Another Fibonacci... yes but with other kinds of result.
The function is named `aroundFib` or `around_fib`, depending of the language.
Its parameter is `n` (positive integer).
First you have to calculate `f` the value of `fibonacci(n)` with `fibonacci(0) --> 0` and
`fibonacci(1) --> 1` (see: )
- 1) Find the count of each digit `ch` in `f` (`ch`: digit from 0 to 9), call this count `cnt` and find the maximum
value of `cnt`, call this maximum `maxcnt`. If there are ties, the digit `ch` to consider is the first one - in natural digit order - giving `maxcnt`.
- 2) Cut the value `f` into chunks of length at most `25`. The last chunk may be 25 long or less.
```
Example: for `n=100` you have only one chunk `354224848179261915075`
Example: for `n=180` f is `18547707689471986212190138521399707760` and you have two chunks 
`1854770768947198621219013` and `8521399707760`. First length here is 25 and second one is 13.
```
- At last return a string in the following format:
"Last chunk ...; Max is ... for digit ..."
where Max is `maxcnt` and digit the first `ch` (in 0..9) leading to `maxcnt`.
```
Example: for `n=100` -> "Last chunk 354224848179261915075; Max is 3 for digit 1" 
Example: for `n=180` -> "Last chunk 8521399707760; Max is 7 for digit 7"
Example: for `n=18000` -> "Last chunk 140258776000; Max is 409 for digit 1"
```
# Beware: 
`fib(18000)` has `3762` digits. Values of `n` are between `500` and `25000`.
# Note
Translators are welcome for all languages, except for Ruby since the Bash tests needing Ruby a Ruby reference solution is already there though not yet published.
def around_fib(n):
    f = str(fib(n))
    ch = ['9', '8', '7', '6', '5', '4', '3', '2', '1', '0']
    maxcnt = 0
    chunk = ''

    for digit in ch:
        cnt = f.count(digit)
        if cnt >= maxcnt:
            maxcnt = cnt
            maxdig = digit

    l = len(f) % 25
    if l == 0:
        start = len(f) - 25
    else:
        start = len(f) - l
    for i in range(start, len(f)):
        chunk += f[i]

    return 'Last chunk ' + chunk + '; Max is %d for digit ' % maxcnt + maxdig


def fib(n):
    a = 0
    b = 1
    if n == 0:
        return a
    elif n == 1:
        return b
    else:
        for i in range(2,n+1):
            c = a + b
            a = b
            b = c
        return b
def around_fib(n):
    f = str(fib(n))
    ch = ['9', '8', '7', '6', '5', '4', '3', '2', '1', '0']
    maxcnt = 0
    chunk = ''

    for digit in ch:
        cnt = f.count(digit)
        if cnt >= maxcnt:
            maxcnt = cnt
            maxdig = digit

    l = len(f) % 25
    if l == 0:
        start = len(f) - 25
    else:
        start = len(f) - l
    for i in range(start, len(f)):
        chunk += f[i]

    return 'Last chunk ' + chunk + '; Max is %d for digit ' % maxcnt + maxdig


def fib(n):
    a = 0
    b = 1
    if n == 0:
        return a
    elif n == 1:
        return b
    else:
        for i in range(2,n+1):
            c = a + b
            a = b
            b = c
        return b