Square pyramidal numbers

Question

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.

Fun observation from OEIS: This sequence contains exactly two perfect squares: f(1) == 1 * 1 (1), and f(24) == 70 * 70 (4900). — DJMcMayhem
– DJMcMayhem, Commented Oct 17, 2017 at 16:31
@Emigna sorry but no, you need to start from f(0) = 0. i've pointed out that to the few answers that failed that requirement — Felipe Nardi Batista
– Felipe Nardi Batista, Commented Oct 18, 2017 at 9:44

Sean · Accepted Answer · 2023-06-12 15:09:57Z

Raku, 16 bytes

[\+] {$++²}...*

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This is an expression for the lazy infinite sequence of numbers.

{ $++² } ... * is the infinite sequence of square numbers. $ is an anonymous state variable whose postincremented value is squared to obtain each new element.
[\+] is the sequence of partial sums of that sequence.

Okx · Accepted Answer · 2017-10-17 12:09:57Z

Neim, 3 bytes

𝐈ᛦ𝐬

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

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Emigna · Accepted Answer · 2017-10-17 12:10:37Z

1

Oasis, 4 bytes

nk+0

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Explanation

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

edited Oct 17, 2017 at 12:10

answered Oct 17, 2017 at 12:04

Emigna

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Mr. Xcoder · Accepted Answer · 2017-10-17 12:17:14Z

1

Gaia, 3 bytes

s¦Σ

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answered Oct 17, 2017 at 12:17

Mr. Xcoder

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G B · Accepted Answer · 2017-10-17 13:32:18Z

Ruby, 18 bytes

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

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

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Community · Accepted Answer · 2020-06-17 09:04:33Z

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.

Nahuel Fouilleul · Accepted Answer · 2017-10-18 10:42:36Z

1

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

edited Oct 18, 2017 at 10:42

answered Oct 17, 2017 at 14:24

Nahuel Fouilleul

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\$\begingroup\$ You can shorten this by factoring out the $_ and replacing = with *=: Try it online! \$\endgroup\$

Xcali
– Xcali

2017-10-17 15:44:00 +00:00
Commented Oct 17, 2017 at 15:44
\$\begingroup\$ This looks to be 22 bytes (21 + 1 for -p)... But here's one for 21 bytes using a different approach entirely: Try it online! \$\endgroup\$

Dom Hastings
– Dom Hastings

2017-10-18 10:13:47 +00:00
Commented Oct 18, 2017 at 10:13
1

\$\begingroup\$ good point, there is also $\+=$_--**2while$_}{ \$\endgroup\$

Nahuel Fouilleul
– Nahuel Fouilleul

2017-10-18 10:40:28 +00:00
Commented Oct 18, 2017 at 10:40

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Community · Accepted Answer · 2020-06-17 09:04:33Z

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.

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Befunge, 30 bytes

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

Works for pretty much any input.

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ATaco · Accepted Answer · 2017-10-21 02:45:20Z

1

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

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edited Oct 21, 2017 at 2:45

answered Oct 19, 2017 at 5:17

ATaco

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\$\begingroup\$ 26 bytes \$\endgroup\$

Dennis
– Dennis

2017-10-19 05:53:13 +00:00
Commented Oct 19, 2017 at 5:53

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pCozmic · Accepted Answer · 2017-11-01 12:30:20Z

1

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

edited Nov 1, 2017 at 12:30

answered Nov 1, 2017 at 11:01

pCozmic

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\$\begingroup\$ Welcome to PPCG! Here are a few ideas: x**2 is x*x, you can avoid the parentheses if you just put the condition on top and avoid the <1 if you switch the true and false branches (because 0 is falsy). However, y if x else 0 can be expressed more concisely as x and y, due to the short-circuiting behaviour of and. So you end up with g=lambda x:x and x*x+g(x-1). :) \$\endgroup\$

Martin Ender
– Martin Ender

2017-11-01 11:20:31 +00:00
Commented Nov 1, 2017 at 11:20
\$\begingroup\$ @Martin Ender Thank you for the tips. I am going to simplify the **2 and the <1, but I won't replace the ternary operator with the "and" statement, since it will be the same as Mr. Xcoder's second answer. \$\endgroup\$

pCozmic
– pCozmic

2017-11-01 12:16:59 +00:00
Commented Nov 1, 2017 at 12:16

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Mr. Xcoder · Accepted Answer · 2018-10-19 22:12:34Z

1

Jelly, 3 bytes

Rḋ`

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answered Oct 19, 2018 at 22:12

Mr. Xcoder

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Bubbler · Accepted Answer · 2020-08-14 01:01:53Z

Flurry, 30 bytes

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

Run example

$ ./flurry -nin -c "{}{({})[<><<>()>]<[][]>}[<>()]" 0 0 $ ./flurry -nin -c "{}{({})[<><<>()>]<[][]>}[<>()]" 1 1 $ ./flurry -nin -c "{}{({})[<><<>()>]<[][]>}[<>()]" 3 14 $ ./flurry -nin -c "{}{({})[<><<>()>]<[][]>}[<>()]" 6 91

The computation is done via the stack, but the final result is outputted from the return value.

Given that we can keep track of iteration indexes using the stack height and multiplication is cheap in Flurry, there's really no "clever" alternative that can lead to shorter code.

{}{...}[<>()] Repeat the lambda n times with the start value of zero (implicit) push argument x to the stack ({}) Pop and re-push x; return x [<><<>()>] succ <[][]> (height ∘ height) In effect, push x and return x + height^2

\$\begingroup\$ "succ" hmmm.... \$\endgroup\$

Razetime
– Razetime

2020-09-22 08:14:42 +00:00
Commented Sep 22, 2020 at 8:14 — Razetime
– Razetime, Commented Sep 22, 2020 at 8:14

DialFrost · Accepted Answer · 2022-10-11 07:18:14Z

Vyxal, 3 bytes

ɾ²∑

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Explained

ɾ²∑ ɾ # range 1 inclusive [1...n] ² # square ∑ # sum

Kip the Malamute · Accepted Answer · 2022-12-17 15:23:09Z

Pyt, 3 bytes

ř²Ʃ

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ř²Ʃ implicit input ř creates array of 1 to n ² squares each element Ʃ sums the list implicit print

chunes · Accepted Answer · 2023-06-10 20:31:06Z

1

Arturo, 19 bytes

$=>[∑map&=>[&^2]]

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answered Jun 10, 2023 at 20:31

chunes

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Nahuel Fouilleul · Accepted Answer · 2025-04-01 09:15:38Z

1

AWK, 25 23 bytes

Thanks to xrs

1,$0=$1^3/3+$1^2/2+$1/6

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edited Apr 1 at 9:15

answered Oct 17, 2017 at 14:41

Nahuel Fouilleul

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\$\begingroup\$ I like the math. 1,$0=$1^3/3+$1^2/2+$1/6 shaves a couple bytes. \$\endgroup\$

xrs
– xrs

2025-03-31 17:53:36 +00:00
Commented Mar 31 at 17:53

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steenbergh · Accepted Answer · 2017-10-17 12:04:12Z

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

hyperneutrino · Accepted Answer · 2017-10-17 12:10:29Z

Proton, 16 bytes

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

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

-2 bytes thanks to Arnauld('s answer)

Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Pyke, 4 bytes

hLXs

Try it here!

or...

SMXs

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 17, 2017 at 12:20

Mr. Xcoder

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Uriel · Accepted Answer · 2017-10-17 12:23:28Z

4, 39 bytes

3.7006110180020100000030301100001195034

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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

Roberto Graham · Accepted Answer · 2017-10-17 12:36:45Z

0

Java 8, 78 57 35 16 bytes

@Arnauld port

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

edited Oct 17, 2017 at 12:36

answered Oct 17, 2017 at 12:18

Roberto Graham

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Octave, 14 bytes

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

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*Dot product.

Or

@(k)sumsq(0:k)

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edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 17, 2017 at 12:29

rahnema1

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user45941 · Accepted Answer · 2017-10-17 12:56:12Z

Haskell, 21 bytes

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

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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.

\$\begingroup\$ Same byte count: f n=sum$map(^2)[1..n]. \$\endgroup\$

Laikoni
– Laikoni

2017-10-17 13:37:29 +00:00
Commented Oct 17, 2017 at 13:37 — Laikoni
– Laikoni, Commented Oct 17, 2017 at 13:37

sergiol · Accepted Answer · 2017-10-17 13:03:57Z

0

Tcl, 36 bytes

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

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answered Oct 17, 2017 at 13:03

sergiol

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\$\begingroup\$ Longer approach, that does not work well with zero: tio.run/##K0nO@f@/oCg/… \$\endgroup\$

sergiol
– sergiol

2017-10-17 13:20:16 +00:00
Commented Oct 17, 2017 at 13:20
\$\begingroup\$ Another longer approach, that does not work well with zero: tio.run/##K0nO@f@/oCg/… \$\endgroup\$

sergiol
– sergiol

2017-10-17 13:21:08 +00:00
Commented Oct 17, 2017 at 13:21
\$\begingroup\$ Yet another longer approach, that does not work well with zero: tio.run/##K0nO@f@/oCg/… \$\endgroup\$

sergiol
– sergiol

2017-10-17 13:35:36 +00:00
Commented Oct 17, 2017 at 13:35
\$\begingroup\$ The same approach as the Javascript answer has one byte more: tio.run/##K0nO@f@/oCg/… Tks, @Arnauld \$\endgroup\$

sergiol
– sergiol

2017-10-17 14:32:18 +00:00
Commented Oct 17, 2017 at 14:32
\$\begingroup\$ Alternative approach, 50 bytes. ##### Requires Tcl >= 8.7 to allow use of lseq. 50 bytes: proc S n {expr [lmap i [lseq 0 $n] {list +$i**2}]} \$\endgroup\$

sergiol
– sergiol

2025-11-13 22:20:30 +00:00
Commented Nov 13 at 22:20

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Martin Ender · Accepted Answer · 2017-10-17 13:32:36Z

Retina, 20 bytes

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

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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.

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

alephalpha · Accepted Answer · 2017-10-17 13:42:21Z

0

Pari/GP, 16 bytes

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

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answered Oct 17, 2017 at 13:42

alephalpha

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

cQuents 0, 3 bytes

;$$

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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

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 17, 2017 at 13:48

Stephen

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\$\begingroup\$ does not work for input 0 \$\endgroup\$

Felipe Nardi Batista
– Felipe Nardi Batista

2017-10-18 09:39:23 +00:00
Commented Oct 18, 2017 at 9:39

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0xffcourse · Accepted Answer · 2017-10-17 14:18:08Z

0

Recursiva, 8 bytes

smBa'Sa'

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answered Oct 17, 2017 at 14:18

0xffcourse

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Martin Ender · Accepted Answer · 2017-10-17 15:11:21Z

Alice, 17 12 bytes

./ O \d2E+.

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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.

Ian H. · Accepted Answer · 2017-10-17 15:27:14Z

Pyth, 10 bytes

VhQ=+Z^N2Z

Outputs the first N elements of the sequence.

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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

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Square pyramidal numbers

Task

Example:

Specification:

78 Answers 78

Raku, 16 bytes

Neim, 3 bytes

Oasis, 4 bytes

Gaia, 3 bytes

Ruby, 18 bytes

dc, 15 12 bytes

Perl 5, 24, 21 bytes

Befunge, 28 bytes

Befunge, 30 bytes

Funky, 33 24 19 bytes

Python, 38 33 bytes

Jelly, 3 bytes

Flurry, 30 bytes

Run example

Vyxal, 3 bytes

Explained

Pyt, 3 bytes

Arturo, 19 bytes

AWK, 25 23 bytes

QBIC, 13 bytes

Explanation

Proton, 16 bytes

Pyke, 4 bytes

4, 39 bytes

Java 8, 78 57 35 16 bytes

Octave, 14 bytes

Haskell, 21 bytes

Tcl, 36 bytes

Retina, 20 bytes

Explanation

Pari/GP, 16 bytes

cQuents 0, 3 bytes

Explanation

Recursiva, 8 bytes

Alice, 17 12 bytes

Explanation

Pyth, 10 bytes

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