Calculate the Super-Logarithm

Question

This should be a simple challenge.

Given a number \$n \ge 0\$, output the super-logarithm (or the \$\log^*\$, log-star, or iterated logarithm, which are equivalent since \$n\$ is never negative for this challenge.) of \$n\$.

$$\log^*(n) = \begin{cases} 0 & \text{ if } n \le 1 \\ 1 + \log^*(\log(n)) & \text{ if } n > 1 \end{cases}$$

This is one of the two inverse functions to tetration. The other is the super-root, which is in a related question.

Examples

Input       Output
0           0
1           0
2           1
3           2
4           2
...
15          2
16          3
...
3814279     3
3814280     4

Rules

• You do not need to support decimals, though you may.
• You need to support input of at least \$3814280 = \left\lceil e^{e^e} \right\rceil\$.
• You may not hard-code the values like 3814280. (Your program must theoretically support higher numbers.) I want an algorithm to be implemented.
• Shortest code wins.

Related OEIS

Answer 1

Jelly, 8 bytes

ÆlÐĿĊḊi1

Try it online! or verify all test cases.

Background

We start by successively taking natural logarithms of the input and the subsequent results until the result no longer changes. This works because the extension of the natural logarithm to the complex plane has a fixed point; if z = e^-W(-1) ≈ 0.318 + 1.337i – where W denotes the Lambert W function – we have log(z) = z.

For input n, after computing [n, log(n), log(log(n)), …, z], we first apply the ceiling function to each of the results. Jelly's implementation (Ċ) actually computes the imaginary part of complex number instead^†, but we're not interested in these anyway.

Once the k^th application of log yields a value less than or equal to 1, Ċ will return 1 for the first time. The 0-based index of that first 1 is the desired result.

The straightforward implementation (compute 1-based index, decrement) fails because of edge case 0, which does not have a 1 in its list of logarithms. In fact, for input 0, the sequence of logarithms is

[0, None]

This is because Jelly's logarithm (Æl) is overloaded; it first tries math.log (real logarithm), then cmath.log (complex logarithm), and finally "gives up" and returns None. Fortunately, Ċ is similarly overloaded and simply returns it argument if it cannot round up or take an imaginary part.

Likewise, input 1 returns

[1, 0, None]

which may create problems in other approaches that do or do not involve Ċ.

One way to fix this problem is apply Ḋ (dequeue; removes first element) to the array of logarithms. This maps

0ÆlÐĿ -> [0, None]    -> [None]
1ÆlÐĿ -> [1, 0, None] -> [0, None]

so neither list has a 1 now. This way, finding the index of the first 1 will return 0 (not found), which is the desired output for inputs 0 and 1.

How it works

ÆlÐĿĊḊi1  Main link. Argument: n (non-negative integer)

  ÐĿ      Apply the following link until the results are no longer unique.
Æl          Natural logarithm.
          Return the array of all unique results.
    Ċ     Round all resulting real numbers up to the nearest integer. This takes
          the imaginary part of complex numbers and does nothing for non-numbers.
     Ḋ    Dequeue; remove the first item (n) of the array of results.
      i1  Find the first index of 1 (0 if not found).

---

^† _{This is one of the only three atoms in Jelly that are overloaded in a non-obvious manner.}

Answer 2

Jelly, 9 bytes

Æl>1$пL’

Try it online!

Test suite. (Slightly modified.)

Explanation

Æl>1$пL’
     п    while loop, collect all intermediate results.
  >1$      condition: z>1
Æl         body: natural logarithm.
       L   length of the array containing all intermediate results,
           meaning number of iterations
        ’  minus one.

Answer 3

Javascript, 45 27 26 bytes

l=a=>a>1&&1+l(Math.log(a))

Here is test suite (3rd rev)

Thanks @LeakyNun for saving 1 byte with conditional and then converting function to lambda, and @Neil for pointing out false is ok return value for <=1 (changed test to be == instead of ===)

Answer 4

C, 38 bytes

f(double n){return n>1?1+f(log(n)):0;}

Pretty self-explanatory.

Try it on ideone.

Answer 5

Mathematica, 21 bytes

If[#>1,1+#0@Log@#,0]&

Recursive anonymous function. Takes an integer as input and returns its super-logarithm as output. Just uses the given definition.

Answer 6

Dyalog APL, 13 bytes

Direct translation of OP:

{⍵≤1:0⋄1+∇⍟⍵}

TryAPL online!

Answer 7

Pyth, 10 bytes

L&>b1hy.lb

Test suite.

This defines a function.

Answer 8

Haskell, 23 bytes

l x|x>1=1+l(log x)|1<2=0

Usage example: l 3814280 -> 4.

Answer 9

Python 3, 45 bytes

import math
s=lambda x:x>1and-~s(math.log(x))

For x <= 1, this returns False (which is == 0 in Python).

Answer 10

05AB1E, 16 13 bytes

[Dî2‹#¼žr.n]¾

Explanation

              # implicit input n
[          ]  # infinite loop
 Dî2‹#        # break if n rounded up is less than 2
      ¼       # else, increase counter
       žr.n   # set next n = log(n)
            ¾ # push counter and implicitly print

Try it online

Answer 11

J, 21 19 18 16 bytes

Saved 2 bytes to Leaky Nun, 1 byte to Galen Ivanov, and 2 bytes to FrownyFrog!

2#@}.(0>.^.)^:a:

Try it online!

Test cases

ls =: >:@$:@^.`0:@.(<:&1)
   ls 0
0
   ls 1
0
   ls 2
1
   ls 3
2
   ls 4
2
   ls 15
2
   ls 16
3
   ls 3814280
4

Answer 12

MATL, 15 12 bytes

0`ZetG>~}x@q

Try it online! Or verify all test cases (slightly modified version to handle several inputs).

How it works

Starting with 0, apply iterated exponentiation until exceeding the input. The output is the number of iterations minus 1.

0       % Push 0
`       % Do...while loop
  Ze    %   Exponential
  t     %   Duplicate
  G     %   Push input
  >~    %   Is current value less than or equal to the input? If so: next iteration
}       % Finally (code executed at the end of the last iteration)
  x     %   Delete
  @q    %   Iteration index minus 1
        % Implicitly end loop
        % Implicitly display stack

Answer 13

Julia, 17 bytes

!x=x>1&&1+!log(x)

Try it online!

Answer 14

Java 7, 47 bytes

int c(double n){return n>1?1+c(Math.log(n)):0;}

Try it online.

The recursive Java 7 style method above is 2 bytes shorter than an iterative Java 8 style lambda:

n->{int c=0;for(;n>1;c++)n=Math.log(n);return c;}

Try it online.

Explanation:

int c(double n){      // Method with double parameter and integer return-type
  return n>1?         //  If the input is larger than 1:
    1+                //   Return 1 +
      c(Math.log(n))  //   A recursive call with log(input)
   :                  //  Else:
    0;                //   Return 0 instead

n->{                  // Method with double parameter and integer return-type
  int c=0;            //  Create a counter, starting at 0
  for(;n>1;           //  Loop as long as the input is still larger than 1:
    c++)              //   Increase the counter by 1
    n=Math.log(n);    //   And update the input to log(input)
  return c;}          //  After the loop: return the counter as result

Answer 15

Jelly, 8 bytes

-Ælß$Ị?‘

Straightforward implementation of the definition. Try it online! or verify all test cases.

How it works

-Ælß$Ị?‘  Main link. Argument: x

     Ị    Insignificant; test if |x| ≤ 1.
      ?   If the result is 1:
-           Return -1.
          Else:
   $        Execute the monadic chain formed by the two links to the left.
Æl            Apply natural logarithm to x.
  ß           Recursively call the main link.
       ‘  Increment the result.

Answer 16

MATLAB / Octave, 44 bytes

function a=g(n);a=0;if n>1;a=1+g(log(n));end

Tried to do it all as one anonymous function, but I forgot that MATLAB/Octave continues to evaluate expressions even if they are multiplied by a boolean false (zero) value:

f=@(n)(n>1)*(1+f(log(n)))

Answer 17

Mathematica, 29 bytes

Simple as all hell, and works for comically large as well as negative inputs:

f[x_]:=If[x>1,1+f[Log[x]],0]

Answer 18

R, 38 37 bytes

f=function(x)if(x>1)1+f(log(x))else 0

Thanks @user5957401 for the extra byte!

Test cases:

> f(0)
[1] 0
> f(1)
[1] 0
> f(2)
[1] 1
> f(3)
[1] 2
> f(4)
[1] 2
> f(3814279)
[1] 3
> f(3814280)
[1] 4

Answer 19

R, 34 bytes

f=pryr::f(`if`(n>1,1+f(log(n)),0))

Try it online!

A non-recursive approach is possible: 36 bytes and takes input from stdin.

n=scan()
while((n=log(n))>0)F=F+1
+F

Answer 20

Emacs Lisp, 38 bytes

(defun l(n)(if(> n 1)(1+(l(log n)))0))

Testcases:

(mapcar 'l '(0 1 2 3 4 15 16 3814279 3814280))
;; (0 0 1 2 2 2 3 3 4)

Answer 21

Perl 5, 35 bytes

Very simple, requires -M5.016 (which is free) to enable the __SUB__ keyword for anonymous recursion.

sub{$_[0]>1?1+__SUB__->(log pop):0}

Another alternative is

sub{$_[0]>1?1+__SUB__->(log pop):0}

which is 34 bytes, and gives the same output for all inputs > 1, but returns the special false value for inputs <= 1. False is numerically equal to zero, but prints as "" (empty string), so it probably doesn't qualify.

Answer 22

CJam (16 bytes)

rd{_1>}{_ml}w],(

Online demo

Simple while loop with pre-condition. (What I really want here is a Golfscript-style unfold operation, but CJam doesn't have one, and floating point in GolfScript is messy and not at all golfy).

Answer 23

PARI/GP, 24 bytes

Just the straightforward recursion.

f(n)=if(n>1,1+f(log(n)))

Answer 24

Racket, 61 bytes

(λ(x)(letrec([a(λ(b)(if(> b 1)(+ 1 (a(log b)))0))])(a x)))

Answer 25

Maple, 32,30 29 bytes

f:=x->`if`(x>1,1+f(log(x)),0)

Test cases:

> f(0.);
  0
> f(1.);
  0
> f(2.);
  1
> f(3.);
  2
> f(4.);
  2
> f(3814279.);
  3
> f(3814280.);
  4

Answer 26

R, 36 bytes

Slightly different approach from Plannapus

->n;a=0;while(n>1){a=a+1;n=log(n)};a

Uses a right assign to run the code -- so the desired number must precede it. i.e.

10->n;a=0;while(n>1){a=a+1;n=log(n)};a

Answer 27

Whispers v3, 9 bytes

>> ln*(L)

Try it online!

Note that I had to push a bug fix in order to make this work. The ln* function had already been added before September 1st 2020, but I hadn't noticed a bug in the parsing that was fixed today

This is a function, going by the standards defined here for Whispers functions. It can be called by appending the code in the Footer to the file.

Whispers v3 also has the slog, lg* (uses \$\log_2\$) and log* (uses \$\log_{10}\$) functions

Answer 28

Ruby, 29 bytes

l=->n{n<=1?0:1+l[Math.log n]}

Answer 29

Perl 6, 21 bytes

{($_,*.log...1>=*)-1}

Try it online!

The parenthesized expression is a sequence. $_, the argument to the function, is the first element. *.log generates each successive element by taking the log of the previous element. The sequence continues until the ending condition, 1 >= *, is true: 1 is greater than or equal to the current element. Subtracting 1 from the sequence coerces it to a number: its length.