[Question]
  [
## Background
The greatest common divisor (*gcd* for short) is a convenient mathematical function, since it has many useful properties.
One of them is [B√©zout's identity](https://en.wikipedia.org/wiki/B%C3%A9zout's_identity): if `d = gcd(a, b)`, then there exist integers `x` and `y` such that `d = x*a + y*b`.
In this challenge, your task is to visualize this property with simple ASCII art.
## Input
Your inputs are two positive integers `a` and `b`, given in any reasonable format.
You may also take unary inputs (repetitions of a single printable ASCII character of your choice), but you must be consistent and use the same format for both inputs.
The inputs may be in any order, and they may be equal.
## Output
Your output is a string `s` of length `lcm(a, b) + 1` (*lcm* stands for lowest common multiple).
The characters of `s` represent integers from `0` to `lcm(a, b)`.
The character `s[i]` is a lowercase `o` if `i` is a multiple of `a` or `b`, and a period `.` otherwise.
Note that zero is a multiple of every number.
Now, because of B√©zout's identity, there will be at least one pair of characters `o` in `s` whose distance is exactly `gcd(a, b)`.
The leftmost such pair is to be replaced by uppercase `O`s; this is the final output.
## Example
Consider the inputs `a = 4` and `b = 6`.
Then we have `gcd(a, b) = 2` and `lcm(a, b) = 12`, so the length of `s` will be `13`.
The multiples of `a` and `b` are highlighted as follows:
```
0  1  2  3  4  5  6  7  8  9 10 11 12
o  .  .  .  o  .  o  .  o  .  .  .  o
```
There are two pairs of `o`s with distance two, but we will only replace the leftmost ones with `O`s, so the final output is
```
o...O.O.o...o
```
## Rules and scoring
You can write a full program or a function.
The lowest byte count wins, and standard loopholes are disallowed.
## Test cases
```
 1  1 -> OO
 2  2 -> O.O
 1  3 -> OOoo
 4  1 -> OOooo
 2  6 -> O.O.o.o
 2  3 -> o.OOo.o
10  2 -> O.O.o.o.o.o
 4  5 -> o...OO..o.o.o..oo...o
 8  6 -> o.....O.O...o...o.o.....o
12 15 -> o...........O..O........o.....o.....o........o..o...........o
19 15 -> o..............o...o..........o.......o......o...........o..o..............OO.............o....o.........o........o.....o............o.o..............o.o............o.....o........o.........o....o.............oo..............o..o...........o......o.......o..........o...o..............o
```
  
[Answer]
# Julia, ~~111~~ ~~110~~ ~~107~~ ~~103~~ 96 bytes
```
f(a,b)=replace(join([i%a*(i%b)<1?"o":"."for i=0:lcm(a,b)]),"o$(d="."^(gcd(a,b)-1))o","O$(d)O",1)
```
This is a function that accepts two integers and returns a string.
Ungolfed:
```
function f(a::Int, b::Int)
    # Construct an array of dots and o's
    x = [i % a * (i % b) < 1 ? "o" : "." for i = 0:lcm(a, b)]
    # Join it into a string
    j = join(x)
    # Replace the first pair with distance gcd(a, b) - 1
    replace(j, "o$(d = "."^(gcd(a, b) - 1))o", "O$(d)O", 1) 
end
```
Saved a byte thanks to nimi!
[Answer]
## [Retina](https://github.com/mbuettner/retina), ~~112~~ ~~109~~ ~~99~~ ~~94~~ 91 bytes
```
^
. 
+r`(?<!^\1+). (.+) 
$'$0
.(?=.* (.+) (.+))(?=\1* |\2* )
o
o(\.*)o((\1\.*o)*) .*
O$1O$2
```
Not very competitive, I think, but number theory in Retina is always quite fun. :)
Takes input as unary numbers using `.` as the unary digit.
[Try it online.](http://retina.tryitonline.net/#code=XgouIAorcmAoPzwhXlwxKykuICguKykgCiQnJDAKLig_PS4qICguKykgKC4rKSkoPz1cMSogfFwyKiApCm8KbyhcLiopbygoXDFcLipvKSpcLiopIC4qCk8kMU8kMg&input=Li4uLi4uLi4uLi4uIC4uLi4uLi4uLi4uLi4uLg)
### Explanation
```
^
. 
```
This inserts a `.` and a space in front of the input. This will ultimately become the output.
```
+r`(?<!^\1+). (.+) 
$'$0
```
This prepends the LCM of `a` and `b` to the string. Since we already have a `.` there, we'll end up with `lcm(a,b)+1`. This is accomplished by repeatedly prepending `b` as long as `a` does not divide this new prefix. We capture `a` into a group one and then check if we can reach the beginning of the string by matching that capture at least once. `b` is then inserted into the string via the rarely used `$'` which inserts everything *after* the match into the substitution.
```
.(?=.* (.+) (.+))(?=\1* |\2* )
o
```
This one matches characters at positions which are divided by `a` or `b`. It makes use of the fact that the result is symmetric: since `lcm(a,b)` is divided by both `a` and `b` going left by subtracting instances of `a` or `b` yields the same pattern as going right from `0` by adding them. The first lookahead simply captures `a` and `b`. The second lookahead checks that there is a multiple of each `a` or `b` characters before the first space.
```
o(\.*)o((\1\.*o)*) .*
O$1O$2
```
As stated on Wikipedia, in addition to B√©zout's identity it is also true that
> 
> The greatest common divisor `d` is the smallest positive integer that can be written as `ax + by`.
> 
> 
> 
This implies that the GCD will correspond to the shortest gap between two `o`s in the output. So we don't have to bother finding the GCD at all. Instead we just look for first instance of the shortest gap. `o(\.*)o` matches a candidate gap and captures its width into group 1. Then we try to reach the first space by alternating between a backreference to group 1 and `o`s (with optional additional `.`s). If there is a shorter gap further to the right, this will fail to match, because we cannot get past that gap with the backreference. As soon as all further gaps are at least as wide as the current one, this matches. We capture the end of the LCM-string into group 2 and match the remainder of the string with `.*`. We write back the uppercase `O`s (with the gap in between) as well as the remainder of the LCM string, but discard everything starting from the space, to remove `a` and `b` from final result.
[Answer]
# Jolf, 52 bytes
```
on*'.wm9jJŒ°R m*Yhm8jJDN?<*%Sj%SJ1'o'.}"'o%o"n"O%O"n
```
I will split this code up into two parts.
```
on*'.wm9jJ
on         set n
  *'.       to a dot repeated
      m9jJ  the gcd of two numeric inputs
Œ°R m*Yhm8jJDN?<*%Sj%SJ1'o'.}"'o%o"n"O%O"n
    *Y                                    multiply (repeat) Y (Y = [])
      hm8jJ                                by the lcm of two inputs + 1
  _m       DN              }              and map the array of that length
             ?<*%Sj%SJ1'o'.               "choose o if i%a*(i%b)<1; otherwise choose ."
 R                          "'            join by empty string
Œ°                            'o%o"n        replace once (capital Rho, 2 bytes): "o"+n+"o"
                                   "O%O"n   with "O"+n+"O"
                                          implicit printing
```
[Try it here!](http://conorobrien-foxx.github.io/Jolf/#code=b24qJy53bTlqSs6hUiBtKllobThqSkROPzwqJVNqJVNKMSdvJy59IidvJW8ibiJPJU8ibg&input=NAoKNg)
[Answer]
# ùîºùïäùïÑùïöùïü, 50 chars / 90 bytes
```
‚©•–ú≈´‚ÅΩ√Æ√≠+1)‚ìú$%√Æ‚Öã$%√≠?‚çò.:‚çòo)‚®ùƒã…º(`o‚¶É‚çò.ƒò–ú≈©‚ÅΩ√Æ√≠-1)}o‚Äù,‚Ü™$√∫‚¨Æ
```
`[Try it here (Firefox only).](http://molarmanful.github.io/ESMin/interpreter.html?eval=true&input=%5B4%2C6%5D&code=%E2%A9%A5%D0%9C%C5%AB%E2%81%BD%C3%AE%C3%AD%2B1%29%E2%93%9C%24%25%C3%AE%E2%85%8B%24%25%C3%AD%3F%E2%8D%98.%3A%E2%8D%98o%29%E2%A8%9D%C4%8B%C9%BC%28%60o%E2%A6%83%E2%8D%98.%C4%98%D0%9C%C5%A9%E2%81%BD%C3%AE%C3%AD-1%29%7Do%E2%80%9D%2C%E2%86%AA%24%C3%BA%E2%AC%AE)`
There must be a way to golf this further!
# Explanation
This is a basic two-phase algorithm. It's actually quite simple.
### Phase 1
```
‚©•–ú≈´‚ÅΩ√Æ√≠+1)‚ìú$%√Æ‚Öã$%√≠?‚çò.:‚çòo)‚®ù
```
First, we create a range from 0 to the LCM+1. Then we map over it, checking if either of the inputs is a factor of the current item in the range. If so, we replace that item with an `o`; otherwise, we replace it with a `.` . Joining it gives us a series of o's and dots that we can pass to phase two.
### Phase 2
```
ƒã…º(`o‚¶É‚çò.ƒò–ú≈©‚ÅΩ√Æ√≠-1)}o‚Äù,‚Ü™$√∫‚¨Æ
```
This is just one big replace function. A regex is created as `o[dots]o`, where the amount of dots is determined by the GCD-1. Since this regex is not global, it will only match the first occurrence. After, the match is replaced by `O[dots]O` using a toUpperCase function.
[Answer]
# [MATL](https://esolangs.org/wiki/MATL), 72 bytes
Uses [version 6.0.0](https://github.com/lmendo/MATL/releases/tag/6.0.0), which is earlier than this challenge. The code runs in Matlab and in Octave.
```
2$tZm1+:1-bbvtbw\A~otbZ}ZdXK1+ltb(3X53$X+1K2$lh*t2=f1)tK+hwg1+Ib('.oO'w)
```
### Example
```
>> matl
 > 2$tZm1+:1-bbvtbw\A~otbZ}ZdXK1+ltb(3X53$X+1K2$lh*t2=f1)tK+hwg1+Ib('.oO'w)
 > 
> 1
> 1
OO
>> matl
 > 2$tZm1+:1-bbvtbw\A~otbZ}ZdXK1+ltb(3X53$X+1K2$lh*t2=f1)tK+hwg1+Ib('.oO'w)
 > 
> 2
> 3
o.OOo.o
>> matl
 > 2$tZm1+:1-bbvtbw\A~otbZ}ZdXK1+ltb(3X53$X+1K2$lh*t2=f1)tK+hwg1+Ib('.oO'w)
 > 
> 12
> 15
o...........O..O........o.....o.....o........o..o...........o
```
### Explanation
I have [no idea](https://i.stack.imgur.com/6x5av.jpg) how it works. I just typed characters randomly. I think there is some convolution involved.
**Edit: [Try it online!](http://matl.tryitonline.net/#code=MiR0Wm0xKzoxLWJiJnZ0YndcQX50Ylp9WmRYSzErbHRiKDNYNTMkWSsxSzIkbGgqdDI9ZjEpdEsraHdnMStJYignLm9PJ3cp&input=MTIKMTU)** The code in the link has been slightly modified to conform to changes in the language (as of June 2, 2016).
[Answer]
## [Japt](http://ethproductions.github.io/japt/), 83 bytes
```
'.pD=U*V/(C=(G=@Y?G$($YX%Y :X} $($UV)+1 ¬£Y%U¬©Y%V?".:o"} $.replace($E=`o{'.pC-1}o`Eu
```
Not fully golfed yet... And doesn't want to be golfed :/
[Answer]
## Javascript, ~~170~~ ~~164~~ ~~161~~ ~~153~~ ~~145~~ ~~141~~ 136 bytes
```
(a,b)=>[...Array(a*b/(c=(g=(a,b)=>b?g(b,a%b):a)(a,b))+1)].map((x,i)=>i%a&&i%b?'.':'o').join``.replace(`o${e='.'.repeat(c-1)}o`,`O${e}O`)
```
That's quite lonnnggggg....
[Demo](http://www.es6fiddle.net/iix3s7ty/), explicitly defined variables because the interpreter uses strict mode.
[Answer]
# Python 2, ~~217~~ ~~200~~ 191 bytes
This is a little blunt, but it works. Any golfing tips are appreciated, ~~especially if you know how to fix that `s[i] = s[v] = "o"` problem I encountered, where that would overwrite "O"s~~ Got it!
```
g=lambda a,b:b and g(b,a%b)or a
def f(a,b):
 h=g(a,b);x=1+a*b/h;s=["."]*x;v=k=0
 for i in range(x):
    if(i%a)*(i%b)<1:
     if k:s[i]="o"
     else:k=i==h+v;s[i]=s[v]="oO"[k]
     v=i
 return''.join(s)
```
**Ungolfed:**
```
def gcd(a,b):                           # recursive gcd function
    if b:
        return g(b,a%b)
    else:
        return a
def f(a,b):
    h = gcd(a,b)
    x = 1 + a*b/h                       # 1 + lcm(a,b)
    s = ["."] * x
    v = 0
    k = 0
    for i in range(x):
        if i%a == 0 and i % b == 0:
            if k == 0:
                k = (i == h+v)          # correct distance apart?
                if k:                   # if "O" just found
                    s[i] = s[v] = "O"
                else:
                    s[i] = s[v] = "o"
            else:
                s[i] = "o"              # if "O" already found, always "o"
            v = i                       # If we found an "o" or an "O", i is the new v
    return ''.join(s)
```
]