13
\$\begingroup\$

Definition

In Mathematics, Harmonic Sequence refers to a sequence where

Harmonic Sequence Equation

i.e. the nth term of the sequence equals the reciprocal of n.


Introduction

In this challenge, given a positive integer n as input, output the Partial Sum of first n terms of the Harmonic Sequence.


Input

You'll be given a positive integer (within the range of numbers supported by your language). It can be either of Signed and Unsigned (depends on you), since the challenge requires only positive integers.

You can take the input in any way except assuming it to be present in a predefined variable. Reading from file, terminal, modal window (prompt() in JavaScript) etc. is allowed. Taking the input as function argument is allowed as well.


Output

Your program should output the sum of the first n terms of the Harmonic Sequence as a float (or integer if the output is evenly divisible by 1) with precision of 5 significant figures, where n refers to the input. To convey the same in Mathematical jargon, you need to compute

Harmonic Sequence Partial Sum of first n terms

where n refers to the input.

You can output in any way except writing the output to a variable. Writing to screen, terminal, file, modal window (alert() in JavaScript) etc. is allowed. Outputting as function return value is allowed as well.


Additional Rules


Test Cases

The Test Cases assume the input to be 1-indexed

Input     Output
1         1
2         1.5
3         1.8333
4         2.0833
5         2.2833

Winning Criterion

This is , so the shortest code in bytes wins!

\$\endgroup\$
  • \$\begingroup\$ Could you give us some testcases? \$\endgroup\$ – Kritixi Lithos May 28 '17 at 9:50
  • 2
    \$\begingroup\$ What precision is required? Exact output is generally only possible as a fraction, but in many languages that will have to be separate numbers for numerator and denominator. Can we output a)a float, b)a fraction or integer pair c)either? \$\endgroup\$ – Level River St May 28 '17 at 9:52
  • 2
    \$\begingroup\$ @Arjun The harmonic series grows to infinity so it will get hard to meet 10 decimal places as the number gets into the thousands and millions. I would go for significant figures rather than decimal places, and I see no need to be so precise. 5 significant figures should be enough. so 9.9999E10 rather than 99999999999.9999999999 \$\endgroup\$ – Level River St May 28 '17 at 10:17
  • \$\begingroup\$ Can we go over 5 significant figures? \$\endgroup\$ – Erik the Outgolfer May 28 '17 at 10:25
  • \$\begingroup\$ By the way, it's known that the harmonic sequence does not contain any integers other than the initial a_1 = 1. (Idea of proof that a_n is not an integer for n>1: let 2^k be the largest power of 2 not exceeding n; then 2^k divides the denominator of a_n.) \$\endgroup\$ – Greg Martin May 28 '17 at 19:53

25 Answers 25

4
\$\begingroup\$

Jelly, 3 bytes

İ€S

Try it online!

1-indexed.

Explanation:

İ€S Main link, monadic
İ€         1 / each one of [1..n]
  S Sum of
\$\endgroup\$
9
\$\begingroup\$

Python 3, 27 bytes

h=lambda n:n and 1/n+h(n-1)
\$\endgroup\$
  • \$\begingroup\$ 0-indexing or 1-indexing? \$\endgroup\$ – Arjun May 28 '17 at 11:29
  • 2
    \$\begingroup\$ Throws RuntimeError when handling input greater than the recursion limit, 1000 by default. \$\endgroup\$ – sagiksp May 28 '17 at 15:04
  • \$\begingroup\$ you can do sys.setrecursionlimit(473755252663) but the stack will eventually overflow quite easily \$\endgroup\$ – cat May 29 '17 at 4:31
  • \$\begingroup\$ @Arjun it's 1-indexed \$\endgroup\$ – shooqie May 29 '17 at 20:23
8
\$\begingroup\$

JavaScript, 19 18 bytes

1 byte saved thanks to @RickHitchcock

f=a=>a&&1/a+f(--a)

This is 1-indexed.

f=a=>a&&1/a+f(--a)

for(i=0;++i<10;)console.log(f(i))

\$\endgroup\$
  • \$\begingroup\$ From what I've seen of other posts, you can remove f= from your answer to save 2 bytes. \$\endgroup\$ – Rick Hitchcock May 28 '17 at 10:44
  • 1
    \$\begingroup\$ @RickHitchcock I cannot remove f= because the function is recursive and it references itself in f(--a). But if this was not a recursive solution, I would have been able to do that \$\endgroup\$ – Kritixi Lithos May 28 '17 at 10:45
  • \$\begingroup\$ Ah, makes sense! Save one byte with f=a=>a&&1/a+f(--a). \$\endgroup\$ – Rick Hitchcock May 28 '17 at 10:49
  • \$\begingroup\$ @RickHitchcock Nice one! \$\endgroup\$ – Kritixi Lithos May 28 '17 at 10:52
6
\$\begingroup\$

APL (Dyalog), 5 bytes

+/÷∘⍳

Try it online!

You can add ⎕PP←{number} to the header to change the precision to {number}.

This is 1-indexed.

Explanation

+/÷∘⍳                     Right argument; n
    ⍳                     Range; 1 2 ... n
  ÷                       Reciprocal; 1/1 1/2 ... 1/n
+/                        Sum; 1/1 + 1/2 + ... + 1/n
\$\endgroup\$
6
\$\begingroup\$

Mathematica, 21 20 16 bytes

This solution is 1-indexed.

Sum[1./i,{i,#}]&
\$\endgroup\$
  • \$\begingroup\$ It is 1-indexing \$\endgroup\$ – J42161217 May 28 '17 at 11:31
  • 1
    \$\begingroup\$ > You must not use a built-in to calculate the partial sum of the first n elements. (Yeah, it's for you Mathematica!) \$\endgroup\$ – MCCCS May 28 '17 at 11:32
  • 4
    \$\begingroup\$ OP means I can't use HarmonicNumber[#] & \$\endgroup\$ – J42161217 May 28 '17 at 11:35
  • 4
    \$\begingroup\$ And one can further shorten to Tr[1./Range@#]&. \$\endgroup\$ – Greg Martin May 28 '17 at 19:56
  • 2
    \$\begingroup\$ @Ian Mathematica may display 5 sig fig, but the function returns machine-precision numbers (52 binary bits or just under 16 decimal digits of precision) \$\endgroup\$ – LLlAMnYP May 30 '17 at 9:12
5
\$\begingroup\$

PHP, 33 Bytes

1-indexing

for(;$i++<$argn;)$s+=1/$i;echo$s;

Try it online!

\$\endgroup\$
5
\$\begingroup\$

Pari/GP, 18 bytes

n->sum(i=1,n,1./i)

1-indexing.

Try it online!

\$\endgroup\$
5
\$\begingroup\$

CJam, 11 bytes

1.ri,:)f/:+

Try it online!

1-indexed.

\$\endgroup\$
5
\$\begingroup\$

Japt -x, 8 6 5 3 bytes

õpJ

With some thanks to ETHproductions

Try it online

\$\endgroup\$
  • \$\begingroup\$ 0-indexing or 1-indexing? \$\endgroup\$ – Arjun May 28 '17 at 11:27
  • \$\begingroup\$ I think you can save a byte with õ x@1/X \$\endgroup\$ – ETHproductions May 28 '17 at 12:25
  • \$\begingroup\$ ...and another couple bytes by using XpJ instead of 1/X :-) \$\endgroup\$ – ETHproductions May 28 '17 at 12:27
  • \$\begingroup\$ Thanks, @ETHproductions :) I twigged those as soon as I walked away. \$\endgroup\$ – Shaggy May 28 '17 at 12:45
  • \$\begingroup\$ Actually I don't think you even need the _ due to auto-functions. I should really write that tip :P (I should have time today or tomorrow, due to it being Memorial Day) \$\endgroup\$ – ETHproductions May 28 '17 at 14:50
4
\$\begingroup\$

CJam, 11 10 bytes

1 byte removed thanks to Erik the outgolfer

ri),{W#+}*

This uses 1-based indexing.

Try it online!

Explanation

ri            e# Read integer, n
  )           e# Increment by 1: gives n+1
   ,          e# Range: gives [0 1 2 ... n]
    {   }*    e# Fold this block over the array
     W#       e# Inverse of a number
       +      e# Add two numbers
\$\endgroup\$
  • \$\begingroup\$ You can use W instead of -1. \$\endgroup\$ – Erik the Outgolfer May 28 '17 at 12:24
  • \$\begingroup\$ @EriktheOutgolfer has outgolfed himself :-) \$\endgroup\$ – Luis Mendo May 28 '17 at 13:04
  • \$\begingroup\$ @LuisMendo I like my name, it's just a name. And yes I outgolfed myself in the process of helping a fellow golfer golf even further. \$\endgroup\$ – Erik the Outgolfer May 28 '17 at 13:10
  • \$\begingroup\$ @Erik It was meant as a joke. Thanks for the help \$\endgroup\$ – Luis Mendo May 28 '17 at 13:57
3
\$\begingroup\$

Haskell, 20 bytes

f 0=0
f n=1/n+f(n-1)

Original solution, 22 bytes

f n=sum[1/k|k<-[1..n]]

These solutios assumes 1-indexed input.

\$\endgroup\$
3
\$\begingroup\$

R, 15 bytes

sum(1/1:scan())

Try it online!

\$\endgroup\$
3
\$\begingroup\$

Tcl 38 bytes

proc h x {expr $x?1./($x)+\[h $x-1]:0}

That's a very dirty hack, and the recursive calls pass literal strings like "5-1-1-1..." until it evaluates to 0.

\$\endgroup\$
  • \$\begingroup\$ Thanks @Christopher for the formatting. With that, the duplication of backslash was no longer necessary. \$\endgroup\$ – avl42 May 28 '17 at 20:30
  • \$\begingroup\$ No problem! It looks better \$\endgroup\$ – Christopher May 28 '17 at 21:30
3
\$\begingroup\$

05AB1E, 3 bytes

LzO

Try it online!

1-indexed.

\$\endgroup\$
2
\$\begingroup\$

MATL, 5 bytes

:l_^s

This solution uses 1-based indexing.

Try it at MATL Online

Explanation

    % Implicitly grab input (N)
:   % Create an array from [1...N]
l_^ % Raise all elements to the -1 power (take the inverse of each)
s   % Sum all values in the array and implicitly display the result
\$\endgroup\$
2
\$\begingroup\$

Axiom, 45 34 bytes

f(x:PI):Any==sum(1./n,n=1..x)::Any

1-Indexed; It has argument one positive integer(PI) and return "Any" that the sys convert (or not convert) to the type useful for next function arg (at last it seems so seeing below examples)

(25) -> [[i,f(i)] for i in 1..9]
   (25)
   [[1,1.0], [2,1.5], [3,1.8333333333 333333333], [4,2.0833333333 333333333],
    [5,2.2833333333 333333333], [6,2.45], [7,2.5928571428 571428572],
    [8,2.7178571428 571428572], [9,2.8289682539 682539683]]
                                                      Type: List List Any
(26) -> f(3000)
   (26)  8.5837498899 591871142
                                        Type: Union(Expression Float,...)
(27) -> f(300000)
   (27)  13.1887550852 056117
                                        Type: Union(Expression Float,...)
(29) -> f(45)^2
   (29)  19.3155689383 88117644
                                                   Type: Expression Float
\$\endgroup\$
2
\$\begingroup\$

Pyth, 5 bytes

scL1S

Try it here.

1-indexed.

\$\endgroup\$
1
\$\begingroup\$

C, 54 bytes

i;float f(n){float s;for(i=n+1;--i;s+=1./i);return s;}

Uses 1-indexed numbers.

\$\endgroup\$
1
\$\begingroup\$

Brachylog, 6 bytes

⟦₁/₁ᵐ+

Try it online!

This is 1-indexed.

Explanation

⟦₁         Range [1, …, Input]
    ᵐ      Map:
  /₁         Inverse
     +     Sum
\$\endgroup\$
1
\$\begingroup\$

QBIC, 13 bytes

[:|c=c+1/a]?c

Explanation

[ |        FOR a = 1 to
 :            the input n
   c=c+    Add to c (starts off as 0)
   1/a     the reciprocal of the loop iterator
]          NEXT
?c         PRINT c
\$\endgroup\$
0
\$\begingroup\$

Haskell, 21 bytes

f n=sum$map(1/)[1..n]
\$\endgroup\$
0
\$\begingroup\$

C (gcc), 35 bytes

float f(n){return n?1./n+f(--n):0;}

Try it online!

\$\endgroup\$
0
\$\begingroup\$

Braingolf, 20 bytes [non-competing]

VR1-1[1,!/M,$_1+]v&+

This won't actually work due to braingolf's inability to work with floats, however the logic is correct.

Explanation:

VR1-1[1,!/M,$_1+]v&+   Implicit input
VR                     Create new stack and return to main stack
  1-                   Decrement input
    1                  Push 1
     [..........]      Loop, always runs once, then decrements first item on stack at ]
                       Breaks out of loop if first item on stack reaches 0
      1,!/             Push 1, swap last 2 values, and divide without popping
                       Original values are kept on stack, and result of division is pushed
          M,$_         Move result of division to next stack, then swap last 2 items and
                       Silently pop last item (1)
              1+       Increment last item on stack
                 v&+   Move to next stack, sum entire stack 
                       Implicit output of last item on current stack

Here's a modified interpreter that supports floats. First argument is input.

\$\endgroup\$
0
\$\begingroup\$

Tcl, 61 bytes

proc h {x s\ 0} {time {set s [expr $s+1./[incr i]]} $x;set s}

Try it online!

\$\endgroup\$
0
\$\begingroup\$

Gol><>, 8 bytes

F1LP,+|B

Try it online!

Example full program & How it works

1AGIE;GN
F1LP,+|B

1AGIE;GN
1AG       Register row 1 as function G
   IE;    Take input as int, halt if EOF
      GN  Call G and print the result as number
          Repeat indefinitely

F1LP,+|B
F     |   Repeat n times...
 1LP,       Compute 1 / (loop counter + 1)
     +      Add
       B  Return
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.