APL (31 19)

Now tests all possible bases.

⊃⍒{+/4=K⊤⍨K⍴⍵}¨⍳K←⎕ 

Explanation:

• ⍳K←⎕: read user input, store in K. Make a list from 1 to K, which are the bases to try.
• {...}¨: for each of these, run the following function
• K⊤⍨K⍴⍵: encode K into that base giving a list of digits (as numbers) per base. Use K digits (a big overestimate, but it doesn't matter because the unused ones will all be zero anyway).
• 4=: see which of these are equal to 4
• +/: sum these, now we know how many fours per base
• ⊃⍒: give the indices of the list if it were sorted downwards, so the index of the biggest one is at the front. Take the first item of this list.

GolfScript, 30 characters

.,{[2+.2$\base{4=},,\]}%$)~p]; 

Works for any base - test the code online.

Comment: This solution was based on the original version of the question. It thus may return a base larger than the input, e.g. for the input 4 it correctly returns base 5 - which is no longer valid by the new rules.

Python 2.x, 77 chars

F=lambda x:max((sum(x/b**d%b==4for d in range(99)),b)for b in range(5,99))[1] 

Works up to base 98 and numbers at most 98 digits long.

J, 38 characters

f=.[:(i.>./)[:+/[:|:4=(10#"0(i.37))#:] 

Usage:

 p 624 5 p 444 10 p 68 16 

VBA, 121

Function k(a) For w=5 To a Z=0:q=a:Do:c=q Mod w:Z=Z-(c=4):q=Int(q/w):Loop Until q=0 If Z>x Then x=Z:k=w Next End Function 

usage:

• direct window: ?k(num)
• Excel formula: =k(A1)

GolfScript (23 chars)

~:^,2>{^\base[4]/,~}$0= 

or

~:^,2>{^\base[4]/,}$-1= 

or

~:^,2>{^\base[4]/,}$)\; 

Note that this takes input from stdin: for a fair comparison with Howard's GolfScript version subtract one char.

Howard points out that the rules have changed, and it's not very logical that they now exclude 4 as a possible input when it has a valid output (any integer greater than 4). To cover that case as well requires an extra 2 characters, which can be added in all kinds of ways:

~:^)),2>{^\base[4]/,}$)\; 

or

~:^,{))^\base[4]/,}$)))\; 

being a couple of the obvious ones.

Jelly, 6 bytes

bⱮċ€4M 

Try it online!

Outputs "all" bases up to N which gives the most 4's. If you want maximal or minimal base, add Ṁ (max) or Ṃ (min) respectively.

How it works

bⱮċ€4M Main link (monad). Input: integer N. bⱮ Convert N to each base of 1..N ċ€4 Count 4's in each representation M Take maximal indices 

Mathematica 59

Code

Sort[{Count[IntegerDigits[n, #], 4], #} & /@ Range[5, 36]][[-1, 2]] 

Let's give the above function a name.

whichBase[n_] := Sort[{Count[IntegerDigits[n, #], 4], #} & /@ Range[2, 36]][[-1, 2]] 

Explanation

1. Count[IntegerDigits[n,k],4]: Count the number of fours in the base k representation of n.
2. Sort the bases from fewest to most 4s.
3. Return the base from the last item in the list, that is, the base that had the representation with the most 4's.

Some special numbers

Now let's apply whichBase to the following special numbers.

numbers= {1953124, 8062156, 26902404, 76695844, 193710244, 444444444, 943179076, 1876283764, 3534833124, 6357245164, 10983816964, 18325193796, 29646969124, 46672774204, 71708377284, 107789473684, 158856009316, 229956041484, 327482302084, 459444789604, 635782877604, 868720588636, 1173168843844, 1567178659764, 2072449425124, 2714896551724, 3525282954756, 4539918979204, 5801435550244, 7359635486844, 9272428079044, 11606852190676} 

{5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, \ 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36}

If you convert each number to the corresponding base, you will see what is special about them.

Japt -h, 10 bytes

444 in base 10 is [4,4,4] which contains the number and digit 4 3 times but 444 in base 100 is [4,44] which also contains the digit 4 3 times, but only as a number once. Given the expected output in the challange for the 444 test case, I'd guess we're supposed to be counting the number 4:

õ ñ@ìX è¥4 

Try it

But if we are counting the digit 4 then:

õ ñ@ìX ¬è4 

Try it

õ :Range [1,...,input] ñ@ :Sort by passing each X through a function ìX : Convert the input to a base X digit array :(VERSION 1) è : Count the elements ¥4 : Equal to 4 :(VERSION 2) ¬ : Join to a string è4 : Count the occurrences of "4" :Implicitly output the last element in the sorted array 

C - (114 characters)

In all it's golfy glory:

x,k,c,d,n;main(v){scanf("%d",&v);for(k=5;v/k;++k){x=v;c=0;while(x)c+=x%k==4,x/=k;c>=d?n=k,d=c:0;}printf("%d",n);} 

And somewhat ungolfed:

x,k,c,d,n; // declare a bunch of ints, initialized to 0 main(v){ // declare one more, without using an extra comma scanf("%d",&v); // get the input (v) for(k=5;v/k;++k){ // loop over each base (k) greater than or equal to (/) // our input (v) x=v; // temp value (x) set to input (v) c=0; // number of 4s in the current base (c) re-initialized while(x) // loop over our temp until it's used up c+=x%k==4, // if the next digit (x%k) is 4 (==4) increment the // current count (c+=) x/=k; // remove the current digit c>=d?n=k,d=c:0; // if the number of 4s in this base (c) is greater // than the current maximum number of 4s (d), then // save the new best base (n), and new maximum // number of 4s } printf("%d",n); // output the result } 

Just for fun here's the output for the numbers [0,127] (these are the largest bases under the input number itself).

0, 0, 0, 0, 0, 5, 6, 7, 8, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 5, 21, 22, 23, 6, 25, 26, 27, 7, 29, 30, 31, 8, 33, 34, 35, 9, 37, 38, 39, 10, 41, 42, 43, 11, 5, 46, 47, 12, 49, 50, 51, 13, 53, 54, 55, 14, 57, 58, 59, 15, 61, 62, 63, 16, 65, 66, 67, 17, 69, 5, 71, 18, 73, 74, 75, 19, 7, 78, 79, 20, 81, 82, 83, 21, 85, 86, 87, 22, 89, 90, 91, 23, 93, 94, 5, 24, 97, 98, 99, 25, 101, 102, 103, 26, 5, 106, 107, 27, 109, 5, 111, 28, 113, 114, 5, 29, 9, 5, 5, 5, 121, 122, 123

R - 148 137 chars

(so, far away from the rest of the competition but still)

f=function(n){s=sapply;which.max(s(lapply(strsplit(s(4:n,function(x){q=n;r="";while(q){r=paste(q%%x,r);q=q%/%x};r})," "),`==`,4),sum))+3} 

Basically transform the input from base 10 to all bases from 4 to n (using modulo %% and integer division %/%) and pick the index of the first one having the most 4s.

f(624) [1] 5 f(444) [1] 10 

J translation of @marinus' APL solution:

NB. Expression form (22 characters, not including "n" - the "argument"): {.\:(+/@(4=$#:[)"0 i.)n NB. Function form (24 characters, not including "f=:"): f=:{.@\:@(+/@(4=$#:[)"0 i.) 

Just for interest, here are some values:

(,.f"0)9+i.24 9 5 10 6 11 7 12 8 13 9 14 5 15 11 16 6 17 13 18 7 19 5 20 5 21 5 22 5 23 5 24 5 25 6 26 6 27 6 28 6 29 5 30 7 31 7 32 7 

It outputs the smallest base that gives a fouriest transform. For the last few values in the table, the representations look like “4n” (e.g 31 in base 7 is “43”).

05AB1E, 10 9 bytes

LBε4¢}Zk> 

-1 byte thanks to @Cowabunghole.

If multiple bases have the same amount of 4s, it will output the smallest one (i.e. 16 will result in 6, but 12 would also have been a possible output).

Try it online or verify all test cases.

Explanation:

L # Create a list in the range [1, (implicit) input] # i.e. 624 → [1,2,3,...,622,623,634] B # Convert the (implicit) input integer to Base-N for each N in this list # i.e. 624 and [1,2,3,...,622,623,624] # → ["1","1001110000","212010",...,"12","11","10"] ε } # Map this list to: 4¢ # Count the number of 4s in the number # → [0,0,0,0,4,0,0,0,0,1,0,2,...,0,0,0] Z # Take the max (without popping the list) # i.e. [0,0,0,0,4,0,0,0,0,1,0,2,...,0,0,0] → 4 k # Get the index of this max in the list # i.e. [0,0,0,0,4,0,0,0,0,1,0,2,...,0,0,0] and 4 → 4 > # Increase it by to convert the 0-index to our base (and output implicitly) # i.e. 4 → 5 

k, 18 bytes

{*>+/'4=x{y\x}'!x} 

Try it online!

{ } /function(x) x '!x /for every y in 0, 1, ..., (x-1): {y\x} / do x in base y 4= /see which digits equal four +/' /sum, to get number of 4s per base *> /get index (which equals base) of largest number of 4s 

C# with Linq 273

using System;using System.Linq;class P{static void Main(){int r,z,k=int.Parse(Console.ReadLine());if(k<=4) return;Console.WriteLine(Enumerable.Range(4, k).Select(x =>{r = 0;z = k;while (z > 0){if(z % x==4){r++;}z/=x;}return new[]{r, x};}).OrderBy(n => n[0]).Last()[1]);}} 

or

using System; using System.Linq; class P { static void Main() { int r, z, k = int.Parse(Console.ReadLine()); if (k <= 4) return; Console.WriteLine( Enumerable.Range(4, k).Select(x => { r = 0; z = k; while (z > 0) { if (z % x == 4) { r++; } z /= x; } return new[] { r, x }; }).OrderBy(n => n[0]).Last()[1]); } } 

Pretty sure the number of variables can be reduced and the if's can be converted to ?s. Oh well...

C# (482 ~423 Bytes)

First attempt at a 'golfed' solution. I used basically the same algorithm as the VBA above. I could probably save some bytes inlining the conversion function, or shortening the name. Like I said this is a first attempt, so please be gentle.

With whitespace:

using System; class Program { static void Main(string[] args) { int n = int.Parse(args[0]); int c=0, m=0; string r=""; int t = 0; for (int i = 5; i < 37; i++) { while (n > 0) { r = (char)((int)(n % i) + 48 + (7 * ((int)(n % i) > 9 ? 1 : 0))) + r; n = (int)(n / i); } t = r.Length - r.Replace("4", "").Length; if (t > c) { c = t; m = i; } } Console.WriteLine("Base: " + m); } } 

Burlesque - 28 bytes

Jbcjro{dg}Z]J{4CN}Cmsb[~Fi?i Jbcjro create a list 1..input and convert input to an infinite list. {dg}Z] convert number to base (zipWith operation) J duplicate {4CN}Cm create comparison function 4CN count the number of fours. sb sort by [~ take the last element (which is going to be the base-n representation where count of fours is highest) Fi Find the index of this last element in the original unsorted list ?i increment (because base 1 is index 0) 

Try it online.

Perl 6, 44 bytes

{max 5..$^a,by=>{grep 4,$a.polymod($_ xx*)}} 

Try it online!

Good old polymod.

Husk, 9 bytes

S€►#4ṠMBḣ 

Try it online!

 € The 1-based index in MB the list of the input's representations in every base Ṡ ḣ from 1 to itself S ► of its element which has the largest # number of 4 fours. 

Julia, 50 bytes

n->sort(2:n+1,by=b->-sum(digits(n,base=b).==4))[1] 

Try it online!