Square pyramidal numbers

Question

A000330 - OEIS

Task

Your task is simple, generate a sequence that, given index i, the value on that position is the sum of squares from 0 upto i where i >= 0.

Example:

Input: 0
Output: 0           (0^2)

Input: 4
Output: 30          (0^2 + 1^2 + 2^2 + 3^2 + 4^2)

Input: 5
Output: 55          (0^2 + 1^2 + 2^2 + 3^2 + 4^2 + 5^2)

Specification:

• You may take no input and output the sequence indefinitely;
• You may take input N and output the Nth element of the sequence;
• You may take input N and output the first N elements of the sequence.

Answer 1

Jelly, 3 bytes

R²S

Try it online!

FGITW

Explanation

R²S  Main Link
R    Generate Range
 ²   Square (each term)
  S  Sum

Answer 2

Python 2, 22 bytes

lambda n:n*~n*~(n*2)/6

Try it online!

This uses the closed-form formula n * (n+1) * (2*n+1) / 6. The code performs the following operations:

• Multiplies n by (n*):

  • The bitwise complement of n (~n), which essentially means -1-n.
  • And by the bitwise complement of 2n (*~(n*2)), which means -1-2n.

• Divides by 6 (/6).

Python 2, 27 bytes

f=lambda n:n and f(n-1)+n*n

Try it online!

_{Saved 1 byte thanks to Rod and 1 thanks to G B.}

Answer 3

MATL, 3 bytes

:Us

... or them?

Try it online!

Explanation

:    % Implicit input n. Push range [1 2 ... n]
U    % Square, element-wise
s    % Sum of array. Implicit display

Answer 4

JavaScript (ES6), 16 bytes

n=>n*(n+++n)*n/6

Demo

let f =

n=>n*(n+++n)*n/6

for(n = 0; n < 10; n++) {
  console.log('a(' + n + ') = ' + f(n))
}

How?

The expression n+++n is parsed as n++ + n ⁽¹⁾. Not that it really matters because n + ++n would also work in this case.

Therefore:

n*(n+++n)*n/6 =
n * (n + (n + 1)) * (n + 1) / 6 =
n * (2 * n + 1) * (n + 1) / 6

which evaluates to sum(k=0...n)(k²).

---

_{(1) This can be verified by doing n='2';console.log(n+++n) which gives the integer 5, whereas n + ++n would give the string '23'.}

Answer 5

05AB1E, 3 bytes

ÝnO

Try it online!

Explanation

  O    # sum of
 n     # squares of
Ý      # range [0 ... input]

Answer 6

Brain-Flak, 36 bytes

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

Try it online!

# Main algorithm
(                                  )  # Push the sum of:
                {({})({}[()])}{}      #   The square of:
 {                              }     #     0 to i 

# Stuff for the loop
  <(({}[()])())>                      # Push i-1, i without counting it in the sum
                                 {}   # Pop the counter (0)

Answer 7

Brain-Flak, 34 bytes

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

Try it online!

How does it work?

Initially I had the same idea as Riley¹ but it felt wrong to use a zeroer. I then realized that

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

Calculates n² - n.

Why? Well we know

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

Calculates n² and loops n times. That means if we switch the order of the two pushes we go from increasing the sum by n + (n-1) each time to increasing the sum by (n-1) + (n-1) each time. This will decrease the result by one per loop, thus making our result n² - n. At the top level this -n cancels with the n generated by the push that we were zeroing alleviating the need for a zeroer and saving us two bytes.

Brain-Flak, 36 bytes

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

Try it online!

Here is another solution, its not as golfy but it's pretty strange so I thought I would leave it as a challenge to figure out how it works.

If you are not into Brain-Flak but you still want the challenge here it is as a summation.

---

^{1: I came up with my solution before I looked at the answers here. So no plagiarism here.}

Answer 8

Husk, 3 bytes

ṁ□ḣ

Try it online!

Answer 9

Ohm v2, 3 bytes

#²Σ

Try it online!

Explanation

#    Range
 ²   Square
  Σ  Sum

Answer 10

Japt, 3 bytes

ô²x

Try it here.

-1 thanks to Shaggy.

Explanation:

ò²x 
ô²  Map square on [0..input]
  x Sum

Answer 11

Wolfram Language (Mathematica), 14 bytes

Tr[Range@#^2]&

Try it online!

Answer 12

Brain-Flak, 46 bytes

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

Try it online!

Answer 13

APL (Dyalog), 7 5 bytes

2 bytes saved thanks to @Mego

+.×⍨⍳

Try it online!

How?

⍳ - range

+.× - dot product

⍨ - with itself

Answer 14

CJam, 10 bytes

ri),{_*+}*

Try it online!

ri            e# Input integer n
  )           e# Add 1
   ,          e# Range [0 1 ... n]
    {   }*    e# Fold (reduce)
     _        e# Duplicate
      *       e# Multiply
       +      e# Add

Answer 15

Julia, 16 14 bytes

2 bytes saved thanks to @MartinEnder

!n=(x=1:n)⋅x

Try it online!

How?

(x=1:n) creates a range of 1 to n and assign to x, ⋅ dot product with x.

Answer 16

Actually, 3 bytes

R;*

Try it online!

Takes N as input, and outputs the Nth element in the sequence.

Explanation:

R;*
R    range(1, N+1) ([1, 2, ..., N])
 ;*  dot product with self

Answer 17

Befunge, 16 bytes

&::2*1+*\1+*6/.@

Using the closed-form formula n*(n+1)*(2n+1)/6.

Try it online!

Befunge, 38 bytes

v>::*\1-:v0+_$.@
&^ <     _\:^
>:#^_.@

Using a loop.

Try it online!

Answer 18

R, 17 bytes

sum((0:scan())^2)

Pretty straightforward, it takes advantage from the fact that ^ (exponentiation) is vectorized in R.

Answer 19

CJam, 9 bytes

ri),_.*:+

Try it online!

Explanation

ri        e# Read input and convert to integer N.
  ),      e# Get range [0 1 2 ... N].
    _     e# Duplicate.
     .*   e# Pairwise products, giving [0 1 4 ... N^2].
       :+ e# Sum.

Alternatively:

ri),2f#:+

This squares each element by mapping 2# instead of using pairwise products. And just for fun another alternative which gets inaccurate for large inputs because it uses floating-point arithmetic:

ri),:mh2#

Answer 20

Cubix, 15 bytes

Iu):^\+*p*6u@O,

Try it online!

My code is a bit sad ):

Computes n*(n+1)*(2n+1)/6

    I u
    ) :
^ \ + * p * 6 u
@ O , . . . . .
    . .
    . .

^Iu : read in input, u-turn
    : stack  n
:)\ : dup, increment, go right..oh, hey, it cheered up!
    : stack: n, n+1
+   : sum
    : stack: n, n+1, 2*n+1
*   : multiply
    : stack: n, n+1, 2*n+1, (n+1)*(2*n+1)
p   : move bottom of stack to top
    : stack: n+1, 2*n+1, (n+1)*(2*n+1), n
*   : multiply
6   : push 6
u   : right u-turn
,   : divide
O   : output
@   : terminate

Answer 21

><>, 15 13 11 bytes

Saved 2 bytes thanks to Not a tree

0:n:l1-:*+!

Try it online!

Outputs the sequence indefinitely.

Answer 22

Labyrinth, 11 bytes

:!\
+ :
*:#

Try it online!

Prints the sequence indefinitely.

Explanation

The instruction pointer just keeps running around the square of code over and over:

:!\    Duplicate the last result (initially zero), print it and a linefeed.
:      Duplicate the result again, which increases the stack depth.
#      Push the stack depth (used as a counter variable).
:*     Square it.
+      Add it to the running total.

Answer 23

Pyth, 7 5 bytes thanks to Steven H

s^R2h

Explanation:

s^R2h       Full program - inputs from stdin and outputs to stdout
s           output the sum of
    h       range(input), with
 ^R2         each element squared

---

My first solution

sm*ddUh

Try it online!

Explanation:

sm*ddUh    Full program - inputs from stdin and outputs to stdout
s          sum of
 m   Uh    each d in range(input)
  *dd      squared

Answer 24

Haskell, 20 bytes

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

Try it online!

Answer 25

Hexagony, 23 bytes

?'+)=:!@/*"*'6/{=+'+}/{

Try it online!

Explanation

Unfolded:

   ? ' + )
  = : ! @ /
 * " * ' 6 /
{ = + ' + } /
 { . . . . .
  . . . . .
   . . . .

This is really just a linear program with the / used for some redirection. The linear code is:

?'+){=+'+}*"*'6{=:!@

Which computes n (n+1) (2n+1) / 6. It uses the following memory edges:

Where the memory point (MP) starts on the edge labelled n, pointing north.

?   Read input into edge labelled 'n'.
'   Move MP backwards onto edge labelled 'n+1'.
+   Copy 'n' into 'n+1'.
)   Increment the value (so that it actually stores the value n+1).
{=  Move MP forwards onto edge labelled 'temp' and turn around to face
    edges 'n' and 'n+1'.
+   Add 'n' and 'n+1' into edge 'temp', so that it stores the value 2n+1.
'   Move MP backwards onto edge labelled '2n+1'.
+   Copy the value 2n+1 into this edge.
}   Move MP forwards onto 'temp' again.
*   Multiply 'n' and 'n+1' into edge 'temp', so that it stores the value
    n(n+1).
"   Move MP backwards onto edge labelled 'product'.
*   Multiply 'temp' and '2n+1' into edge 'product', so that it stores the
    value n(n+1)(2n+1).
'   Move MP backwards onto edge labelled '6'.
6   Store an actual 6 there.
{=  Move MP forwards onto edge labelled 'result' and turn around, so that
    the MP faces edges 'product' and '6'.
:   Divide 'product' by '6' into 'result', so that it stores the value
    n(n+1)(2n+1)/6, i.e. the actual result.
!   Print the result.
@   Terminate the program.

In theory it might be possible to fit this program into side-length 3, because the / aren't needed for the computation, the : can be reused to terminate the program, and some of the '"=+*{ might be reusable as well, bringing the number of required commands below 19 (the maximum for side-length 3). I doubt it's possible to find such a solution by hand though, if one exists at all.

Answer 26

Excel, 19 bytes

=A1^3/3+A1^2/2+A1/6

Answer 27

Oasis, 4 bytes

nk+0

Try it online!

Explanation

   0      # a(0) = 0
nk+       # a(n) = a(n-1)+n^2

Answer 28

Neim, 3 bytes

ᛦ

This could've been a challenge to show off Neim's polygonal number builtins, but apparently not.

Try it online!

Answer 29

Brachylog, 5 bytes

⟦^₂ᵐ+

Try it online!

Explanation

⟦        Range
 ^₂ᵐ     Map square
    +    Sum

Answer 30

Gaia, 3 bytes

s¦Σ

Try it online!

Answer 31

Perl 5, 24, 21 bytes

Thanks to Dom, solutions with 21 bytes

$\+=$_--**2while$_}{

or

map$\+=$_**2,1..$_}{

or

$\+=$_**2for 1..$_}{

previous were 21 + 1 -p flag, 3 bytes saved thanks to Xcali

$_*=($_+1)*(2*$_+1)/6

and 23 +1

$_=$_*($_+1)*(2*$_+1)/6

or

$_=$_**3/3+$_**2/2+$_/6

Answer 32

awk, 25 bytes

{print$1^3/3+$1^2/2+$1/6}

Answer 33

J, 10 9 bytes

4%~3!2*>:

Try it online!

Saved 1 byte using an explicit formula, thanks to miles!

---

J, 10 bytes

1#.]*:@-i.

J has a range function, but this gives us number from 0 to N-1. To remedy this, we can just take the argument and subtract the range from it, giving us a range from N to 1. This is done with ] -i.. The rest of the code simply square this list argument (*:@) and then sums it (1#.).

Other contenders

11 bytes: 1#.*:@i.@>:

13 bytes: 1#.[:,@:*/~i.

Answer 34

dc, 15 12 bytes

d1+dd+1-**6/

This is simply the factorised Faulhaber polynomial: n(n+1)(2n+1)/6, except that the (2n+1) term is calculated as 2(n+1)-1.

Answer 35

Funky, 33 24 19 bytes

-9 bytes thanks to Dennis

-5 bytes thanks to ASCII-only and bugfixes.

f=n=>n?f(n-1)+n*n:0

Try it online!

Answer 36

Befunge, 28 bytes

&00pg#v_.@
-1:g00<^g00+*:p00

Works for inputs in the range [0, 128). Due to befunge being entirely stack-based, and yet having limited stack manipulation operations available, the only way to work with three values (sum, partial sum, and counter) is to store a value temporarily by modifying the program itself using the p instruction. Since p assigns a value as ASCII, the stored value wraps around at 128, storing a negative value instead of a positive value.

Try it online!

Befunge, 30 bytes

& >::*\:v
$<^-1_v#<@.
_^#:\+<\

Works for pretty much any input.

Try it online!

Answer 37

Python, 38 33 bytes

-5 bytes thanks to Martin Ender

g=lambda x:x*x+g(x-1) if x else 0

Haven't seen any solution with recursion yet, so I thought I'd post this one. It may not be really competitive, but this is my first time posting.

Edit: It would've been -11 bytes thanks to Martin Ender, but I would've ended up with the same answer as Mr. Xcoder's

Answer 38

Proton, 16 bytes

x=>x*(x+++x)*x/6

Try it online!

Straightforward Solution: range+map(x=>x**2)+sum

-2 bytes thanks to Arnauld('s answer)

Answer 39

QBIC, 13 bytes

[:|p=p+a^2]?p

Explanation

[:|    FOR a = 1 to <input from cmd line>
p=p+   increment p by
a^2    a squared
]      NEXT
?p     PRINT p

Answer 40

Java 8, 78 57 35 16 bytes

@Arnauld port

i->i*(i+++i)*i/6

Answer 41

Pyke, 4 bytes

hLXs

Try it here!

or...

SMXs

Answer 42

4, 39 bytes

3.7006110180020100000030301100001195034

Try it online!

How?

3.
7 00        grid[0] = input()
6 11 01     grid[11] = 1
8 00        while grid[0] != 0:
2 01 00 00     grid[1] = grid[0] * grid[0]
0 03 03 01     grid[3] = grid[3] + grid[1]
1 00 00 11     grid[0] = grid[0] - grid[11]
9         
5 03        print(grid[3]) 
4

Answer 43

Octave, 14 bytes

@(k)(a=0:k)*a'

Try it online!

*Dot product.

Or

@(k)sumsq(0:k)

Try it online!

Answer 44

Haskell, 21 bytes

f n=n*(n+1)*(2*n+1)/6

Try it online!

Boring, straightforward implementation of the closed-form formula on the OEIS page. Expanding into polynomial form doesn't save any bytes: f n=(2*n^3+3*n^2+n)/6.

Answer 45

Tcl, 36 bytes

proc S n {expr $n*($n+1)*(2*$n+1)/6}

Try it online!

Answer 46

ARBLE, 14 13 bytes

-1 bytes thanks to Mr. Xcoder

n*~n*~(n+n)/6

Try it online!

Answer 47

Alice, 17 12 bytes

./ O \d2E+.

Try it online!

Prints the sequence indefinitely.

The trailing linefeed is significant.

Explanation

.   Duplicate the current total. Initially this pushes two zeros onto the
    previously empty stack.
/   Switch to Ordinal.
O   Print the current total with a trailing linefeed.
\   Switch back to Cardinal.
d   Push the stack depth, which acts as a counter variable.
2E  Square it.
+   Add it to the current total.
.   Duplicate it to increment the stack depth for the next iteration.

Answer 48

Ruby, 18 bytes

->n{n*~n*~(n+n)/6}

Using the sama formula as everybody else, saved 1 byte with a double negative multiplication.

Try it online!

Answer 49

Retina, 20 bytes

.+
$*
M!&`.+
1
$%_
1

Try it online!

Explanation

.+
$*

Convert input to unary.

M!&`.+

Get all overlapping matches of .+ which turns the input into a range of unary numbers from 1 to n.

1
$%_

Replace each 1 with the entire line its on, squaring the unary value on each line.

1

Count the number of 1s, summing all lines and converting them back to decimal.

Answer 50

Pari/GP, 16 bytes

n->(v=[0..n])*v~

Try it online!

Answer 51

cQuents 0, 3 bytes

;$$

Try it online!

Take that, Oasis.

Explanation

;     Mode: Sum - given n, output the sum of the sequence up to n
 $$   Each term in the sequence equals the index * the index

Answer 52

Recursiva, 8 bytes

smBa'Sa'

Try it online!

Answer 53

Bash, 48 30 bytes

echo "$1*($1+1)*(2*$1+1)/6"|bc

Try it online!

Answer 54

Pyth, 10 bytes

VhQ=+Z^N2Z

Outputs the first N elements of the sequence.

Try it online!

---

How?

VhQ=+Z^N2Z              Full program

VhQ                     Loop until Q + 1 is reached
   =+Z                  Assign and increment Z by ...
      ^N2               ... N squared
         Z              Implicity prints Z

---

11 byte alternative.

VhQ aY^N2sY

Try it online!

Answer 55

Pushy, 6 bytes

R2KeS#

Try it online!

Explanation

R     Push integers from 1 to implicit n, where n is implicit input
2     Push 2
K     Next command will use the entire stack
e     Pop 2. Raise each of the remaing stack entries to that
S     Sum the entire stack
#     Print as integer

Answer 56

Jq 1.5, 20 bytes

[range(.+1)|.*.]|add

Input is N. Output is N'th element of sequence. Expanded:

[
    range(.+1)            # generate series 0, 1, 2, 3, ... N
  | .*.                   # square each term
]                         # collect into an array
| add                     # compute the sum

Try it online!

Answer 57

K (oK), 9 bytes

Solution:

+/x*x:!1+

Try it online!

Examples:

> +/x*x:!1+4
30
> +/x*x:!1+5
55

Explanation:

Evaluated right-to-left:

+/x*x:!1+ / solution
       1+ / add 1 to input, 1+4 => 5
      !   / til, !5 => 0 1 2 3 4
    x:    / save as variable x
  x*      / vectorised multiplication, x*x, 0 1 2 3 4*0 1 2 3 4 => 0 1 3 9 16
+/        / addition over (sum), +/0 1 3 9 16 => 30

Answer 58

C#(.NET Core), 20 bytes

a=>a*(a+1)*(2*a+1)/6

Try it online!

Answer 59

R, 16 bytes

(n=0:scan())%*%n

Try it online!

Answer 60

TacO, 10 bytes

@i i
+%+*i

Try it online!

Answer 61

Threead, 40 bytes

I[     -]_
   _^>1
  _2 _r

[<

 lo
 ]
+

Try it online!

De-sliced

This is actually 4 bytes larger, due to whitespace.

I[     -]_   lo
   _^>1   [< ]
  _2 _r     +

Process.

1. Read an integer into the first value
2. While the first value is non-zero.
   1. Blank the third value, push 2 onto it.
   2. Blank the second value, then calculate the first value to the power of the third, (i^2)
   3. Move the second value to the right and place a 1 onto it.
   4. Put the value of the first value onto the third value.
   5. Calculate the third value minus the second value (i-1)
3. Now the second tape contains the squares of all numbers from inp to 1 squared.
4. Blank the first value
5. While the second value is non-zero
   1. Move the second tape to the left.
   2. Add the first value to the second value into the third value.
   3. Move the third value into the first value
6. Now, all the i^2 values are summed together, output the first value.

Answer 62

ReRegex, 53 bytes

#import math
^k([1-9]\d*)/k($1-1)+$1*$1/k0+/0/k#input

Explained

#import math                # Import the math library.
^k([1-9]\d*)/k($1-1)+$1*$1/ # Search the buffer for "k" followed by a non-zero number. Replace it with "k<n-1>+n*n"
k0+/0/                      # Replace k0 with just 0
k#input                     # Default the buffer to k<input>

Try it online!

Answer 63

TI-BASIC, 10 bytes

sum(seq(X²,X,0,Ans

Quite simple. The sum of the sequence of the first Ans squares.

Answer 64

MaybeLater, 36 bytes

n=(readn)/1
write(n*(n+1)*(1+n*2)/6)

Using the closed form formula.

Try it online!

Answer 65

Common Lisp, 30 bytes

(loop as i to(read)sum(* i i))

Try it online!

Answer 66

JavaScript (ES6), 18 bytes

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

---

Try it

o.innerText=(
f=n=>n&&n*n--+f(n)
)(i.value=8);oninput=_=>o.innerText=f(+i.value)

<input id=i type=number><pre id=o>

Answer 67

TI-Basic, 10 bytes

sum(seq(OO,O,0,Ans

sp00ky

Answer 68

Milky Way, 10 bytes

'L§{:*}G!

Try it online!

Explanation:

'          push input to stack
 L         push inclusive range
  ${  }    map over every element ...
    :*     ... duplicate and multiply (square it)
       G   sum of list
        !  output

Answer 69

Add++, 13 bytes

D,f,@,RdBcB*s

Try it online!

D,f,@,   - Create a monadic function called f. Argument n (example: 5)
      R  - Generate range;   STACK = [[1 2 3 4 5]]
      d  - Duplicate;        STACK = [[1 2 3 4 5] [1 2 3 4 5]]
      Bc - Push the columns; STACK = [[1 1] [2 2] [3 3] [4 4] [5 5]]
      B* - Product of each;  STACK = [1 4 9 16 25]
      s  - Push the sum;     STACK = [1 4 9 16 25 55]
         - Implicitly return 55

Answer 70

Jelly, 3 bytes

Rḋ`

Try it online!

Answer 71

Postscript, 149 bytes

/d {def}def
/a {
0 i l 1 0 l
i 2 ge {/i i 1 sub d a}if
} d
/l {rlineto}d
/m {moveto}d
/b {/j exch d /i j d 0 0 m j a j 0 lineto closepath fill}d
5 b

Answer 72

I'm a beginner in programming. I just do that for fun and I'm thankful for any help.

I tried in JavaScript:

var x = 9;

var y = 0;

var z;

for(var i=0; i<=x; i++){z = i*i; y += z;}

console.log(y)