Japt, 11 10 bytes

>0©ÒßUÉ/°T

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>0©ÒßUÉ/°T     :Implicit input of integer U
>0             :Greater than 0?
  ©            :Logical AND with
   Ò           :Negate the bitwise NOT of (i.e., increment)
    ß          :Recursive call with argument
     UÉ/       :  U minus 1 divided by ...
        °T     :  T (initially 0) prefix incremented

posted over 2 years ago

CC BY-SA 4.0

2y ago

Shaggy‭

1951 reputation 4 104 198 153

Python 3, 35 bytes

f=lambda x,i=1:x>0and-~f(x-1/i,i+1)

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Subtracts each 1/i in turn from the initial value until the result is no longer positive.

posted over 2 years ago

CC BY-SA 4.0

2y ago

xnor‭

261 reputation 0 7 26 16

AppleScript, 178 bytes

set n to text returned of (display dialog "" default answer "") as number
set k to 0
set h to 0
repeat
set k to k + 1
set h to h + (1 / k)
if h >= n then exit repeat
end repeat
k

It's really a shame that AppleScript doesn't support explicit stdin.

posted over 2 years ago

CC BY-SA 4.0

DonielF‭

101 reputation 0 3 10 0

Raku, 27 bytes

{+([\+](1 X/1..*)...*>=$_)}

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{                         }  # Anonymous code block
   [\+](        )            # Get the partial sum of
        1 X/                 # 1 over each of
            1..*             # All positive integers
                 ...*>=$_    # Take from the list until it is bigger than the input
 +(                      )   # And return the length of the list

This will start to suffer precision issues, but you can add .FatRat to work correctly (but much slower)

posted over 2 years ago

CC BY-SA 4.0

2y ago

Jo King‭

251 reputation 0 6 25 3

Scala, 55 bytes

(Stream from 1 map 1.0./scanLeft.0)(_+_)indexWhere _.<=

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Written in a more sane manner:

n => Stream.from(1).map(1.0/_).scanLeft(0.0)(_+_).indexWhere(_ >= n)

Explanation:

(
  Stream from 1       //Make an infinite list of integers, starting at 1
    map 1.0./         //Find the reciprocal of each (divide 1.0)
    scanLeft .0)(_+_) //Sum each partial sequence
indexWhere            //Find the index where
_.<=                  //n (underscore) is less than or equal to a partial sum

posted about 2 years ago

CC BY-SA 4.0

user‭

1526 reputation 9 59 172 89

Stax, 9 bytes

Ç≈f♠É↔X+ö

Run and debug it

Explanation(Unpacked):

wii{um|+;<
w          iterate until a falsy result is reached:
 ii        push iteration number i twice 
   {um     map range [1..i] to their reciprocals
      |+   sum that list
        ;< compare to the input(gets popped)
           implicit output of top of stack

posted over 2 years ago

CC BY-SA 4.0

Razetime‭

2780 reputation 42 62 313 82

JavaScript (Node.js), 47 bytes

f=(a,b=0,c=a=>a&&1/a+c(a-1))=>c(b)<a?f(a,b+1):b

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For large numbers, a rounding error will occur and yield incorrect result. The solution below works for any integer (no rounding errors):

93 bytes

f=(a,i=0n,p=a=>g(a)/c(a),g=(a,b=a)=>a&&c(b)/a+g(~-a,b),c=a=>a?a*c(~-a):1n)=>p(i)<a?f(a,-~i):i

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posted over 2 years ago

CC BY-SA 4.0

Hakerh400‭

1581 reputation 4 55 163 63

Jelly, 8 bytes

İ€S<¬ʋ1#

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How it works

İ€S<¬ʋ1# - Main link. Takes n on the left
     ʋ   - Group the previous 4 links into a dyad f(k, n):
 €       -   Over each integer 1 to k:
İ        -     Get its reciprocal
  S      -   Sum
   <     -   Less than n?
    ¬    -   Logical not
      1# - Find the first integer k ≥ n such that f(k, n) is true

posted almost 2 years ago

CC BY-SA 4.0

caird coinheringaahing‭

516 reputation 3 13 61 12

JavaScript, 29 bytes

f=(n,x=0)=>n>0?f(n-1/++x,x):x

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posted almost 2 years ago

CC BY-SA 4.0

2y ago

Shaggy‭

1951 reputation 4 104 198 153

J, 23 bytes

>:@]^:(>1#.1%1+i.)^:_&1

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>:@]^:(...)^:_&1   Increments from 1 while
>1#.1%1+i.         the harmonic sum up to that is less than the input

posted over 1 year ago

CC BY-SA 4.0

Bubbler‭

11 reputation 0 1 1 0

Goruby, 38 bytes

->n{dw{$.+=1;$..mp{1.0/-~_1}.su<n};$.}

This was not as much fun as I thought it would be, but it was fun enough to use once.

posted 10 months ago

CC BY-SA 4.0

south‭

311 reputation 0 44 31 22