# Exact generalised harmonic numbers

The generalised harmonic number of order $$\m\$$ of $$\n\$$ is

$$H_{n,m} = \sum_{k=1}^n \frac 1 {k^m}$$

For example, the harmonic numbers are $$\H_{n,1}\$$, and $$\H_{\infty,2} = \frac {\pi^2} 6\$$. These are related to the Riemann zeta function as

$$\zeta(m) = \lim_{n \to \infty} H_{n,m}$$

Given two positive integers $$\n > 0\$$, $$\m > 0\$$, output the exact rational number $$\H_{n,m}\$$. The fraction should be reduced to its simplest term (i.e. if it is $$\\frac a b\$$, $$\\gcd(a, b) = 1\$$). You may output as a numerator/denominator pair, a rational number or any clear value that distinguishes itself as a rational number. You may not output as a floating point number.

This is , so the shortest code in bytes wins

## Test cases

n, m -> Hₙ,ₘ
3, 7 -> 282251/279936
6, 4 -> 14011361/12960000
5, 5 -> 806108207/777600000
4, 8 -> 431733409/429981696
3, 1 -> 11/6
8, 3 -> 78708473/65856000
7, 2 -> 266681/176400
6, 7 -> 940908897061/933120000000
2, 8 -> 257/256
5, 7 -> 2822716691183/2799360000000

• Brownie points for beating/matching my 4 byte M answer May 18, 2021 at 0:40
• May 18, 2021 at 0:41

# M, 4 bytes

Rİ*S


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Rİ*S  Main Link; takes n on the left and m on the right
R     Range: [1, 2, 3, ..., n]
İ    Inverse/Reciprocal: [1, 1/2, 1/3, ..., 1/n]
*   Exponent: [1, 1/2^m, 1/3^m, ..., 1/n^m]
S  Sum

• What kind of language is M? The README just has an.... M. May 18, 2021 at 2:55
• @Jonah it's Jelly but using Python's sympy for infinite precision math. Or rather, an extremely outdated revision of Jelly lol May 18, 2021 at 3:09

# Wolfram Language (Mathematica), 14 bytes

HarmonicNumber


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Naturally.

Alternatively, 16 bytes without the built-in:

Tr[Range@#^-#2]&


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• I get the feeling there have been Wolfram answers where the built-in is longer than the non-built-in answer. May 18, 2021 at 2:14
• @chunes One example given in the Mathematica golfing tips thread is EuclideanDistance: it's always shorter to write (Norm[#-#2]&).
– att
May 18, 2021 at 2:22

# Raku, 29 bytes

{sum((1/1..$^n)X**-$^m).nude}


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This function takes two arguments, $^m and $^n. 1/1 .. $^n is a sequence of rational numbers from 1 to the second argument. X** -$^m produces the exponentiated cross product of that list with the negative of the first argument. sum sums those rational numbers, and .nude produces a two-element list of the numerator and denominator of the sum.

# JavaScript (ES7), 68 bytes

A 1-byte shorter, slightly less readable version with a single recursive call. This is otherwise identical to the commented version below.

m=>g=(n,N=0,D=1)=>D?g(n-!!n,n?p=N*n**m+D:D,n?q=D*n**m:N%D):[p/N,q/N]


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# JavaScript (ES7), 69 bytes

Expects (m)(n). Returns [numerator, denominator].

m=>g=(n,N=0,D=1)=>n?g(n-1,p=N*n**m+D,q=D*n**m):D?g(0,D,N%D):[p/N,q/N]


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### How?

The recursive function $$\g\$$ first computes the unreduced numerator and denominator $$\(N,D)\$$ of $$\H_{n,m}\$$ and saves a copy of the final result into $$\(p,q)\$$:

n ? g(n - 1, p = N * n**m + D, q = D * n**m) : ...


When $$\n=0\$$, it enters its 2nd phase where the GCD of $$\(p,q)\$$ is computed in $$\N\$$:

... : D ? g(0, D, N % D) : ...


When $$\D=0\$$, it eventually returns:

... : [p / N, q / N]


# J, 13 bytes

1#.(%@^~1+i.)


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# Jelly, 13 12 bytes

*€:@S,:gɗɗP$ Try it online! A byte less than *€µP:¹S,P:g/$, but the abuse of ɗ leads me to believe this can be shorter yet.

*€              Raise each 1 .. n to the power of m.
ɗP$For that list and the product of its elements: :@ divide each by the product S and sum; ɗ for that sum: , pair it with the product, : and divide both by g the GCD of the sum and product.  • I feel like there's some ð chaining able to be done here, but I can't figure anything out :/ May 18, 2021 at 1:39 • @cairdcoinheringaahing Outputting reversed pairs, this is the same length, so maybe you're on to something May 18, 2021 at 2:20 • I've not manged to find a save here but I do have *€µṭP:\§µ:g/ and half a dozen variants (all outputting [denominator, numerator]) if it helps anyone. May 18, 2021 at 13:10 # Java, 155 138 bytes int g(int a,int b){return b<1?a:g(b,a%b);} m->n->{int p=1,d=0,t;for(;n>0;d=d*t+p,p*=t)t=(int)Math.pow(n--,m);return d/g(d,p)+"/"+p/g(d,p);}  Try it online! $$Denominator_{n, m} = \prod_{k=1}^n k^m$$ $$Numerator_{n,m} = \sum_{k=1}^n \frac {Denominator_{n,m}} {k^m} = Numerator_{n-1,m} \times k^m + Denominator_{n-1,m}$$ # Java + Commons Lang 2, 127 bytes m->n->{org.apache.commons.lang.math.Fraction x=null;for(x=x.ZERO;n>0;)x=x.add(x.getFraction(1,(int)Math.pow(n--,m)));return x;}  # Factor + math.unicode, 35 24 bytes [ [1,b] 0 rot - v^n Σ ]  Try it online! -11 thanks to @Bubbler! ## Explanation: It's a quotation (anonymous function) that takes two integers from the data stack as input and leaves a fully-reduced mixed fraction (it's simply the way that Factor is) on the data stack as output. • [1,b] Make a range from 1 to n. • 0 rot Push 0 and bring m to the top of the stack. • - Subtract m from 0. • v^n Raise every element in the range to the -m power. • Σ Sum. • 24 bytes May 18, 2021 at 2:32 • @Bubbler Wow, very nice! May 18, 2021 at 2:38 # J, 11 bytes 1#.#\@$^-@]


Try it online!

TFW J beats Jelly...

Used as n f m, where n and m are given as extended-precision integers.

### How it works

1#.#\@$^-@] NB. dyadic train; left = n, right = m #\@$        NB. 1..n in a dyadic context:
@$NB. reshape m into dimension n, and then #\ NB. get the lengths of prefixes ^ NB. each raised to the power of -@] NB. -m 1#. NB. sum of them  # Vyxal, 14 bytes ɾ$eDΠ$/∑$Π":ġ/


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# Nim, 81 bytes

import math,rationals
func H(n,m:int):any=
var r=0//1;for i in 1..n:r+=1//i^m
r


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n#m=sum\$map((1%).(^m))[1..n]

• I think this requires an import to Data.Ratio to get %: Try it online
– xnor
May 18, 2021 at 1:53
• Though you can also force a rational output with normal arithmetic by specifying the type explicitly: Try it online
– xnor
May 18, 2021 at 2:07
• I admit it does. Does that cost any extra bytes, and if so, how many? May 18, 2021 at 3:49
• With the import, you just include it as part of the code like this.
– xnor
May 18, 2021 at 7:53
• A few tips. 1) You can import GHC.Real instead of Data.Ratio for -2 bytes. 2) Using a list comprehension, you can shorten your code a bit: n#m=sum[1%k^m|k<-[1..n]]. 3) Instead of importing Data.Ratio or  GHC.Real, you can actually use the Prelude function toRational: n#m=sum[1/toRational k^m|k<-[1..n]], for a total of 35 bytes. May 18, 2021 at 8:33

# Stax, 6 bytes

ê☺σ;vù


Run and debug it

# Ruby, 27 bytes

->n,m{(1..n).sum{|x|x**-m}}


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• You can update your answer to ruby 2.7+ to make use of numbered arguments and save 2 bytes ->n,m{(1..n).sum{_1**-m}}
– user100752
May 18, 2021 at 9:36
• It would not work on TIO, so I prefer to keep it this way.
– G B
May 18, 2021 at 11:36

# Python 3.8, 91 bytes

import math
f=lambda n,m,N=0,D=1:n and f(n-1,m,N*n**m+D,D*n**m)or(N/(G:=math.gcd(N,D)),D/G)


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Inputs $$\n\$$ and $$\m\$$ and returns the numerator and denominator of $$\H_{n,m}\$$ as a tuple.

Uses the formula from Arnauld's JavaScript answer.

# APL(Dyalog Unicode), 31 27 bytes SBCS

{(⊢÷∨/){⍵,⍨+/⍵÷x}∧/x←⍵*⍨⍳⍺}


Try it on APLgolf!

(⊢÷∨/){⍵,⍨+/⍵÷x}∧/x←⍵*⍨⍳⍺ is equivelent to x←(⍳⍺)*⍵⋄(⊢÷∨/){⍵,⍨+/⍵÷x}∧/x. APL translated to Python:

from math import *

def H(n, m):
x = [i**m for i in range(1,n+1)] # x←(⍳⍺)*⍵
tmp = lcm(*x) # ∧/x
tmp = (sum(tmp//i for i in x), tmp) # {⍵,⍨+/⍵÷x}
tmp = [e//gcd(*tmp) for e in tmp] # (⊢÷∨/)
return tmp


First I make a list from 1 to n -- ⍳⍺. Then I raise every element of the list to the power of m to calculate the value of all the denominators, and assign it to the variable x -- x ← (⍳⍺)*⍵.

I then calculate the denominator after all the fractions are added together by taking the least common multiple of all the elements of x (all the denominators) -- ∧/x.

Next I create a two element tuple, the second element of which is the denominator -- {⍵,⍨...}. The first element is the numerator, calculated as the sum of the denominator divided by each of the original denominators -- +/⍵÷x.

Lastly, I simplify the fraction by dividing it by it's gcd -- (⊢÷∨/).

# Excel, 70 bytes

=LET(k,SEQUENCE(A1)^B1,d,PRODUCT(k),n,SUM(d/k),g,GCD(n,d),n/g&"/"&d/g)


# Charcoal, 40 bytes

≔Ｘ…·¹ＮＮθ≔Πθη≔Σ÷ηθζ⊞υζ⊞υηＷζ«≔﹪ηιζ≔ιη»Ｉ÷υη


Try it online! Link is to verbose version of code. Explanation:

≔Ｘ…·¹ＮＮθ


Generate the first n powers of m.

≔Πθη≔Σ÷ηθζ


Calculate the product, then divide that by each power and take the sum.

⊞υζ⊞υη


Save the values for later.

Ｗζ«≔﹪ηιζ≔ιη»


Find the GCD.

Ｉ÷υη


Divide the saved values by the GCD.

# Core Maude, 111 bytes

fmod H is pr RAT . op __ : Nat Nat -> Rat . vars N M : Nat . eq 0 M = 0 . eq(s N)M = 1 / s N ^ M +(N M) . endfm


### Example Session

             \||||||||||||||||||/
--- Welcome to Maude ---
/||||||||||||||||||\
Maude 3.1 built: Oct 12 2020 20:12:31
Tue May 18 23:37:06 2021
Maude> fmod H is pr RAT . op __ : Nat Nat -> Rat . vars N M : Nat . eq 0 M = 0 . eq(s N)M = 1 / s N ^ M +(N M) . endfm
Maude> red 3 7 .
reduce in H : 3 7 .
rewrites: 17 in 0ms cpu (0ms real) (17206 rewrites/second)
result PosRat: 282251/279936
Maude> red 6 4 .
reduce in H : 6 4 .
rewrites: 40 in 0ms cpu (0ms real) (~ rewrites/second)
result PosRat: 14011361/12960000
Maude> red 3 1 .
reduce in H : 3 1 .
rewrites: 17 in 0ms cpu (0ms real) (~ rewrites/second)
result PosRat: 11/6
Maude> red 2 8 .
reduce in H : 2 8 .
rewrites: 10 in 0ms cpu (0ms real) (~ rewrites/second)
result PosRat: 257/256
Maude> red 5 7 .
reduce in H : 5 7 .
rewrites: 32 in 0ms cpu (0ms real) (~ rewrites/second)
result PosRat: 2822716691183/2799360000000


### Ungolfed

fmod GENERALIZED-HARMONIC-NUMBERS is
protecting RAT .
op H : Nat Nat -> Rat .
vars N M : Nat .
eq H(0, M) = 0 .
eq H(s N, M) = 1 / (s N ^ M) + H(N, M) .
endfm


H(N, M) is the naive recursive definition of $$\H_{n, m}\$$. Maude's built-in rational number module automatically reduces fractions to lowest common denominator.

There's not much golfing we can do because only a handful of characters can separate identifiers ((, ), [, ], {, }, ,, and space). I've renamed the function operator H to __ (juxtaposition of two values) in the golfed version to save a few bytes.

# MATLAB/Octave, 25 bytes

@(n,m)sum(1./sym(1:n).^m)


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Anonymous function. Returns accurate value of symbolic type, which can represent real numbers (so rational numbers including).
Unfortunatelly, to my suprise, MATLAB doesn't implement that function, only non-generalized harmonic numbers (so only for ).