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

Jelly,  5  4 bytes

Ḋ²ḋṖ

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How?

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

Ḋ²ḋṖ - Link: non-negative integer, n
Ḋ    - dequeue (implicit range)       [2,3,4,5,...,n]
 ²   - square (vectorises)            [4,9,16,25,...,n*n]
   Ṗ - pop (implicit range)           [1,2,3,4,...,n-1]
  ḋ  - dot product                    4*1+9*2+16*3+25*4+...+n*n*(n-1)

Answer 2

Python 3,  28  27 bytes

-1 thanks to xnor

lambda n:(n**3-n)*(n/4+1/6)

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Implements \$n(n-1)(n+1)(3n+2)/12\$

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Python 2,  29  28 bytes: lambda n:(n**3-n)*(3*n+2)/12

Answer 3

APL (Dyalog Unicode), 10 bytes

1⊥⍳×⍳×1-⍨⍳

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How it works

1⊥⍳×⍳×1-⍨⍳
  ⍳×⍳×1-⍨⍳  Compute (x^3 - x^2) for 1..n
1⊥          Sum

Uses the fact that "square of sum" is equal to "sum of cubes".

Answer 4

TI-Basic (TI-83 series), 12 11 bytes

sum(Ans² nCr 2/{2,3Ans

Implements \$\binom{n^2}{2}(\frac12 + \frac1{3n})\$. Takes input in Ans: for example, run 10:prgmX to compute the result for input 10.

Answer 5

Brain-Flak, 74 72 68 64 bytes

((([{}])){({}())}{})([{({}())({})}{}]{(({}())){({})({}())}{}}{})

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Pretty simple way of doing it with a couple of tricky shifts. Hopefully someone will find some more tricks to make this even shorter.

Answer 6

Japt -x, 9 8 5 4 bytes

õ²í*

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Explanation

õ        :Range [1,input]
 ²       :Square each
  í      :Interleave with 0-based indices
   *     :Reduce each pair by multiplication
         :Implicit output of the sum of the resulting array

Answer 7

JavaScript, 20 bytes

f=n=>n&&n*n*--n+f(n)

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Answer 8

Charcoal, 12 10 bytes

ＩΣＥＮ×ιＸ⊕ι²

Try it online! Link is to verbose version of code. Explanation: \$ ( \sum_1^n x )^2 = \sum_1^n x^3 \$ so \$ ( \sum_1^n x )^2 - \sum_1^n x^2 = \sum_1^n (x^3 - x^2) = \sum_1^n (x - 1)x^2 = \sum_0^{n-1} x(x + 1)^2 \$.

   Ｎ        Input number
  Ｅ         Map over implicit range i.e. 0 .. n - 1
        ι   Current value
       ⊕    Incremented
         ²  Literal 2
      Ｘ     Power
     ι      Current value
    ×       Multiply
 Σ          Sum
Ｉ           Cast to string
            Implicitly print

Answer 9

Perl 6, 22 bytes

{sum (1..$_)>>²Z*^$_}

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

Answer 10

dc, 16 bytes

?dd3^r-r3*2+*C/p

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

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Answer 11

APL(Dyalog), 17 bytes

{+/(¯1↓⍵)×1↓×⍨⍵}⍳

(Much longer) Port of Jonathan Allan's Jelly answer.

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Answer 12

APL (Dyalog), 16 bytes

((×⍨+/)-(+/×⍨))⍳

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 (×⍨+/)            The square (× self) of the sum (+ fold)
       -           minus
        (+/×⍨)     the sum of the square
(             )⍳   of [1, 2, … input].

Answer 13

Mathematica, 21 17 bytes

^{-4 bytes thanks to alephalpha.}

(3#+2)(#^3-#)/12&

Pure function. Takes an integer as input and returns an integer as output. Just implements the polynomial, since Sums, Ranges, Trs, etc. take up a lot of bytes.

Answer 14

Java (JDK), 23 bytes

n->(3*n+2)*(n*n*n-n)/12

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Answer 15

MathGolf, 6 bytes

{î²ï*+

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

Explanation:

{       Loop (implicit) input times
 î²     1-index of loop squared
    *   Multiplied by
   ï    The 0-index of the loop
     +  And add to the running total

Answer 16

05AB1E, 8 bytes

ÝDOnsnO-

Explanation:

ÝDOnsnO-     //Full program
Ý            //Push [0..a] where a is implicit input
 D           //Duplicate top of stack
  On         //Push sum, then square it
    s        //Swap top two elements of stack
     nO      //Square each element, then push sum
       -     //Difference (implicitly printed)

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Answer 17

C, C++, 46 40 37 bytes ( #define ), 50 47 46 bytes ( function )

-1 byte thanks to Zacharý

-11 bytes thanks to ceilingcat

Macro version :

#define F(n)n*n*~n*~n/4+n*~n*(n-~n)/6

Function version :

int f(int n){return~n*n*n*~n/4+n*~n*(n-~n)/6;}

Thoses lines are based on thoses 2 formulas :

Sum of numbers between 1 and n = n*(n+1)/2
Sum of squares between 1 and n = n*(n+1)*(2n+1)/6

So the formula to get the answer is simply (n*(n+1)/2) * (n*(n+1)/2) - n*(n+1)*(2n+1)/6

And now to "optimize" the byte count, we break parenthesis and move stuff around, while testing it always gives the same result

(n*(n+1)/2) * (n*(n+1)/2) - n*(n+1)*(2n+1)/6 => n*(n+1)/2*n*(n+1)/2 - n*(n+1)*(2n+1)/6 => n*(n+1)*n*(n+1)/4 - n*(n+1)*(2n+1)/6

Notice the pattern p = n*n+1 = n*n+n, so in the function, we declare another variable int p = n*n+n and it gives :

p*p/4 - p*(2n+1)/6

For p*(p/4-(2*n+1)/6) and so n*(n+1)*(n*(n+1)/4 - (2n+1)/6), it works half the time only, and I suspect integer division to be the cause ( f(3) giving 24 instead of 22, f(24) giving 85200 instead of 85100, so we can't factorize the macro's formula that way, even if mathematically it is the same.

Both the macro and function version are here because of macro substitution :

F(3) gives 3*3*(3+1)*(3+1)/4-3*(3+1)*(2*3+1)/6 = 22
F(5-2) gives 5-2*5-2*(5-2+1)*(5-2+1)/4-5-2*(5-2+1)*(2*5-2+1)/6 = -30

and mess up with the operator precedence. the function version does not have this problem

Answer 18

JavaScript (ES6), 22 bytes

n=>n*~-n*-~n*(n/4+1/6)

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Answer 19

Pyth, 7 bytes

sm**hdh

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Uses the formula in Neil's answer.

sm**hdhddQ   Implicit: Q=eval(input())
             Trailing ddQ inferred
 m       Q   Map [0-Q) as d, using:
    hd         Increment d
   *  hd       Multiply the above with another copy
  *     d      Multiply the above by d
s            Sum, implicit print 

Answer 20

SNOBOL4 (CSNOBOL4), 70 69 bytes

 N =INPUT
I X =X + N ^ 3 - N ^ 2
 N =GT(N) N - 1 :S(I)
 OUTPUT =X
END

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Answer 21

Pari/GP, 21 bytes

n->(3*n+2)*(n^3-n)/12

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Answer 22

05AB1E, 6 bytes

LnDƶαO

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Explanation

L         # push range [1 ... input]
 n        # square each
  D       # duplicate
   ƶ      # lift, multiply each by its 1-based index
    α     # element-wise absolute difference
     O    # sum

Some other versions at the same byte count:

L<ān*O
Ln.āPO
L¦nā*O

Answer 23

R, 28 bytes

x=1:scan();sum(x)^2-sum(x^2)

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Answer 24

Clojure, 58 bytes

(fn[s](-(Math/pow(reduce + s)2)(reduce +(map #(* % %)s))))

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Edit: I misunderstood the question

Clojure, 55, 35 bytes

#(* %(+ 1 %)(- % 1)(+(* 3 %)2)1/12)

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Answer 25

cQuents, 17 15 bytes

b$)^2-c$
;$
;$$

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Explanation

 b$)^2-c$     First line
:             Implicit (output nth term in sequence)
 b$)          Each term in the sequence equals the second line at the current index
    ^2        squared
      -c$     minus the third line at the current index

;$            Second line - sum of integers up to n
;$$           Third line - sum of squares up to n

Answer 26

MATLAB/Octave, 21 bytes

@(x)(x^3-x)*(x/4+1/6)

Anonymous function. We could save another 4 bytes if we assume x can be given as variable in workspace.
Implements \$n(n−1)(n+1)(3n+2)/12\$

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One of my approaches was @(x)-sum([sum(1:x)*i,1:x].^2) which is few bytes longer but I found it interesting because It used imaginary unit to change the sign.
I also found out I can generate sum of squares by multiplying normal and transposed vector with (1:x)*(1:x)'. Might be useful for future challenges!

Answer 27

Labyrinth, 20 bytes

?(:):):_3*(***_12/!@

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Straightforward closed-form formula of \$(x-1)x(x+1)(3x+2)/12\$.

?      Take input [x]
(:):)  Decrement, dup, increment, dup, increment [x-1 x x+1]
:_3*(  Dup, 3 times, decrement [x-1 x x+1 3x+2]
***    Multiply four values
_12/   Divide by 12
!@     Print it and halt

Labyrinth, 21 bytes

?
::(";;;!@
{  :
+**}

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Labyrinth, 22 bytes

   ?
:(::
}  "
**+{;!@

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Two versions using a loop to evaluate \$\sum_{i=1}^{x}{i^2(i-1)}\$. Both versions test the loop exit with i-1 == 0, discard the top values ; as necessary, print the sum and exit.

?     Take input [x]; enter loop [sum i] with initial values of sum=0, i=x
::(   Dup, dup, decrement [sum i i i-1]
:}    Dup and move i-1 to aux. store [sum i i i-1 | i-1]
**+   Multiply top three values and add to sum [sum' | i-1]
{     Move i-1 back to the main stack

      Loop exit test can be done anywhere in the last iteration
      (i==1 or i==0), since that iteration does not affect the sum.

Answer 28

ARM (Thumb), 18 bytes

.section .text
.global func
.thumb
// int func(int dummy, int x)
// returns sum(i*i*(i-1) for i in 1..x)
// r1 = i (x->1), r2 = i*i*(i-1), r0 = sum
func:
    mov r0, #0       // 2000   int sum = 0;
    mov r2, #0       // 2200   int prod = 0;
.loop:               //        do {
    add r0, r0, r2   // 1880   sum += prod;
    mov r2, r1       // 1c0a   prod = x;
    mul r2, r1       // 434a   prod *= x;
    sub r1, #1       // 3901   x -= 1;
    mul r2, r1       // 434a   prod *= x;
    bne .loop        // d1f9   } while (prod != 0);
    bx lr            // 4770   return sum;

A function that takes the input integer via r1 and returns the value via r0.

Uses the iterative formula \$\sum_{i=1}^{x}{i^2(i-1)}\$, iterating backwards. The loop terminates when the value of \$i^2(i-1)\$ is 0.

For the purposes of writing test cases, the C-side function signature takes a dummy value for r0.

Test cases written in C:

#include <stdio.h>

int func(int dummy, int x);
int main(void) {
    int dummy;
    int inputs[] = {1, 2, 3, 10, 24, 100};
    for(int i = 0; i < 6; ++i) {
        int v = inputs[i];
        printf("input = %d -> ans = %d\n", v, func(dummy, v));
    }
    return 0;
}

Build, run, objdump commands:

arm-linux-gnueabihf-gcc main_c.c func_c.S -static -o main_c
qemu-arm -L /usr/arm-linux-gnueabihf ./main_c
arm-linux-gnueabihf-objdump -d main_c | awk -v RS= '/^[[:xdigit:]]+ <(func|.loop)>/'

(The objdump + awk trick comes from this SO answer.)

Answer 29

Vyxal, 8 bytes

ɾ∑²⁰ɾ²∑-

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literal solution, squares the sum and subtracts the sum of the squares.

Answer 30

Factor + math.unicode, 32 bytes

[ [1,b] 3 dupn Σ sq -rot v. - ]

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word   | data stack                                     | comment
-------+------------------------------------------------+--------------
       | 10                                             | example input
[1,b]  | { 1 2 ... 10 }                                 |
3      | { 1 2 ... 10 } 3                               |
dupn   | { 1 2 ... 10 } { 1 2 ... 10 } { 1 2 ... 10 }   |
Σ      | { 1 2 ... 10 } { 1 2 ... 10 } 55               | sum
sq     | { 1 2 ... 10 } { 1 2 ... 10 } 3025             | square
-rot   | 3025 { 1 2 ... 10 } { 1 2 ... 10 }             |
v.     | 3025 385                                       | dot product
-      | 2640                                           | output