Calculate the lowest number where the sum of the sequence of numbers exceeds a given value

Question

Given you have an infinite sequence of numbers defined as follows:

1: 1 = 1
2: 1 + 2 = 3
3: 1 + 3 = 4
4: 1 + 2 + 4 = 7
5: 1 + 5 = 6
6: 1 +たす 2 +たす 3 +たす 6 =わ 12
7: 1 + 7 = 8
...

The sequence is the sum of the divisors of n, including 1 and n.

Given a positive integer x as input, calculate the lowest number n which will produce a result greater than x.

Test cases

f(100) = 48, ∑ = 124
f(25000) = 7200, ∑ = 25389
f(5000000) = 1164240, ∑ = 5088960

Expected Output

Your program should return both n and the sum of its divisors, like so:

$ ./challenge 100
48,124

Rules

This is code-golf so the shortest code in bytes, in each language wins.

• Martin Ender: Is that sequence just the sum of ns divisors? You'll probably want to state that explicitly.
• Martin Ender: Also, judging by your "expected output" you want both n and f(n), but you don't say so anywhere in the specification.
• Mr. Xcoder: Bonuses are bad, especially when they are vague. I decided to remove it, in order to protect this from being downvoted.
• caird coinheringaahing: Could you recheck f(1000) = 48? The divisor sum of 48 is 124
• Zgarb: It's good to wait at least a week before accepting an answer, otherwise you might discourage new solutions.

Answer 1

Jelly, (削除) 18 (削除ここまで) (削除) 12 (削除ここまで) (削除) 11 (削除ここまで) 10 bytes

1Æs>\#ḢṄÆs

Try it online!

-1 byte thanks to Mr. Xcoder!

How it works

1Æs>\#ḢṄÆs - Main link. Argument: n (integer)
1 \# - Find the first n integers where...
 Æs - the divisor sum
 > - is greater than the input
 Ṅ - Print...
 Ḣ - the first element
 Æs - then print the divisor sum

• dylnan: Could you explain why the 1 is necessary and how the ¥ acts?
• caird coinheringaahing: @dylnan The 1 tells # to start counting from 1 and the ¥ takes the previous two links (Æs and >) and applies them as a dyad (i.e. with two arguments), with the left argument being the iteration, and the right argument being the input.
• dylnan: Oh, that makes sense now. # had been a bit confusing to me before in some cases.

Answer 2

Brachylog, 9 bytes

∧;S?hf+S>

This program takes input from the "output variable" ., and outputs to the "input variable" ?. Try it online!

Explanation

∧;S?hf+S>
∧;S There is a pair [N,S]
 ? which equals the output
 h such that its first element's
 f factors'
 + sum
 S equals S,
 > and is greater than the input.

The implicit variable N is enumerated in increasing order, so its lowest legal value is used for the output.

Answer 3

Wolfram Language (Mathematica), 53 bytes

{#,f@#}&@@Select[Range[x=#]+1,(f=Tr@*Divisors)@#>x&]&

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Tries all values between 2 and x+1, where x is the input.

(The Select returns a list of all values that work, but the function {#,f@#}& takes all of these as inputs, and then ignores all its inputs but the first.)

Answer 4

R, 71 bytes

function(x)for(n in 1:x){z=sum(which(n%%1:n==0));if(z>x)return(c(n,z))}

Try it online!

• Giuseppe: 59 bytes but it will stack overflow for large x.

Answer 5

Husk, (削除) 12 (削除ここまで) 11 bytes

§eVḟ>0moΣḊN

-1 byte, thanks to @Zgarb!

Try it online!

• ბიმო: Clever! Weird though how , doesn't work (or inference takes too long?).
• Zgarb: It does infer a type, but outputs an infinite list. It may be caused by the overloading of ḟ that takes an integer as the second argument, but that's just a guess.

Answer 6

R, 73 bytes

n=scan();while(1){d=(x=1:T)[!T%%x];if(sum(d)>n)break;T=T+1};cat(T,sum(d))

Try it online!

Outgolfed by duckmayr.

• duckmayr: 71 bytes codegolf.stackexchange.com/a/149747/76266

Answer 7

Japt, 15 bytes

[@<(V=Xâ x}a V]

Try it

---

Explanation

Implicit input of integer U. [] is our array wrapper. For the first element, @ }a is a function that run continuously until it returns a truthy value, passing itself an incrementing integer (starting at 0) each time, and outputting the final value of that integer. â gets the divisors of the current integer (X), x sums them and that result is assigned to variable V. < checks if U is less than V. The second element in the array is then just V.

Answer 8

Clojure, 127 bytes

(defn f[n](reduce +(filter #(zero?(rem n %))(range 1(inc n)))))
(defn e[n](loop[i 1 n n](if(>(f i)n){i,(f i)}(recur(inc i)n))))

Try it online!

thanks to @steadybox for -4 bytes!

• caird coinheringaahing: Welcome to the site!
• Steadybox: Some whitespace can be removed to save a few bytes. Try it online!
• NikoNyrh: In this case reduce can be replaced by apply, also the function e could be expressed as an anonymous function via the #(...) syntax, you don't need to name it at Code Golf. #(=(rem n %)0) is shorter than #(zero?(rem n %)). And remember that , is whitespace, and can be removed in this case as it is followed by (, so it will be parsed correctly.
• Alonoaky: @NikoNyrh nice to meet a fellow clojurist, i'll edit this post soon

Answer 9

Ruby, 58 bytes

Full program because I'm not sure if lambdas are allowed. /shrug

gets
$.+=1until$_.to_i.<v=(1..$.).sum{|n|$.%n<1?n:0}
p$.,v

Try it online!

Explanation

gets # read line ($_ is used instead of v= because it cuts a space)
$.+=1 # $. is "lines read" variable which starts at 1 because we read 1 line
 until # repeat as long as the next part is not true
$_.to_i # input, as numeric
 .<v= # is <, but invoked as function to lower operator prescedence
 (1..$.) # Range of 1 to n
 .sum{|n| # .sum maps values into new ones and adds them together
 $.%n<1?n:0 # Factor -> add to sum, non-factor -> 0
 }
p$.,v # output n and sum

• Giuseppe: Lambdas are certainly allowed.

Answer 10

JavaScript (ES6), (削除) 61 (削除ここまで) 58 bytes

f=(n,i=1,s=j=0)=>j++<i?f(n,i,i%j?s:s+j):s>n?[i,s]:f(n,++i)

<input type=number min=0 oninput=o.textContent=f(this.value)><pre id=o>

Edit: Saved 3 bytes thanks to @Arnauld.

• StudleyJr: I get "Script error." when inputting a value over 545
• Neil: Try using Safari; apparently it supports Tail Call Optimisation. (Or if you can find them, some versions of Chrome enable it via the "Experimental JavaScript features".)

Answer 11

05AB1E, 11 bytes

>LʒÑO‹}нDÑO

Try it online!

Leaves the output on the stack, as allowed per meta consensus. I added ) for the sake of visualization, but the program also implicitly prints the top of the stack.

Answer 12

APL (Dyalog), 32 bytes

{⍺<o←+/o/⍨0=⍵|⍨o←⍳⍵:⍵,o⋄⍺∇⍵+1}∘0

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⍵o⍺⍵.

Answer 13

SOGL V0.12, 14 bytes

1[:Λ∑:A.>?ao←I

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Explanation:

1 push 1
 [ while ToS != 0
 :Λ get the divisors
 ∑ sum
 :A save on variable A without popping
 .>? ← if greater than the input
 ao output the variable A
 ← and stop the program, implicitly outputting ToS - the counter
 I increment the counter

Answer 14

C, (削除) 79 (削除ここまで) 78 bytes

i,n,s;f(x){for(i=n=s=0;x>s;s+=n%++i?0:i)i-n||(++n,i=s=0);printf("%d %d",n,s);}

Try it online!

• ceilingcat: Suggest i=s=!++n instead of ++n,i=s=0

Answer 15

MATL, 12 bytes

`@Z\sG>~}@6M

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Explanation

` % Do...while
 @ % Push iteration index (1-based)
 Z\ % Array of divisors
 s % Sum of array
 G % Push input
 >~ % Greater than; logical negate. This is the loop condition
} % Finally (execute on loop exit)
 @ % Push latest iteration index
 6M % Push latest sum of divisors again
 % End (implicit). Run new iteration if top of the stack is true
 % Display stack (implicit)

Answer 16

Gaia, 11 bytes

dΣ@>
↑#(:dΣ

Try it online!

Leaves the output on the stack, as allowed per meta consensus. I added €. for the sake of visualization, but the program also implicitly prints the top of the stack.

Answer 17

Haskell, 59 bytes

f x=[(i,s)|i<-[1..],s<-[sum[d|d<-[1..i],i`mod`d<1]],s>x]!!0

-1 byte, thanks to @nimi!

Try it online!

Answer 18

Ohm v2, 11 bytes

£^DVΣD3>‽;«

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Answer 19

Factor, 88

USE: math.primes.factors [ 0 0 [ drop 1 + dup divisors sum pick over > ] loop rot drop ]

Brute-force search. It's a quotation (lambda), call it with x on the stack, leaves n and f(n) on the stack.

As a word:

: f(n)>x ( x -- n f(n) )
 0 0 [ drop 1 + dup divisors sum pick over > ] loop rot drop ;

Answer 20

Python 3, 163 bytes

def f(x):
 def d(x):return[i for i in range(1,x+1) if x%i==0]
 return min(i for i in range(x) if sum(d(i)) >x),sum(d(min(i for i in range(x) if sum(d(i)) >x)))

• Jonathan Frech: Hello and welcome to PPCG; nice first post! From a golfing aspect, you could save some bytes by removing whitespace, using lambda functions, collapsing everything onto one line and not repeating yourself. We also usually link to an online testing environment, like for example TIO (105 bytes, using the techniques described above.)
• Eric Duminil: @JonathanFrech: Excellent comment. Thanks for your patience with noobies in general and noob in particular ;)

Answer 21

Python 3, 100 bytes

d=lambda y:sum(i+1for i in range(y)if y%-~i<1)
f=lambda x:min((j,d(j))for j in range(x+1)if x<=d(j))

Try it online!

Thanks to Jonathan Frech's comment on the previous python 3 attempt, I have just greatly expanded my knowledge of python syntax. I'd never have thought of the -~i for i+1 trick, which saves two characters.

However, that answer is 1) not minimal and 2) doesn't work for x=1 (due to an off-by-one error which is easy to make while going for brevity; I suggest everyone else check their answers for this edge case!).

Quick explanation: sum(i+1for i in range(y)if y%-~i<1) is equivalent to sum(i for i in range(1,y+1)if y%i<1) but saves two characters. Thanks again to Mr. Frech.

d=lambda y:sum(i+1for i in range(y)if y%-~i<1) therefore returns the divisors of y.

f=lambda x:min((j,d(j))for j in range(x+1)if x<=d(j)) is where I really did work. Since comparing a tuple works in dictionary order, we can compare j,d(j) as easily as we can compare j, and this lets us not have to find the minimal j, store it in a variable, and /then/ compute the tuple in a separate operation. Also, we have to have the <=, not <, in x<=d(j), because d(1) is 1 so if x is 1 you get nothing. This is also why we need range(x+1) and not range(x).

I'd previously had d return the tuple, but then I have to subscript it in f, so that takes three more characters.

• caird coinheringaahing: Welcome to the site and nice first post. You can get to 98 bytes by removing the f= as anonymous functions are perfectly acceptable here!
• Sam Howard: You can't call an anonymous function from another line of code, is the problem -- I have a separate print(f(100)) statement to test that the function works.
• caird coinheringaahing: That's not a problem here. It's perfectly acceptable and works to not include the f= in your byte count, and is a good way to golf in Python. Check this for more golfing tips in Python!
• Sam Howard: Hm. I can equal, but not better, my submission by appending q=range and replacing range with q in both existing instances. Sadly, this doesn't improve it and since lambda is a keyword I can't use it for that, I'd have to do exec() tricks wasting too many characters.
• Jonathan Frech: @MichaelBoger Well, you can call an anonymous function in Python; lambda expressions do not have to be assigned to a variable.

Answer 22

Python 2, 81 bytes

def f(n):
 a=b=0
 while b<n:
	a+=1;i=b=0
	while i<a:i+=1;b+=i*(a%i<1)
 return a,b

Try it online!

• Jonathan Frech: 77 bytes.
• Sam Howard: Replacing the tabs with two spaces makes this work in python 3 at 83 bytes, although to try it I had to put parentheses in the print statement. You can also replace the return statement with a print statement and not need an auxiliary function to print it; the bytes stay the same.

Answer 23

Java (OpenJDK 8), 91 bytes

i->{for(int j=0;j++<i;)for(int k=0,l=0;k++<j;)if((l+=j%k<1?k:0)>i)return k+","+l;return"";}

Try it online! (timeout on third test case)

• ceilingcat: 89 bytes

Answer 24

Perl 5, 60 + 1 (-p) = 61 bytes

$"=$_;until($_>$"){$_=$/=0;$\--;$_+=$\%$/?0:$/until++$/>-$\}

Try it online!

Output format:

sum-n

Answer 25

Clojure, 102 bytes

#(loop[i 1](let[s(apply +(for[j(range 1(inc i)):when(=(mod i j)0)]j))](if(> s %)[i s](recur(inc i)))))

Answer 26

PHP, 69 bytes

for(;$argv[1]>=$t;)for($t=$j=++$i;--$j;)$t+=$i%$j?0:$j;echo$i,',',$t;

Answer 27

Perl 6, 48 bytes

{first :kv,*>$_,map {sum grep $_%%*,1..$_},0..*}

Try it online!