Can this number be written in (3^x) - 1 format?

Question

Challenge:

Create a program that accepts a positive integer and checks if it can be written in the form of (3^x)-1, where X is another positive integer.

If it can, output X

If it can't, output -1 or a falsy statement.

Example inputs/outputs

Input:

It can be written as (3^1) - 1, so we output x which is 1

Output:

Input:

26 can be written as (3^3) - 1, so we output x (3)

Output:

Input:

1024 can't be written in the form of (3^x) - 1, so we output -1

Output:

-1

This is code-golf so least amount of bytes wins

Related OEIS: A024023

I ask to output X because I believe it's more challenging that way. Simply finding if it is of format 3^x - 1 would be too easy for a challenge, in my opinion. — Offtkp, Commented Jan 6, 2017 at 14:56
Unless if it's a falsy statement in your programming language, then no. — Offtkp, Commented Jan 6, 2017 at 16:57
having to handle non-negative intergers would make 0 3^0-1 a valid output and thus not useable as false, — Jasen, Commented Jan 7, 2017 at 7:40
anyone thinking of using log() in their answer should confirm it giives the correct answer 5 when 242 is input. — Jasen, Commented Jan 7, 2017 at 9:57

Fatalize · Accepted Answer · 2023-04-05 11:07:32Z

2

Brachylog, 6 bytes

;3^₍-₁

Try it online!

Takes input through the Output variable.

Explanation

;3         [?, 3]
  ^₍       3^?
    -₁     3^? - 1 = Input

This will unify ? with the right value if possible, and output false. otherwise.

answered Apr 5, 2023 at 11:07

Fatalize

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Python, 64 bytes

Outputs False if the number cannot be written in that format.

def f(n):L=[3**x-1for x in range(n)];print n in L and L.index(n)

This also works in 64 bytes, and prints empty string as a falsy output:

def f(n):
 try:print[3**x-1for x in range(n)].index(n)
 except:0

A creative solution for 65 bytes, outputting 0 for falsy:

lambda n:-~",".join(`3**x-1`for x in range(n+1)).find(',%s,'%n)/2

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jan 6, 2017 at 15:13

mbomb007

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\$\begingroup\$ Does not output x nor -1. \$\endgroup\$
– dfernan
Commented Jan 6, 2017 at 15:17
\$\begingroup\$ The program should output x instead of n in case of a match. \$\endgroup\$
– dfernan
Commented Jan 6, 2017 at 15:20
\$\begingroup\$ No, it should output the positive integer that when replaced with X, you get the input. The question refers to X as a variable, not as a string \$\endgroup\$
– Offtkp
Commented Jan 6, 2017 at 15:22
\$\begingroup\$ @P.Ktinos Fixed it. \$\endgroup\$
– mbomb007
Commented Jan 6, 2017 at 15:31

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Pyth, 10 bytes

*J@hQ3!%J1

Try it here!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jan 6, 2017 at 15:56

Blue

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Glen O · Accepted Answer · 2017-01-06 16:00:08Z

1

Julia, 30 bytes

n->findfirst(n.==3.^(0:n)-1)-1

It's a simple function - it creates a vector that has a true only in the corresponding position in 3^a-1, where a is a vector containing integers between 0 and n. It finds the "first" position that is true and subtracts 1 (if it's all false, the find evaluates to zero, and it returns -1).

As 0:n has 0 in the first spot, the subtract 1 corrects for indexing and also enables the -1 false response.

answered Jan 6, 2017 at 16:00

Glen O

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Pyke, 9 6 bytes

3m^Qh@

Try it here!

3m^    -  map(3**i, range(input))
     @ - V in ^
   Qh  -  input + 1

Old 9 byte version:

b3'l}\2q*

Try it here!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jan 6, 2017 at 15:59

Blue

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Pyth 8 bytes

xm^3dUQh

     UQ  # generate all values 1..Q (Q is the input)
 m^3d    # map 3^d over this ^ list
x      h # find the input+1 (hQ) in the result of the last command

Try here

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jan 6, 2017 at 16:16

Rod

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

R, 34 25 bytes

a=log(scan()+1,3);a*!a%%1

Calculate the base 3 logarithm of the input + 1. Test if the result is an integer, if it is it outputs it, if not it outputs 0 as falsey value. Thanks to @Billywob for the extra 9 bytes off!

Test cases:

> a=log(scan()+1,3);a*!a%%1
1: 1024
2: 
Read 1 item
[1] 0

> a=log(scan()+1,3);a*!a%%1
1: 26
2: 
Read 1 item
[1] 3

Old version at 34 bytes which outputs -1 as falsey value.

a=log(scan()+1,3);`if`(!a%%1,a,-1)

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jan 6, 2017 at 15:51

plannapus

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\$\begingroup\$ If you do a*!a%%1 it will output a if true and 0 otherwise and you can skip the if thing. \$\endgroup\$
– Billywob
Commented Jan 6, 2017 at 16:07
\$\begingroup\$ The spec says "If it can't, output -1 or a falsy statement." and 0 is interpreted as FALSE in R so I would say it's valid. \$\endgroup\$
– Billywob
Commented Jan 6, 2017 at 16:19

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Steadybox · Accepted Answer · 2017-01-06 18:06:26Z

1

C, 81 bytes

i,j,k;f(n){for(i=0;++i<n;){for(k=3,j=0;++j<i;k*=3);if(n==k-1)return i;}return-1;}

answered Jan 6, 2017 at 18:06

Steadybox

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\$\begingroup\$ I think you save bytes by using pow(3,i) instead of defining your own. Gcc complains about the missing #include <math.h> but compiles it anyway. I did have to cast to int. You also may be able to gain some by adding an r variable, initializing to -1, and then if(n==k-1)r=i;}return r;} \$\endgroup\$
– nmjcman101
Commented Jan 6, 2017 at 18:54
2

\$\begingroup\$ @nmjcman101 Thanks, but pow() produces some incorrect results because of floating point inaccuracy. (When cast to int, 2.9999 will be 2, not 3). Adding a variable r sounds like a good idea, but it actually results in a 2 bytes longer code. \$\endgroup\$
– Steadybox
Commented Jan 6, 2017 at 22:33
\$\begingroup\$ yeah, log doesn't work, pow probably won't either. \$\endgroup\$
– Jasen
Commented Jan 7, 2017 at 9:28

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ETHproductions · Accepted Answer · 2017-01-07 01:36:55Z

1

Japt, 8 bytes

o m!³a°U

Try it online!

This code expands into the following:

Uo m!p3 a++U

Uo            // Create the range [0...U).
   m!p3       // Map each item X to 3**X.
        a++U  // Take the index of U+1. Returns -1 if it doesn't exist.
              // Implicit: output result of last expression

answered Jan 7, 2017 at 1:36

ETHproductions

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\$\begingroup\$ Very nice solution. \$\endgroup\$
– Oliver
Commented Jan 7, 2017 at 6:04

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Ilmari Karonen · Accepted Answer · 2017-01-08 05:26:10Z

GolfScript, 11 bytes

~..3\?\)%!*

Try it online!

Uses the fact that 3ⁿ is divisible by n+1 if and only if n+1 itself is a power of 3. Outputs its input n if n+1 is a power of 3, otherwise outputs 0 (which is falsy in GolfScript).

De-golfed:

~             # eval the input, converting it from string to integer
 ..           # make two copies of the input number
   3\?        # raise 3 to the power of the input number
      \)%     # reduce the result modulo the input number plus one
         !    # boolean negate the result, mapping 0 to 1 and all other values to 0
          *   # multiply the input number with the result

Ps. Here's a simple test harness that runs the code above (minus the initial ~, which is not needed since the inputs are already numbers) on all integers from 0 to 9999 and prints those for which it returns a truthy result:

10000,{ ..3\?\)%!* },`

The output of this program should be:

[2 8 26 80 242 728 2186 6560]

(The output doesn't include 0 because, even though the formula used does correctly detect it as one less than a power of 3, the result is still 0 × 1 = 0, and thus falsy. Fortunately, 0 is not a positive integer, and thus isn't a valid input for this challenge anyway.)

Community · Accepted Answer · 2020-06-17 09:04:33Z

Octave, 23 bytes

@(x)find(3.^(1:x)-1==x)

Verify all test cases!

Explanation:

This is an anonymous function that takes a positive integer x as input. .^ is element-wise power in Octave, so 3.^(1:x) is 3^1, 3^2, 3^3 .... Subtracting 1 gives 3^1-1, 3^2-1, 3^3-1 ... which can be compared to x.

find(a,b) takes a vector a as input, and attempts to find the scalar b in that vector and returns its index. If it's not found then it will output an empty matrix []. An empty matrix is a falsey value in Octave.

find(3.^(1:x)-1==x) searches for x in the vector 3^1-1, 3^2-1, 3^3-1 ... and attempts to return its index. If it's not in the vector then it returns an empty (falsey) matrix.

Etaoin Shrdlu · Accepted Answer · 2017-01-09 16:30:28Z

1

C, 76 bytes

main(i,a,c){scanf("%d",&a);for(c=0,++a;i<a;i*=3,++c);printf("%d",(i==a)*c);}

answered Jan 9, 2017 at 16:30

Etaoin Shrdlu

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Federico Poloni · Accepted Answer · 2017-01-10 15:06:32Z

1

Sagemath, 45 bytes

This is simply @dfernan's solution repackaged as Sagemath (which is basically Python + some math libraries loaded by default and syntactic sugar).

In Sagemath, we can avoid the import math and we can use ^ for exponentiation, so we save a few chars.

def f(n):x=ceil(log(n,3));print((3^x-1==n)*x)

Test it online

edited Jan 10, 2017 at 15:06

answered Jan 10, 2017 at 13:35

Federico Poloni

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mreff555 · Accepted Answer · 2017-04-29 16:52:24Z

1

c++ 60 bytes

int f(n){float o=log1p(n)/log(3);return o/floor(o)!=1?-1:o;}

explanation:

int f(n){             
  float o=log1p(n)/log(3);       // eval for x using log3 function 
  return o/floor(o)!=1?-1:o;     // if no remainder output X 
}

edited Apr 29, 2017 at 16:52

answered Jan 8, 2017 at 15:44

mreff555

1114 bronze badges

\$\begingroup\$ I think you don't have to use uint16_t. You can use normal int. \$\endgroup\$
– Linnea Gräf
Commented Jan 8, 2017 at 19:55
\$\begingroup\$ I guess I could to save a few bytes but it seemed to me that if I wasn't safeguarding against negative integer input then I wasn't following the guidelines. \$\endgroup\$
– mreff555
Commented Jan 14, 2017 at 21:04
\$\begingroup\$ this compile to me: int f(n){float o=log1p(n)/log(3);return o/floor(o)!=1?-1:o;} \$\endgroup\$
– user58988
Commented Apr 29, 2017 at 15:53
1

\$\begingroup\$ @mreff555 You don't need to safeguard against negative input, you can assume positive input. \$\endgroup\$
– Erik the Outgolfer
Commented Apr 29, 2017 at 16:05

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xnor · Accepted Answer · 2017-12-25 05:52:33Z

1

Python 2, 34 bytes

lambda n:(~n>>3**n%-~n*n)**4/80%80

Try it online!

Works for all Python ints, up to at least 2^100.

edited Dec 25, 2017 at 5:52

answered Dec 25, 2017 at 5:14

xnor

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DLosc · Accepted Answer · 2017-12-25 08:29:25Z

1

Pip, 13 bytes

aTB:3MNa=2&#a

Try it online!

Explanation

               a is 1st command-line argument (implicit)
aTB:3          Convert a to base 3 and assign back to a
     MNa=2     Does the min of a's digits equal 2?
          &    Logical-and
           #a  Length of a
               If there are non-2 digits, we get the falsey value 0; otherwise, we get
               the number of digits

answered Dec 25, 2017 at 8:29

DLosc

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Mr. Xcoder · Accepted Answer · 2017-12-25 08:54:12Z

1

05AB1E, 6 bytes

>3.n.ï

Try it online!

> increments, 3 pushes a 3 to the stack, .n find the logarithm with base 3, .ï checks if it is equal to its integer part.

Returns 0 for falsy: If it can't, output -1 or a falsy statement.

answered Dec 25, 2017 at 8:54

Mr. Xcoder

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Uriel · Accepted Answer · 2017-12-25 12:49:18Z

1

APL (Dyalog), 11 bytes

⊢|⊢⍳⍨¯1+3*⍳

Try it online!

Uses ⎕IO←0.

How?

⍳ - range of 0 to n-1.

3* - raise 3 to the power of each element.

¯1+ - decrement each by 1.

⊢⍳⍨ - search the index of n in that list (if not exists, this would return the maximum index plus 1 - which is n.

⊢| - modulo by n. This would keep the index, if found, and zero-out numbers not contained in the list that would produce n % n = 0.

APL (Dyalog), 14 bytes

(∧/×≢)2=3⊥⍣¯1⊢

Try it online!

edited Dec 25, 2017 at 12:49

answered Dec 25, 2017 at 12:33

Uriel

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\$\begingroup\$ "26 can be written as (3^3) - 1" 124 can be written as (5^3)-1 but your code for 124 not print 5 print 0 \$\endgroup\$
– user58988
Commented Feb 2, 2018 at 15:45
\$\begingroup\$ Ok I confuse exponent and base \$\endgroup\$
– user58988
Commented Feb 2, 2018 at 15:52
\$\begingroup\$ @RosLuP you mind deleting the comments? people use to DV without much thinking when seeing these \$\endgroup\$
– Uriel
Commented Feb 3, 2018 at 16:17

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

APL NARS, 14 chars or 28 bytes

{r×r=⌊r←3⍟1+⍵}

Test:

  f←{r×r=⌊r←3⍟1+⍵}
  f 2     
1
  f 26
3
  f 1024
0

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 25, 2017 at 11:56

user58988

\$\begingroup\$ This is 28 bytes in NARS, but exactly the same solution is only 14 bytes in Dyalog APL. Also, you can save two bytes by conversion to tradfn, r×r=⌊r←3⍟1+⎕, letting the program prompt for input. \$\endgroup\$
– Adám
Commented Dec 25, 2017 at 16:15
\$\begingroup\$ @Adám i prefer functions \$\endgroup\$
– user58988
Commented Dec 25, 2017 at 19:47

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Befunge-93, 26 bytes

&1+>:3%v
1+\^v-1_3/\
.@.$_

Try It Online

Prints 0 as the falsey.

How it Works

&1+... Gets the input and adds one
......
......

...>:3%v    Check if the number is divisible by 3
1+\^..._3/\ If not, divide the number by 3 and increment a counter
...         Repeat until the number is not divisible by 3

.........   If the final number is a one, print the counter
....v-1_... Else pop the counter and print a 0
.@.$_       End the program

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 27, 2017 at 9:52

Jo King♦

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Bubbler · Accepted Answer · 2020-11-25 01:44:56Z

Factor, 30 bytes

[ 1 + 3 over [0,b) n^v index ]

Try it online!

Port of Dennis' Jelly answer. The index word conveniently returns f (the only falsy value in Factor) when the item is not found.

[              ! anonymous lambda
  1 + 3 over   ! ( n+1 3 n+1 )
  [0,b)        ! ( n+1 3 {0..n} )
  n^v          ! ( n+1 {3^0..3^n} )
  index        ! 0-based index of n+1 in {3^0..3^n}; false if not found
]

Factor, 45 bytes

[ 3 >base dup [ 50 = ] all? swap length and ]

Try it online!

A base-conversion approach.

[                    ! anonymous lambda, accepting a positive integer n
  3 >base            ! ( str ) convert to base-3 string
  dup [ 50 = ] all?  ! ( str ? ) test if it is all 2's
  swap length and    ! ( len/f ) if true, return the length of str
                     ! otherwise return false
]

Shaggy · Accepted Answer · 2022-07-23 22:04:00Z

1

Japt `-æ`, 6 bytes

Outputs undefined, which is falsey, instead of -1.

NøÓ3pU

Try it

answered Jul 23, 2022 at 22:04

Shaggy

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emanresu A · Accepted Answer · 2022-07-23 23:35:02Z

1

Vyxal, 7 bytes

›3•D⌊=*

Try it Online!

-1 thanks to Steffan.

A different approach that's annoyingly longer.

›       # Increment
 3•     # Log 3
   D⌊=  # Check if it's an integer
      * # Multiply by that

edited Jul 23, 2022 at 23:35

answered Jul 23, 2022 at 23:18

emanresu A

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\$\begingroup\$ @Steffan I guess, although it's kinda ambiiguous as 0 is an allowed output. \$\endgroup\$
– emanresu A
Commented Jul 23, 2022 at 23:33

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DLosc · Accepted Answer · 2022-07-23 23:48:22Z

tinylisp, 68 bytes

(d F(q((N A)(i(l N A)()(i(e A N)0(a 1(F N(a(a A A)A
(q((N)(F(a N 1)1

The solution is the anonymous function on the second line. Try it online!

Explanation

The function F does most of the work. It takes a target number N and an accumulator A. If N is a power of three, it returns the exponent; otherwise, it returns nil, which is a falsey value.

(d F               ; Define F
 (q                ; to be a function
  ((N A)           ; that takes arguments N and A:
   (i(l N A)       ; If N is less than A,
    ()             ; return nil
    (i(e A N)      ; Else, if A is equal to N,
     0             ; return 0
     (a 1          ; Else, add 1 to the result of
      (F N         ; a recursive call with the same N
       (a(a A A)A  ; and three times the A (A+A+A)

If N equals three to the X, the 0 base case will be reached at the Xth level of recursion, meaning that 1 will be added to it X times for a result of X. If the () base case is reached instead, 1 will be added to it some number of times, but adding a number to nil gives nil (and an error message).

The submission function is then just a wrapper around F:

(q          ; A function
 ((N)       ; that takes argument N:
  (F        ; Call F with arguments
   (a N 1)  ; N+1
   1        ; and initial accumulator 1

The Thonnu · Accepted Answer · 2023-03-02 16:24:20Z

1

Thunno `D`, \$ 7 \log_{256}(96) \approx \$ 5.76 bytes

R3@1-Ah

Attempt This Online!

Explanation

R3@1-Ah  # D flag duplicates implicit input
R        # Push range(0, input)
 3@      # 3 ** each
   1-    # Decrement
     Ah  # Index of input in this list
         # (0-indexed, -1 if not found)

answered Mar 2, 2023 at 16:24

The Thonnu

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emirps · Accepted Answer · 2023-04-04 14:43:22Z

1

Stax, 6 bytes

ö♥ô·ô₧

Run and debug it

This unpacks to the following:

Stax, 7 bytes

f3s#vx=

Run and debug it

f       # filter over 1..input
 3s#    # 3^n
    v   # -1
     x= # is equal to the input

Outputs nothing if falsy

answered Apr 4, 2023 at 14:43

emirps

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alephalpha · Accepted Answer · 2023-04-11 07:32:26Z

1

Nekomata, 5 bytes

3D←P∑

Attempt This Online!

3D      Convert to base 3
  ←     Decrement every digit
   P    Check if all digits are positive
    ∑   Sum

answered Apr 11, 2023 at 7:32

alephalpha

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Neil · Accepted Answer · 2017-01-06 19:59:14Z

0

JavaScript (ES6), 40 bytes

f=(n,p=0,k=1)=>n<k?n>k-2&&p:f(n,p+1,k*3)

Returns false or the power. A simple port of @Arnauld's ES7 answer would have taken 43 bytes.

answered Jan 6, 2017 at 19:59

Neil

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\$\begingroup\$ I think f=(n,p,k=1)=>n<k?n>k-2&&p:f(n,-~p,k*3) works and saves 2 bytes. \$\endgroup\$
– Arnauld
Commented Jan 7, 2017 at 0:24

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

PHP, 36 47 bytes

If log(input+1,3) differs from its integer value, print 0; else print the logarithm:
<?=(0|$x=log($argv[1]+1,3))-$x?0:$x; (36 bytes) fails for 242.
<?=strstr($x=log($argv[1]+1,3),".")?0:$x; and <?=(0|$x=log($argv[1]+1,3))-$x>1e-7?0:$x; (41 bytes) may fail for larger $x.

This version is safe:

for(;3**++$x<$n=1+$argv[1];);echo$n<3**$x?0:$x;

1.Loop $x up from 1 while 3^$x is smaller than argument+1.
2.Print 0 if the expression is larger than input+1, $x else.

Takes input from command line argument. Run with -nr.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jan 7, 2017 at 3:32

Titus

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1

\$\begingroup\$ gives wrong answer for 242 input \$\endgroup\$
– Jasen
Commented Jan 7, 2017 at 9:22
\$\begingroup\$ @Jasen: The downvote was ridiculous, but it´s fixed now. \$\endgroup\$
– Titus
Commented Jan 7, 2017 at 11:36

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hyperneutrino · Accepted Answer · 2017-01-08 03:19:55Z

0

Java 7, 180 bytes

class A{public static void main(String[]q)throws Exception{int a,b=0;while((a=System.in.read()-48)>=0)b=b*10+(a);double k=Math.log(b+1)/Math.log(3);System.out.print(k%1==0?k:-1);}}

Really simple approach. Input, then add one, then log₃ the number, and if it's a integer, print it; otherwise, print -1. Could use some work.

Only works up to (3^19)-1.

edited Jan 8, 2017 at 3:19

answered Jan 7, 2017 at 3:17

hyperneutrino♦

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\$\begingroup\$ Why is this non-competing? The Non-Competing status is only reserved for languages or language features that were added after the challenge was posted. \$\endgroup\$
– user41805
Commented Jan 7, 2017 at 11:10
\$\begingroup\$ As I said, size doesn't matter, so this can compete. \$\endgroup\$
– Offtkp
Commented Jan 7, 2017 at 20:55
\$\begingroup\$ @KritixiLithos Oh okay. I wasn't aware of the exact meaning of Non-Competing. Thanks. Also, I'll update the title. \$\endgroup\$
– hyperneutrino ♦
Commented Jan 8, 2017 at 3:19
1

\$\begingroup\$ You can use interface A{...} and drop the public from main(). Why do you have (a) instead of a in the while loop. I think you can use float k=... instead of double k=.... Should be -5 bytes if I counted right. \$\endgroup\$
– Linnea Gräf
Commented Jan 8, 2017 at 20:04
\$\begingroup\$ @RomanGräf I appreciate your suggestions; however, none of them are of any use for me, unfortunately. Your first suggestion only works in Java 8, and there is already a far better Java 8 solution out there. I have (a) in the while loop because I am doing a comparison of an assignment statement, and assignment has the lowest priority on the order of operations and thus requires a set of brackets around it. Finally, I have double because Math#log returns a double and casting would obviously be much slower. Regardless, thank you for the suggestions, but I will not be incorporating them. \$\endgroup\$
– hyperneutrino ♦
Commented Jan 9, 2017 at 3:20

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Can this number be written in (3^x) - 1 format?

73 Answers 73

Brachylog, 6 bytes

Explanation

Python, 64 bytes

Pyth, 10 bytes

Julia, 30 bytes

Pyke, 9 6 bytes

Pyth 8 bytes

R, 34 25 bytes

C, 81 bytes

Japt, 8 bytes

GolfScript, 11 bytes

Octave, 23 bytes

C, 76 bytes

Sagemath, 45 bytes

Python 2, 34 bytes

Pip, 13 bytes

Explanation

05AB1E, 6 bytes

APL (Dyalog), 11 bytes

APL (Dyalog), 14 bytes

APL NARS, 14 chars or 28 bytes

Befunge-93, 26 bytes

How it Works

Factor, 30 bytes

Factor, 45 bytes

Japt -æ, 6 bytes

Vyxal, 7 bytes

tinylisp, 68 bytes

Explanation

Thunno D, \$ 7 \log_{256}(96) \approx \$ 5.76 bytes

Explanation

Stax, 6 bytes

Stax, 7 bytes

Nekomata, 5 bytes

JavaScript (ES6), 40 bytes

PHP, 36 47 bytes

Java 7, 180 bytes

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Japt `-æ`, 6 bytes

Thunno `D`, \$ 7 \log_{256}(96) \approx \$ 5.76 bytes

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