 14 41
   f 23 59 59
0 0 0
   f 3 14 15
3 14 31

Lua

I've got four different solutions, as I wasn't sure about some of the requirements.

Version 1: removal of 0s, command line input as well as os.time() backup (315)

Minimised:

z=arg if z[1] then y={hour=z[1],min=z[2],sec=z[3],day=1,month=1,year=1}end h,g,e,u=os.date,os.time(y),":",tonumber while 1 do g=g+1 b,c,d=u(h("%H",g)),u(h("%M",g)),u(h("%S",g)) a=b..c..d for x=1,#a/2 do p=1 for _ in a:gmatch(a:sub(1,x))do p=p+1 if p>math.ceil(#a/x) then print(b..e..c..e..d)return 1 end end end end

Full version with comments:

z=arg if z[1] then y={hour=z[1],min=z[2],sec=z[3],day=1,month=1,year=1}end --get command line arguments
h,g,e,u=os.date,os.time(y),":",tonumber --set up references, if command line arguments accepted use y else use current time
while 1 do
    g=g+1
    b,c,d=u(h("%H",g)),u(h("%M",g)),u(h("%S",g)) --get HH:MM:SS seperately (which allows removal of the zeroes with tonumber())
    a=b..c..d  --concat
    for x=1,#a/2 do  --check up to half of the string
        p=1
        for _ in a:gmatch(a:sub(1,x))do --for each match
            p=p+1  --count number of matches
            if p>math.ceil(#a/x) then print(b..e..c..e..d)return 1 end --if matches span entire string, cheer (and print in a pretty format)
        end
    end
end

The other versions are very similar, so I will only post the minimised versions:

Version 2: no command line input (239)

h,g,e,u=os.date,os.time(),":",tonumber while 1 do g=g+1 b,c,d=u(h("%H",g)),u(h("%M",g)),u(h("%S",g)) a=b..c..d for x=1,#a/2 do p=1 for _ in a:gmatch(a:sub(1,x))do p=p+1 if p>math.ceil(#a/x) then print(b..e..c..e..d)return 1 end end end end

Version 3: no 0 removal, with command line input (240)

z=arg if z[1] then y={hour=z[1],min=z[2],sec=z[3],day=1,month=1,year=1}end h,g=os.date,os.time(y) while 1 do g=g+1 a=h("%H%M%S",g) for x=1,3 do p=1 for _ in a:gmatch(a:sub(1,x))do p=p+1 if p>6/x then print(h("%T",g))return 1 end end end end

Version 4: none of the fancy stuff (no 0 removal or command line input) (164)

h,g=os.date,os.time() while 1 do g=g+1 a=h("%H%M%S",g) for x=1,3 do p=1 for _ in a:gmatch(a:sub(1,x))do p=p+1 if p>6/x then print(h("%T",g))return 1 end end end end

Usage instructions

In a terminal, run (versions 1 and 3)

lua interesting.lua HOURS MINUTES SECONDS

or just

lua interesting.lua

If you want it to go off the system clock.

Is there a prize for feature completeness? :P

Mathematica 125

F=Do[#~MatchQ~{#〚-1〛..,_}&&Break@#&@Partition[(i=IntegerDigits)@f[n~i~60,100],m,m,1,a_],
{n,#~(f=FromDigits)~60+1,7^6},{m,3}]&

It returns a pattern of the next interesting time:

F@{11, 11, 11}
F@{4, 14, 14}

{{1, 1, 2}, {1, 1, 2}}

{{4, 1, 4}, {4, 1, a_}}

a_ marks the end of the time.

Haskell - 227223

That's one way of doing it.

    import Data.List
e _ []=False
e f s=let (h,t)=f s in t`isPrefixOf`h||h`isPrefixOf`t&&e f t
i [a,b,s]=head[t|s<-[s+1..],f<-map splitAt[1..3],let m=b+s`div`60;h=a+m`div`60;t=[h`mod`24,m`mod`60,s`mod`60],e f$concatMap show t]

Examples

λ: i [11,11,11]
[11,21,1]
λ: i [4,14,14]
[4,14,41]