R, 73 bytes

paste(names(b<-by((r<-rle(el(strsplit(scan(,""),""))))$l,r$v,max)),"=",b)

Try it online!

I would like to mention that this solution uses rle function, which was introduced in the R version 3.6.2 (Dec. 2019); therefore my answer and the earlier one from 2014 are not directly comparable.

05AB1E, 17 bytes

ÙvIγʒyå}€gZy…ÿ=ÿ,

Try it online!

ÙvIγʒyå}€gZy…ÿ=ÿ,   Argument s
Ùv                  For each c in s uniquified
  Iγ                Split s into chunks of consecutive equal elements
    ʒyå}            Filter: Keep elements that contain c
        €g          Map to element length
          Z         Get max
           y…ÿ=ÿ,   Print "c=max"

Python 3, 69 bytes

s=input()
for c in set(s):
 i=0
 while-~i*c in s:i+=1
 print(c,'=',i)

Try it online!

Even golfed Python can be very readable. I think this code is fully idiomatic except for the -~i for i+1 and the single-letter variables. Thanks to pxeger for saving 1 byte.

Example runs:

>>> helloworld
e = 1
d = 1
h = 1
l = 2
o = 1
r = 1
w = 1
>>> acbaabbbaaaaacc
a = 5
c = 2
b = 3

Bash / GNU core utils (40)

fold -1|uniq -c|sort -k2 -k1r|uniq -f1

Try it online!

Output when input is "acbaabbbaaaaacc":

      5 a
      3 b
      2 c

Explanation:

• fold -1 adds a newline after each character
• uniq -c counts the number of consecutive identical lines
• sort -k2 -k1r sorts the result so as the maximum count for each character comes first
• uniq -f1 keeps only the first line for each character

Jelly, 12 bytes

ŒriƇṀj”=ʋⱮQY

Try it online!

-2 bytes thanks to Unrelated String!

8 bytes to output as a list of pairs

How it works

ŒriƇṀj”=ʋⱮQY - Main link. Takes a string S on the left
        ʋ    - Group the previous 4 links into a dyad f(S, C):
Œr           -   Run-length encode S
   Ƈ         -   For each sublist in rle(S), keep it if:
  i          -     It contains C
    Ṁ        -   Maximum
                 As all elements are in the form [C, n] for integers n,
                 Python considers the maximum as the greatest n
     j”=     -   Join with "="; Yields [C, "=", n]
          Q  - Get the unique characters of S
         Ɱ   - For each unique character C in S, yield f(C, S)
           Y - Join by newlines and output

Husk, 16 bytes

mS:(:'=s▲`mg¹#)U

Try it online!

Explanation:

m              U    # for each of the unique elements of the input
 S:                 # prepend the element to
   (:'=       )     # '=' prepended to
        ▲           # the maximum of
             #      # the number of times it occurs in 
         `mg¹       # each group of identical items in the input,
       s            # converted into a string

I saw that Razetime had answered this in Husk, so I had to have a go too...

Husk, 18 bytes

mȯJ'=§eo;←osL►Lk←g

Try it online!

11 bytes if we ignore the exact input format and display as a list of pairs instead.

Java, 192 bytes

import java.util.*;s->{int p=0,c=0;Map<Character,Integer>m=new HashMap();for(byte b:s.getBytes()){c=p==b?c+1:1;m.merge((char)(p=b),c,Math::max);}m.forEach((k,v)->System.out.println(k+"="+v));}

Try it online!

Java, 251 bytes (with regular expressions)

import java.util.regex.*;import java.util.function.*;s->{Matcher m=Pattern.compile("(.)\\1*").matcher(s);Map<String,Integer>a=new HashMap();while(m.find())a.merge(m.group(1),m.group().length(),Math::max);a.forEach((k,v)->System.out.println(k+'='+v));}

Try it online!

APL (Dyalog), 41 bytes

{(⊃,'=',⍕∘≢)¨l[∪⍳⍨⊃¨l←n[⍒≢¨n←⍵⊂⍨1,2≠/⍵]]}

Try it online!

Ungolfed:

{
  n←⍵⊂⍨1,2≠/⍵
  l←n[⍒≢¨n]
  (⊃,'=',⍕∘≢)¨l[∪⍳⍨⊃¨l]
}

{ anonymous function

 2≠/⍵ pair-wise sliding window inequality of the argument
 1, prepend a one
 ⍵⊂⍨ use that to partition the argument
 n← store in n

 n[…] index n with:
  ⍒ the descending order of
  ≢¨ the length of each element in
  ⍵ argument
 l← store in l

 l[…] index l with:
  ∪ the unique of
  ⍳⍨ the first occurrence of each element of
  ⊃¨l the first element of each of l
 (…)¨ apply the following tacit function on each element
  ⊃ the first element
  , concatenated to
  '=' an equal sign
  , concatenated to
  ⍕∘≢ the formatted length

}

Clojure, 107 bytes

#(apply str(for[p(vals(group-by last(partition-by(fn[i]i)%)))](str(ffirst p)\=(apply max(map count p))\,)))

Returns "a=5,c=2,b=3," for the example input. This would have been 89 bytes, returning ([\a 5] [\c 2] [\b 3]):

#(for[p(vals(group-by last(partition-by(fn[i]i)%)))][(ffirst p)(apply max(map count p))])

Python 3, 88 bytes

from re import*;a=input();[(i[0],len(max(i))) for i in(findall(l+"+",a)for l in set(a))]

Outputs in a format of:

[('b', 3), ('c', 2), ('a', 5)]

BACCHUS, 41

'acbaabbbaaaaacc'j:A=·z#:B=($A,$0h:a(·>0.$0,'=',·+)?),$B¨n

I have discounted the actual parameter String.

Explanation

j:A= Transform the String in a block (some sort of array) and stores it in A variable.

·z#:B= Read last value on the stack (the block) and removes al duplicates, storing it in B.

( Open for each

$A,$0h:a counts how many times the current element of the for each is present in A (read as a String) and pushes this value to Stack (this avoid inmediate printing of the value).

( Open if

·>0. is last value is > 0 then

$0,'=',·+ Concatenates current for each element, '=' and last value in stack (that will be printed out)

)? Close if

,$B¨ Close for each indicating the block to loop over (B).

n Indicates that the output must be space separated

Perl - 65 71 76 characters

My first code golf!

For each answer, copy to golf.pl and run as:

echo acbaabbbaaaaacc | perl golf.pl

My shortest solution prints each character as many times as it appears, since that is not prohibited by the rules.

$_=$i=<>;for(/./g){$l=length((sort$i=~/$_*/g)[-1]);print"$_=$l
"}

My next-shortest solution (85 90 characters) only prints each character once:

<>=~s/((.)\2*)(?{$l=length$1;$h{$2}=$l if$l>$h{$2}})//rg;print"$_=$h{$_}
"for keys %h

J 45

~.,.'=',.":@({.@>>.//.$&>)@(<;.2~2&(~:/\),1:)

As a verb. Though for using it , you'd need to assign it to a name, eg:

f=:~.,.'=',.":@({.@>>.//.$&>)@(<;.2~2&(~:/\),1:)
f 'acbaabbbaaaaacc'
a=5
c=2
b=3

Short explanation:

                                @(        2&(~:/\),1:): find where neighbors are different
                                   <;.2~              : cut in these places
            (              $&> )                      : get the length of the string in the box
             ({.@>)                                   : get the first letter from the box
                   (>./)/.                            : get the maximum for each unique char
~.,.'=',.":@(                  )                      : Output: letter = stringified value

Clojure - 105 bytes

I'm learning Clojure at the moment and what would be a better way to do it than golfing? So, here's my second entry to this challenge.

(def f #(apply str(for[c(distinct %)](str c\=(apply max(map count(re-seq(re-pattern(str c\+))%)))"\n"))))

Examples:

=> (f "acbaabbbaaaaacc")
"a=5\nc=2\nb=3\n"
=> (f "aaaabaa")
"a=4\nb=1\n"
=> (println (f "acbaabbbaaaaacc"))
a=5
c=2
b=3

Scala – 99 as program, 97 as function

Program (reads one line from stdin)

for((k,v)<-"(.)\\1*".r.findAllIn(Console.in.readLine).toSeq.groupBy(_(0)))println(k+"="+v.max.size)

Function:

def f(s:String)=for((k,v)<-"(.)\\1*".r.findAllIn(s).toSeq.groupBy(_(0)))println(k+"="+v.max.size)

It's written in Python, in order to run the code, just call thefunction maxsequence(str). For instance, maxsequence('aaaaaannndmdejlsfnsfsssssnnnnnxxx') or maxsequence("kdkdkdkdjeeeiwwwnnnmdnnsbjdiiiiiiiiiidndbbcbbccbcvcvdddcjdjdjwwwwwwkkkkxlxllllllll")

def maxsequence(str):

    count = 1 # count if the letters are repeat
    step = 0 # once the next letter changed, step = count.
    countArray = [] # put all the sequence numbers in an array
    lettersArray = [] # put all the repeat letters in an array
    max = 0
    # for calculating the max of the array, it needs two indexes to do that
    indexFirst = 0
    indexNext = 1
    i = 0
    while(i<len(str) and i<=(len(str)-2)):
        if(str[i]==str[i+1]):
            count += 1
            step = count
        else:
            lettersArray.append(str[i])
            step = count
            countArray.append(step)
            count = 1
        i += 1
    countArray.append(step)
    lettersArray.append(str[i])

    while(countArray[indexFirst]>=countArray[indexNext] and indexNext<(len(countArray)-1)):
        indexNext += 1
        if(countArray[indexFirst]<=countArray[indexNext]):
            indexFirst = indexNext
    max = countArray[indexFirst]

    for i in range(len(lettersArray)):
        print lettersArray[i],"=",countArray[i]

    print "The max sequence is:", max
    return max

Powershell 80 77 72

$x=$args;[char[]]"$x"|sort -u|%{"$_="+($x-split"[^$_]"|sort)[-1].length}

You need to run it on console...

JavaScript - 91

for(i=0,s=(t=prompt()).match(/(.)\1*/g);c=s[i++];)t.match(c+c[0])||alert(c[0]+'='+c.length)

EDIT: My first solution obeys the rules, but it prints several times single char occurrences like abab => a=1,b=1,a=1,b=1 so I came out with this (101 chars), for those not satisfied with my first one:

for(i=0,s=(t=prompt()).match(/((.)\2*)(?!.*\1)/g);c=s[i++];)t.match(c+c[0])||alert(c[0]+'='+c.length)

R (219, 213, 197, 196, 191, 165, 169, 157, 163 characters)

The data.table version added. There was an error in the previous data.table version.

Golfed data.table version (163)

require(data.table);f=function(x){x=strsplit(x,"")[[1]];data.table(x=x,y=cumsum(c(1,x[-1]!=head(x,-1))))[,.N,list(x,y)][order(-N)][!duplicated(x),paste0(x,"=",N)]}

Ungolfed data.table version

require(data.table)
f <- function(x) {
  x <- strsplit(x, "")[[1]]
  data.table(a=x, y=cumsum(c(1, x[-1] != head(x, -1))))[
  , .N, list(a, y)][order(-N)][!duplicated(a), paste0(a, "=", N)]
}

Golfed data.frame version (174)

f=function(x){x=strsplit(x,"")[[1]];d=data.frame(a=x,y=cumsum(c(1,x[-1]!=head(x,-1))));d=aggregate(d$y,d,length);d=d[order(-d$x),];d=d[!duplicated(d$a),];paste0(d$a,"=",d$x)}

Ungolfed data.frame version

f <- function(x) {
  x <- strsplit(x, "")[[1]]
  d <- data.frame(a = x, y = cumsum(c(1, x[-1] != head(x, -1))))
  d <- aggregate(d$y, d, length)
  d <- d[order(-d$x), ]
  d <- d[!duplicated(d$a), ]
  paste0(d$a, "=", d$x)
}

f("acbaabbbaaaaacc")
f("acbaabbbaaaaaccdee")

Javascript (E6) 103

A javascript solution that cares about the requested output format.

F=a=>(p=q={},m={},[...a].map(x=>(x!=p&&(q[p=x]=0),m[x]>++q[x]?0:m[x]=q[x])),''+[i+'='+m[i]for(i in m)])

Ungolfed

F=a=>(
  p=q={}, m={},
  [...a].map(
    x=>(
      x!=p && (q[p=x]=0),
      m[x] > ++q[x] ? 0 : m[x] = q[x]
    )
  ),
  '' + [i+'='+m[i] for(i in m)]
)

Test

In Firefox console

console.log(F("aaaaaddfffabbbbdb"))

a=5,d=2,f=3,b=4

C# (LinQPad)

146

This is tsavino's answer but shorter. Here, I used Distinct() instead of GroupBy(c=>c). Also the curly braces from the foreach-loop are left out:

void v(string i){foreach(var c in i.Distinct())Console.WriteLine(c+"="+(from Match m in Regex.Matches(i,"["+c+"]+")select m.Value.Length).Max());}

136

I tried using a lambda expression instead of the normal query syntax but since I needed a Cast<Match> first, the code became 1 character longer... Anyhow, since it can be executed in LinQPad, you can use Dump() instead of Console.WriteLine():

void v(string i){foreach(var c in i.Distinct())(c+"="+(from Match m in Regex.Matches(i,"["+c+"]+")select m.Value.Length).Max()).Dump();}

Further study of the code got me thinking about the Max(). This function also accepts a Func. This way I could skip the Select part when using the lambda epxression:

void v(string i){foreach(var c in i.Distinct())(c+"="+Regex.Matches(i,"["+c+"]+").Cast<Match>().Max(m=>m.Value.Length)).Dump();}

Thus, final result:

128

Update:

Thanks to the tip from Dan Puzey, I was able to save another 6 characters:

void v(string i){i.Distinct().Select(c=>c+"="+Regex.Matches(i,"["+c+"]+").Cast<Match>().Max(m=>m‌​.Value.Length)).Dump();}

Length:

122

Groovy - 80 chars

Based on this clever answer by xnor :

t=args[0];t.toSet().each{i=0;
while(t.contains(it*++i));
println "${it}=${i-1}"}

Output:

$ groovy Golf.groovy abbcccdddd
d=4
b=2
c=3
a=1

Ungolfed:

t=args[0]

t.toSet().each { c ->
    i=0
    s=c

    // repeat the char c with length i
    // e.g. "b", "bb", "bbb", etc
    // stop when we find a length that is not in t:
    // this is the max + 1
    while (t.contains(s)) {
        i++
        s=c*i
    }
    println "${c}=${i-1}"
}

JavaScript [83 bytes]

prompt().match(/(.)\1*/g).sort().reduce(function(a,b){return a[b[0]]=b.length,a},{})

^{Run this code in the browser console.}

For input "acbaabbbaaaaacc" the console should output "Object {a: 5, b: 3, c: 2}".

Haskell - 113 120 bytes

import Data.List
main=interact$show.map(\s@(c:_)->(c,length s)).sort.nubBy(\(a:_)(b:_)->a==b).reverse.sort.group

Tested with

$ printf "acbaabbbaaaaacc" | ./sl
[('a',5),('b',3),('c',2)]

8086 machine code, 82 80

Contents of the x.com file:

B7 3D 89 DF B1 80 F3 AA 0D 0A 24 B4 01 CD 21 42
38 D8 74 F7 38 17 77 02 88 17 88 C3 31 D2 3C 0D
75 E9 BF 21 3D B1 5E 31 C0 F3 AE E3 EE 4F BB 04
01 8A 05 D4 0A 86 E0 0D 30 30 89 47 02 3C 30 77
04 88 67 03 43 89 3F 89 DA B4 09 CD 21 47 EB D7

It only supports repetitions of up to 99 characters.

Source code (served as input for the debug.com assembler), with comments!

a
    mov bh, 3d         ; storage of 128 bytes at address 3d00
    mov di, bx
    mov cl, 80
    rep stosb          ; zero the array
    db 0d 0a 24
; 10b
    mov ah, 1
    int 21             ; input a char
    inc dx             ; calculate the run length
    cmp al, bl         ; is it a repeated character?
    je  10b
    cmp [bx], dl       ; is the new run length greater than previous?
    ja  11a
    mov [bx], dl       ; store the new run length
; 11a
    mov bl, al         ; remember current repeating character
    xor dx, dx         ; initialize run length to 0
    cmp al, d          ; end of input?
    jne 10b            ; no - repeat
    mov di, 3d21       ; start printing run lengths with char 21
    mov cl, 5e         ; num of iterations = num of printable characters
; 127
    xor ax, ax
    repe scasb         ; look for a nonzero run length
    jcxz 11b           ; no nonzero length - exit
    dec di
    mov bx, 104        ; address of output string
    mov al, [di]       ; read the run length
    aam                ; convert to decimal
    xchg al, ah
    or  ax, 3030
    mov [bx+2], ax
    cmp al, 30         ; was it less than 10?
    ja  145
    mov [bx+3], ah     ; output only one digit
    inc bx             ; adjust for shorter string
; 145
    mov [bx], di       ; store "x=" into output string
    mov dx, bx         ; print it
    mov ah, 9
    int 21
    inc di
    jmp 127            ; repeat
; 150

rcx 50
n my.com
w
q

Here are some golfing techniques used here that I think were fun:

• array's address is 3d00, where 3d is the ascii-code for =. This way, the address for array's entry for character x is 3d78. When interpreted as a 2-character string, it's x=.
• Output buffer is at address 104; it overwrites initialization code that is no longer needed. End-of-line sequence 0D 0A 24 is executed as harmless code.
• The aam instruction here doesn't provide any golfing, though it could...
• Writing the number twice, first assuming it's greater than 10, and then correcting if it's smaller.
• Exit instruction is at an obscure address 11b, which contains the needed machine code C3 by luck.

C, 126 125 119 bytes

l,n,c[256];main(p){while(~(p=getchar()))n*=p==l,c[l=p]=c[p]>++n?c[p]:n;for(l=256;--l;)c[l]&&printf("%c=%d\n",l,c[l]);}

Running:

$ gcc seq.c 2>& /dev/null
$ echo -n 'acbaabbbaaaaacc' | ./a.out
c=2
b=3
a=5

JavaScript 116

prompt(x={}).replace(/(.)\1*/g,function(m,l){n=m.length
if(!x[l]||x[l]<n)x[l]=n})
for(k in x)console.log(k+'='+x[k])

C 169

Iterates each printable character in ASCII table and counts max from input string.

#define N 128
int c,i,n;
char Y[N],*p;
int main(){gets(Y);
for(c=33;c<127;c++){p=Y;n=0,i=0;while(*p){if(*p==c){i++;}else{n=(i>n)?i:n;i=0;}p++;}
if(n>0) printf("%c=%d\n",c,n);}
}

T-SQL (2012) 189 171

Edit: removed ORDER BY because rules allow any output order.

Takes input from a CHAR variable, @a, and uses a recursive CTE to create a row for each character in the string and figures out sequential occurrences.

After that, it's a simple SELECT and GROUP BY with consideration for the order of the output.

Try it out on SQL Fiddle.

WITH x AS(
    SELECT @a i,''c,''d,0r,1n
    UNION ALL 
    SELECT i,SUBSTRING(i,n,1),c,IIF(d=c,r+1,1),n+1
    FROM x
    WHERE n<LEN(i)+2
)
SELECT d+'='+LTRIM(MAX(r))
FROM x
WHERE n>2
GROUP BY d

Assigning the variable:

DECLARE @a CHAR(99) = 'acbaabbbaaaaacc';

Sample output:

a=5
c=2
b=3

CoffeeScript, 109 bytes

I like regex.

f=(s)->a={};a[t[0]]=t.length for t in s.match(/((.)\2*)(?!.*\1)/g).reverse();(k+'='+v for k,v of a).join '\n'

Here is the compiled JavaScript you can try in your browser's console

f = function(s) {
  var a, t, _i, _len, _ref;
  a = {};
  _ref = s.match(/((.)\2*)(?!.*\1)/g).reverse();
  for (_i = 0, _len = _ref.length; _i < _len; _i++) {
    t = _ref[_i];
    a[t[0]] = t.length;
  }
  return a;
};

Then you can call

f("acbaabbbaaaaacc")

to get

c=2
a=5
b=3

Java Code: Out of interest I just found a solution will further enhance soon.

import java.util.HashMap;

import java.util.Map.Entry; import java.util.Scanner;

public class MaxSequenceLength { private HashMap characterCount = new HashMap<>();

public static void main(String[] args) {
    Scanner inputScanner = new Scanner(System.in);
    new MaxSequenceLength().validateInput(inputScanner.nextLine());
    inputScanner.close();
}

private void validateInput(String inputString) {
    if (inputString.length() <= 1024) {
        scanInput(inputString.toCharArray());
    }
}

private void scanInput(char[] characterArrayInput) {
    for (char tempCharacter : characterArrayInput) {
        if ((int) tempCharacter < 65
                || ((int) tempCharacter > 90 && (int) tempCharacter < 97)
                || (int) tempCharacter > 122) {
            continue;
        } else if (characterCount.containsKey(tempCharacter)) {
            characterCount.put(tempCharacter,
                    characterCount.get(tempCharacter) + 1);
        } else {
            characterCount.put(tempCharacter, 1);
        }
    }
    for (Entry<Character, Integer> tempEntry : characterCount.entrySet()) {
        System.out.println(tempEntry.getKey() + " = "
                + tempEntry.getValue());
    }
}

}

The code would ignore all the other symbols and give only case sensitive character output.

Java 247

import java.util.*;public class a{public static void main(String[]a){Map<Character, Integer> m = new HashMap<>();for(char c:a[0].toCharArray()){Integer v=m.get(c);m.put(c,v==null?1:v+1);}for(char c:m.keySet())System.out.println(c+"="+m.get(c));}}

C# in LINQPad - 159 Bytes

Well, at least I beat T-SQL ;P Won't beat anyone else, but I thought I'd share it anyway.

void v(string i){foreach(var c in i.GroupBy(c=>c)){Console.WriteLine(c.Key+"="+(from Match m in Regex.Matches(i,"["+c.Key+"]+")select m.Value.Length).Max());}}

Usage:

v("acbaabbbaaaaacc");

Suggestions are always welcome!

CJam, 27 26 25 bytes

l:S_&{'=L{2$+_S\#)}g,(N}/

Try it online.

Example

$ cjam maxseq.cjam <<< "acbaabbbaaaaacc"
a=5
c=2
b=3

How it works

l:S       " Read one line from STDIN and store the result in “S”.                   ";
_&        " Intersect the string with itself to remove duplicate characters.        ";
{         " For each unique character “C” in “S”:                                   ";
  '=L     " Push '=' and ''.                                                        ";
  {       "                                                                         ";
    2$+_  " Append “C” and duplicate.                                               ";
    S\#)  " Get the index of the modified string in “S” and increment it.           ";
  }g      " If the result is positive, there is a match; repeat the loop.           ";
  ,       " Retrieve the length of the string.                                      ";
  (       " Decrement to obtain the highest value that did result in a match.       ";
  N       " Push a linefeed.                                                        ";
}/        "                                                                         ";

GolfScript, 26 bytes

:s.&{61{2$=}s%1,/$-1=,n+}%

Try it online.

Explanation:

• :s saves the input string in the variable s for later use.
• .& extracts the unique characters in the input, which the rest of the code in the { }% loop then iterates over.
• 61 pushes the number 61 (ASCII code for an equals sign) on top of the current character on the stack, to act as an output delimiter.
• {2$=}s% takes the string s and replaces its characters with a 1 if they equal the current character being iterated over, or 0 if they don't. (It also leaves the current character on the stack for output.)
• 1,/ takes this string of ones and zeros, and splits it at zeros.
• $ sorts the resulting substrings, -1= extracts the last substring (which, since they all consist of repetitions of the same character, is the longest), and , returns the length of this substring.
• n+ stringifies the length and appends a newline to it.

Ps. If the equals signs in the output are optional, the 61 can be omitted (and the 2$ replaced by 1$), for a total length of 24 bytes:

:s.&{{1$=}s%1,/$-1=,n+}%

Python (62 55)

n=raw_input()
print map(lambda x:[x,n.count(x)],set(n))

Old answer:

n=raw_input()
s=set(n)
print zip(s,map(lambda x:[x,n.count(x)],s))

Pyth, 24 25 26 (or 29)

=ZwFY{Z=bkW'bZ~bY)p(Yltb

Test can be done here: link

Outputs in the format:

('a', 5)
('c', 2)
('b', 3)

Explanation:

=Zw              Store one line of stdin in Z
FY{Z             For Y in set(Z):
=bk              b=''
W'bZ             while b in Z:
~bY              b+=Y
)                end while
p(Yltb           print (Y, len(b)-1)

Python:

k=""
Z=copy(input())
for Y in set(Z):
 b=copy(k)
 while (b in Z):
  b+=Y
 print(_tuple(Y,len(tail(b))))

For proper (a=5) output, use:

=ZwFY{Z=bkW'bZ~bY)p++Y"="`ltb

29 characters

PHP, 104 102 96

<?php function _($s){while($n=$s[$i++]){$a[$n]=max($a[$n],$n!=$s[$i-2]?$v=1:++$v);}print_r($a);}

usage

_('asdaaaadddscc');

printed

Array ( [a] => 4 [s] => 1 [d] => 3 [c] => 2 )

Javascript, 116 bytes

y=x=prompt();while(y)r=RegExp(y[0]+'+','g'),alert(y[0]+'='+x.match(r).sort().reverse()[0].length),y=y.replace(r,'')

Sample output:

lollolllollollllollolllooollo
l=4
o=3

acbaabbbaaaaacc
a=5
c=2
b=3

helloworld
h=1
e=1
l=2
o=1
w=1
r=1
d=1 

Javascript, 109 104 100 98 bytes

function c(s){q=l={};s.split('').map(function(k){q[k]=Math.max(n=k==l?n+1:1,q[l=k]|0)});return q}

Example usage:

console.log(c("aaaaaddfffabbbbdb"))

outputs:

{ a: 5, d: 2, f: 3, b: 4 }

Mathematica, 74 72 69

Print[#[[1,1]],"=",Max[Tr/@(#^0)]]&/@Split@Characters@#~GatherBy~Max&

% @ "acbaabbbaaaaacc"

a=5
c=2
b=3

Not very good but strings are not Mathematica's best area. Getting better though. :-)

JavaScript - 141 137 125

I don't like regex :)

function g(a){i=o=[],a=a.split('');for(s=1;i<a.length;){l=a[i++];if(b=l==a[i])s++;if(!b|!i){o[l]=o[l]>s?o[l]:s;s=1}}return o}

Run

console.log(g("acbaabbbaaaaacc"));

outputs

[ c: 2, a: 5, b: 3 ]

J - 52 bytes

Well, a simple approach again.

f=:([,'=',m=:":@<:@#@[`(]m~[,{.@[)@.(+./@E.))"0 1~~.

Explanation:

f=:([,'=',m=:":@<:@#@[`(]m~[,{.@[)@.(+./@E.))"0 1~~.
                                                 ~~. Create a set of the input and apply it as the left argument to the following.
   ([,'=',m=:":@<:@#@[`(]m~[,{.@[)@.(+./@E.))"0 1    The function that does the work
                                             "0 1    Apply every element from the left argument (letters) with the whole right argument (text).
                                  @.(+./@E.)         Check if the left string is in right string.
                       (]m~[,{.@[)                   If yes, add one letter to the left string and recurse.
             ":@<:@#@[                               If not, return (length of the left string - 1), stringified.
    [,'=',                                           Append it to the letter + '='

Example:

   f 'acbaabbbaaaaacc'
a=5
c=2
b=3
   f 'aaaabaa'
a=4
b=1

If free-form output is allowed (as in many other answers), I have a 45 bytes version too. These boxes represent a list of boxes (yes, they're printed like that, although SE's line-height breaks them).

   f=:([;m=:<:@#@[`(]m~[,{.@[)@.(+./@E.))"0 1~~.
   f 'acbaabbbaaaaacc'
┌─┬─┐
│a│5│
├─┼─┤
│c│2│
├─┼─┤
│b│3│
└─┴─┘
   f 'aaaabaabba'
┌─┬─┐
│a│4│
├─┼─┤
│b│2│
└─┴─┘

Python3 - 111, 126, 115 114 111 bytes

Executable code that will read 1 line (only use lowercase letters a-z)

d={}.fromkeys(map(chr,range(97,123)),0)
for c in input():d[c]+=1
[print("%s=%d"%(p,d[p]))for p in d if d[p]>0]

Edit: Excluded unnecessary output on request from @Therare

The output looks nice

~/codegolf $ python3 maxseq.py 
helloworld
l=3
o=2
h=1
e=1
d=1
w=1
r=1

Julia, 85

f(s)=(l=0;n=1;a=Dict();[c==l?n+=1:(n>get(a,l,1)&&(a[l]=n);n=1;l=c) for c in s*" "];a)
julia> f("acbaabbbaaaaacc")
{'a'=>5,'c'=>2,'b'=>3}

Ruby, 72

(a=$*[0]).chars.uniq.map{|b|puts [b,a.scan(/#{b}+/).map(&:size).max]*?=}

This takes input from command line arguments and outputs to stdout.

Ruby, 58

h={}
gets.scan(/(.)\1*/){h[$1]=[h[$1]||0,$&.size].max}
p h

Takes input from STDIN, outputs it to STDOUT in the form {"a"=>5, "c"=>2, "b"=>3}

F# - 106

let f s=
 let m=ref(Map.ofList[for c in 'a'..'z'->c,0])
 String.iter(fun c->m:=(!m).Add(c,(!m).[c]+1))s;m

In FSI, calling

f "acbaabbbaaaaacc"

gives

val it : Map<char,int> ref =
  {contents =
    map
      [('a', 8); ('b', 4); ('c', 3); ('d', 0); ('e', 0); ('f', 0); ('g', 0);
       ('h', 0); ('i', 0); ...];}

However, to print it without the extra information, call it like this:

f "acbaabbbaaaaacc" |> (!) |> Map.filter (fun _ n -> n > 0)

which gives

val it : Map<char,int> = map [('a', 8); ('b', 4); ('c', 3)]