Sum square difference

Question

The sum of the squares of the first ten natural numbers is, \$1^2 + 2^2 + \dots + 10^2 = 385\$

The square of the sum of the first ten natural numbers is,

\$(1 + 2 + ... + 10)^2 = 55^2 = 3025\$

Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is

\$3025 − 385 = 2640\$

For a given input n, find the difference between the sum of the squares of the first n natural numbers and the square of the sum.

Test cases

1       => 0
2       => 4
3       => 22
10      => 2640
24      => 85100
100     => 25164150

This challenge was first announced at Project Euler #6.

Winning Criteria

• There are no rules about what should be the behavior with negative or zero input.

• The shortest answer wins.

Answer 1

Fig, \$9\log_{256}(96)\approx\$ 7.408 bytes

a
R@-mQJS

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Port of 05AB1E. Literal solution is 2 chars longer.

a
R@-mQJS
-------
a       # Range [1, n]
------- # Use ^ as the input to the next line:
     JS # Prepend the sum to the list
   mQ   # Square each
R@      # Reduce by
  -     # Subtraction

Answer 2

Excel, 21 bytes

=(A1^3-A1)*(A1/4+1/6)

This is the heartless Excel version of the formula figured out by others. There is also the 33 byte solution that feels a little more Excel-y:

=LET(n,SEQUENCE(A1),SUM(n^3-n^2))

... and the 38 byte solution that looks the most like the original formulas in the question:

=LET(n,SEQUENCE(A1),SUM(n)^2-SUMSQ(n))

Answer 3

Husk, 11 8 bytes

-ṁ□ḣ¹□Σḣ

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Explanation

-ṁ□ḣ¹□Σḣ
-              Subtract
 ṁ□ḣ¹          The sum of the first n squares
     □         From the square of
      Σḣ       The sum from 1 to n

First time code golfing so this solution probably can be improved.

Thanks to Dominic van Essen for the -3 bytes.

Answer 4

Julia 1.0, 20 bytes

Using the same formula as bigyihsuan:

~n=(n^3-n)*(3n+2)/12

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---

Naive approach (34 bytes):

~n=(a=[1:n...];sum(a)^2-sum(a.^2))

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Answer 5

APL(NARS), 13 chars, 26 bytes

{+/⍵×⍵×⍵-1}∘⍳

use the formula Sum'w=1..n'(ww(w-1)) possible i wrote the same some other wrote + or - as "1⊥⍳×⍳×⍳-1"; test:

  g←{+/⍵×⍵×⍵-1}∘⍳
  g 0
0
  g 1
0
  g 2
4
  g 3
22
  g 10
2640

Answer 6

Stax, 4 bytes

╡⌠(♠

Run and debug it

For all positive k integers up to the input, add k^2 * (k-1).

Answer 7

QBASIC, 45 44 bytes

Going pure-math saves 1 byte!

INPUT n
?n^2*(n+1)*(n+1)/4-n*(n+1)*(2*n+1)/6

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---

Previous, loop-based answer

INPUT n
FOR q=1TO n
a=a+q^2
b=b+q
NEXT
?b^2-a

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Note that the REPL is a bit more expanded because the interpreter fails otherwise.

Answer 8

JAEL, 13 10 bytes

#&àĝ&oȦ

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Explanation (generated automatically):

./jael --explain '#&àĝ&oȦ'
ORIGINAL CODE:  #&àĝ&oȦ

EXPANDING EXPLANATION:
à => `a
ĝ => ^g
Ȧ => .a!

EXPANDED CODE:  #&`a^g&o.a!

COMPLETED CODE: #&`a^g&o.a!,

#          ,            repeat (p1) times:
 &                              push number of iterations of this loop
  `                             push 1
   a                            push p1 + p2
    ^                           push 2
     g                          push p2 ^ p1
      &                         push number of iterations of this loop
       o                        push p1 * p2
        .                       push the value under the tape head
         a                      push p1 + p2
          !                     write p1 to the tapehead
            ␄           print machine state

Answer 9

PowerShell, 73 39 bytes

1.."$args"|%{$r+=$_;$s+=$_*$_}
$r*$r-$s

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-34 bytes thanks to @mazzy and his genius PowerShell-foo

Answer 10

International Phonetic Esoteric Language, 21 bytes

ɪbwbbbffʍz3ʍf2sf{C}vo

Implements \$(n^3-n)(3n+2)/12\$.

Explanation:

ɪ       (input)
bw      (store n)
bbbffʍz (n^3-n)
3ʍf2s   (3n+2)
f       (multiply)
{C}v    (/12)
o       (print)

Testing against examples:

$ for i in 1 2 3 10 24 100; do echo {$i} | src/interpreter.py 'ɪbwbbbffʍz3ʍf2sf{C}vo'; done
0.0
4.0
22.0
2640.0
85100.0
25164150.0

Answer 11

Arn, 9 bytes

└V0¯„○aÌ$

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Explained

Unpacked: +{^3-^2}\~.

Performs \$\large\sum_{k=1}^{n}{(k^3-k^2)}\$ where n is the input

          \       Fold with
+                 Addition
  {               After mapping with this block
        _         Implied variable (implicit index of block).
      ^           To the power of
        3         Literal three
    -             Minus
        _
      ^
        2         Literal two
  }               End block
            ~     1-range to
              _   Initialized to STDIN; implied

Answer 12

Haskell, 22 bytes

f n=(3*n+2)*(n^3-n)/12

Answer 13

Pip, 21 bytes

($+\,a)**2-$+(\,a)**2

My first self written pip answer!

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Answer 14

Assembly (MIPS, SPIM), 142 114 bytes, 6 * 12 10 = 72 60 assembled bytes

^{Saved 2 instructions (12 bytes) from @Bubbler's golf.}

main:li$v0,5
syscall
li$a0,0
l:mul$t0,$v0,$v0
sub$v0,$v0,1
mul$t0,$t0,$v0
add$a0,$a0,$t0
bgtz$v0,l
li$v0,1
syscall

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Explanation

main:
    li $v0, 5              # Set syscall value 5
    syscall                # Syscall, v0 = input integer

    li $a0, 0              # a0 = 0

    loop:                  # Main loop (v0 is the counter):
        mul $t0, $v0, $v0  #     t0 = v0 * v0
        sub $v0, $v0, 1    #     v0 = v0 - 1
        mul $t0, $t0, $v0  #     t0 = t0 * v0
        add $a0, $a0, $t0  #     a0 = a0 + t0

        bgtz $v0, loop     #     if v0 > 0, jump to loop

    li $v0, 1              # Syscall value 1
    syscall                # Syscall (Output value of a0 as integer)

Answer 15

Flurry -nii, 32 bytes

{}{{}[<><<>()>]<([])[][]>}[<>()]

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Iterates through k=0 .. n-1 using the stack height, adding k(k+1)^2 each iteration.

How it works

main = n add-next-term 0

// Initial height is 0
// Increase the height at (push height), so subsequent `height` calls give k+1
add-next-term = \x. x succ <(push height) height height>
= \x. x + k * (k+1) * (k+1)

Answer 16

JavaScript (Node.js), 59 48 47 bytes

n=>(a=(x,b)=>x&&x*(b||x)+a(--x,b))(n,1)**2-a(n)

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Answer 17

C (gcc), 29 bytes

f(n){n=(3*n+2)*(n*n*n-n)/12;}

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Answer 18

J, 15 bytes

(+/@:**:@>:)@i.

Uses the form \$\sum_{i=0}^{n-1}((i+1)^2\cdot i)\$

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(+/@:**:@>:)@i.
             i. NB. range 0..n-1
            @   NB. atop f (g x)
(          )    NB. monadic hook, y f (g y)
      *:@>:     NB. increment then square
 +/@:*          NB. multiply square with each i then sum the result
                NB. u@:v where v is dyadic and u is always monadic,
                NB. like atop except u will execute once on the entire result

Answer 19

Pip, 15 13 11 bytes

SQ_*D_MS\,a

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Explanation

Adapted from one of the formulas in Engineer Toast's Excel answer:

$$ \sum_{i=1}^{n} i^3-i^2 $$

which is equivalent to

$$ \sum_{i=1}^{n} i^2(i-1) $$

SQ_*D_MS\,a
          a  Command-line argument
        \,   Inclusive range from 1
      MS     Map this function to each and sum the results:
SQ_            The argument squared
   *           Times
    D_         The argument decremented

---

Original 15-byte solution that implements the description directly:

\,:aSQ$+a-$+SQa

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Answer 20

Vyxal, 7 bytes

ɾ:∑p²ƒ-

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A port of Fig which is a port of 05ab1e

Explained

ɾ:∑p²ƒ-
ɾ:∑p     # prepend the sum of the range [1, n] to the range [1, n]
   ²ƒ-  # square everything and reduce by subtraction

Answer 21

C#, 89 Bytes

int x=0,y=0;for(int i=1;i<=Int32.Parse(s[0]);i++){x+=i*i;y+=i;}Console.Write($"{y*y-x}");

ungolfed:

int x=0,y=0;
for(int i=1;i<=Int32.Parse(s[0]);i++){
x+=i*i;
y+=i;
}
Console.Write($"{y*y-x}");

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Answer 22

Axiom, 39 bytes

f(n)==reduce(+,[x^3-x^2 for x in 1..n])

test:

-> [[x,f x]for x in [1,2,3,10]]
     [[1,0],[2,4],[3,22],[10,2640]]

Answer 23

Clojure, 91 bytes

(fn s[n](let[u #(apply + %)q #(Math/pow % 2)m(range 1(inc n))](-(q(u m))(u(map #(q %)m)))))

The naive, literal approach. See below:

(defn sum-sq-diff [n]
    (let [; Shortcut functions to save bytes
          sum #(apply + %)
          square #(Math/pow % 2)
          nums (range 1 (inc n))

          ss1 (sum (map #(square %) nums))
          ss2 (square (sum nums))]

      (- ss2 ss1)))

(mapv sum-sq-diff [1 2 3 10 24 100])
=> [0.0 4.0 22.0 2640.0 85100.0 2.516415E7]

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Answer 24

Ruby, 24 bytes

->n{(n+n+3*n*=n)*~-n/12}

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Answer 25

Python 3, 72 bytes

x=[i+1for i in range(int(input()))]
print(sum(x)**2-sum(i**2for i in x))

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Answer 26

F# (Mono), 57 41 bytes

let f x=Seq.sumBy(fun y->y*y*(y-1))[1..x]

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Answer 27

05AB1E, 6 bytes

LDOšnÆ

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Explanation:

           # implicit input (example: 3)
L          # range ([1, 2, 3])
 DOš       # prepend the sum ([6, 1, 2, 3])
    n      # square each ([36, 1, 4, 9])
     Æ     # reduce by subtraction (22)
           # implicit output

Æ isn't useful often, but this is its time to shine. This beats the naïve LOnILnO- by two whole bytes.

Answer 28

Perl 6, 22 bytes

{($_³-$_)*($_/4+⅙)}

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Answer 29

PHP, 37 bytes

while($i<$argn)$s+=$i++*$i*$i;echo$s;

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Standalone program input number via STDIN.

Answer 30

Keg, 15 bytes

&⑻Ï⑷²⑸⅀⑻:⑨*½²$-

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Calculates the sum of squares then the square of the sum.