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

Mr. Xcoder · Accepted Answer · 2017-11-04 07:03:09Z

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.}

\$\begingroup\$ This is very clever! \$\endgroup\$

Skyler
– Skyler

2017-10-17 16:00:25 +00:00
Commented Oct 17, 2017 at 16:00 — Skyler
– Skyler, Commented Oct 17, 2017 at 16:00

Arnauld · Accepted Answer · 2017-10-17 13:27:31Z

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'.}

DialFrost · Accepted Answer · 2022-10-11 07:16:21Z

MATL, 3 bytes

:Us

... or them?

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Explanation

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

Emigna · Accepted Answer · 2017-10-17 12:08:35Z

6

05AB1E, 3 bytes

ÝnO

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Explanation

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

edited Oct 17, 2017 at 12:08

answered Oct 17, 2017 at 12:03

Emigna

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Riley · Accepted Answer · 2017-10-17 13:28:37Z

6

Brain-Flak, 36 bytes

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

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# 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)

edited Oct 17, 2017 at 13:28

answered Oct 17, 2017 at 13:21

Riley

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\$\begingroup\$ Nicely done! :) I came up with ({<(({}))>{({})({}[()])}{}<({}[()])>}) for 38 \$\endgroup\$

DJMcMayhem
– DJMcMayhem

2017-10-17 16:20:03 +00:00
Commented Oct 17, 2017 at 16:20
\$\begingroup\$ 34 bytes ;) \$\endgroup\$

Wheat Wizard
– Wheat Wizard ♦

2017-10-17 19:00:05 +00:00
Commented Oct 17, 2017 at 19:00

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Wheat Wizard · Accepted Answer · 2017-10-18 02:47:02Z

Brain-Flak, 34 bytes

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

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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.}

I knew there had to be a way to do this, and I had a feeling you would be the one to figure out the math for it. — Riley
– Riley, Commented Oct 17, 2017 at 19:02

hyperneutrino · Accepted Answer · 2017-10-17 11:55:51Z

4

Jelly, 3 bytes

R²S

Try it online!

FGITW

Explanation

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

answered Oct 17, 2017 at 11:55

hyperneutrino♦

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\$\begingroup\$ A longer alternative would be Ræ.R \$\endgroup\$

Mr. Xcoder
– Mr. Xcoder

2017-10-17 12:40:49 +00:00
Commented Oct 17, 2017 at 12:40

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Mr. Xcoder · Accepted Answer · 2017-10-17 12:05:19Z

4

Husk, 3 bytes

ṁ□ḣ

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answered Oct 17, 2017 at 12:05

Mr. Xcoder

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Martin Ender · Accepted Answer · 2017-10-18 12:03:36Z

Hexagony, 23 bytes

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

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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.

Cinaski · Accepted Answer · 2017-10-17 12:47:04Z

3

Ohm v2, 3 bytes

#²Σ

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Explanation

# Range ² Square Σ Sum

edited Oct 17, 2017 at 12:47

answered Oct 17, 2017 at 12:00

Cinaski

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Erik the Outgolfer · Accepted Answer · 2017-10-18 11:21:37Z

3

Japt, 3 bytes

ô²x

Try it here.

-1 thanks to Shaggy.

Explanation:

ò²x ô² Map square on [0..input] x Sum

edited Oct 18, 2017 at 11:21

answered Oct 17, 2017 at 12:13

Erik the Outgolfer

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\$\begingroup\$ 3 bytes using the shortcut for p2. \$\endgroup\$

Shaggy
– Shaggy

2017-10-17 14:06:20 +00:00
Commented Oct 17, 2017 at 14:06
\$\begingroup\$ @FelipeNardiBatista fixed \$\endgroup\$

Erik the Outgolfer
– Erik the Outgolfer

2017-10-18 11:21:42 +00:00
Commented Oct 18, 2017 at 11:21

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hyperneutrino · Accepted Answer · 2017-10-17 12:02:26Z

2

Brain-Flak, 46 bytes

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

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answered Oct 17, 2017 at 12:02

hyperneutrino♦

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\$\begingroup\$ A little friendly competition :) \$\endgroup\$

Riley
– Riley

2017-10-17 13:29:25 +00:00
Commented Oct 17, 2017 at 13:29
\$\begingroup\$ @Riley Ooh nice :) \$\endgroup\$

hyperneutrino
– hyperneutrino ♦

2017-10-17 13:52:19 +00:00
Commented Oct 17, 2017 at 13:52
\$\begingroup\$ Rather than pushing the square onto the alternate stack, you can sum it directly by evaluating and not pushing it, then pushing the whole loop instead. This changes your second half to ({{({})({}[()])}{}}{}), and saves you 10 bytes. (If that doesn't make sense, ping me in the third stack) \$\endgroup\$

DJMcMayhem
– DJMcMayhem

2017-10-17 16:36:30 +00:00
Commented Oct 17, 2017 at 16:36

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Luis Mendo · Accepted Answer · 2017-10-17 12:05:42Z

2

CJam, 10 bytes

ri),{_*+}*

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

answered Oct 17, 2017 at 12:05

Luis Mendo

107k10 gold badges139 silver badges382 bronze badges

\$\begingroup\$ How about ri),_.*:+ or ri),2f#:+? \$\endgroup\$

Martin Ender
– Martin Ender

2017-10-17 13:15:46 +00:00
Commented Oct 17, 2017 at 13:15
\$\begingroup\$ @Martin Good idea! I think you should post it as a different answer \$\endgroup\$

Luis Mendo
– Luis Mendo

2017-10-17 13:25:02 +00:00
Commented Oct 17, 2017 at 13:25

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totallyhuman · Accepted Answer · 2017-10-17 12:10:41Z

2

Wolfram Language (Mathematica), 14 bytes

Tr[Range@#^2]&

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edited Oct 17, 2017 at 12:10

answered Oct 17, 2017 at 12:01

totallyhuman

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4

\$\begingroup\$ #.#&@Range@#& saves a byte \$\endgroup\$

miles
– miles

2017-10-17 12:25:09 +00:00
Commented Oct 17, 2017 at 12:25
5

\$\begingroup\$ #.#&@*Range . \$\endgroup\$

alephalpha
– alephalpha

2017-10-17 13:42:11 +00:00
Commented Oct 17, 2017 at 13:42

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

2

Brachylog, 5 bytes

⟦^₂ᵐ+

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Explanation

⟦ Range ^₂ᵐ Map square + Sum

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 17, 2017 at 12:13

Fatalize

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user45941 · Accepted Answer · 2017-10-17 12:31:49Z

Actually, 3 bytes

R;*

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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

Uriel · Accepted Answer · 2017-10-17 12:55:25Z

2

APL (Dyalog), 7 5 bytes

2 bytes saved thanks to @Mego

+.×⍨⍳

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How?

⍳ - range

+.× - dot product

⍨ - with itself

edited Oct 17, 2017 at 12:55

answered Oct 17, 2017 at 12:03

Uriel

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\$\begingroup\$ @Uriel my bad, i thought the ¨⍳ was necessary \$\endgroup\$

Felipe Nardi Batista
– Felipe Nardi Batista

2017-10-18 10:26:42 +00:00
Commented Oct 18, 2017 at 10:26

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

2

R, 17 bytes

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

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

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 17, 2017 at 13:14

Rudier

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1

\$\begingroup\$ (x=0:scan())%*%x is shorter by a byte, but I believe you need a cat to get output. \$\endgroup\$

Giuseppe
– Giuseppe

2017-10-17 14:21:20 +00:00
Commented Oct 17, 2017 at 14:21
\$\begingroup\$ @Giuseppe I've just tried it and your code works without cat, it outputs a 1x1 matrix. \$\endgroup\$

Rui Barradas
– Rui Barradas

2017-10-19 16:30:14 +00:00
Commented Oct 19, 2017 at 16:30
\$\begingroup\$ @RuiBarradas Current meta consensus is that cat is necessary for this to qualify as a full program. If you wish to change that, answer this question and get some traction among the other R people on the site. \$\endgroup\$

Giuseppe
– Giuseppe

2017-10-19 16:33:53 +00:00
Commented Oct 19, 2017 at 16:33

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Martin Ender · Accepted Answer · 2017-10-17 13:28:32Z

CJam, 9 bytes

ri),_.*:+

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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#

Uriel · Accepted Answer · 2017-10-17 14:19:25Z

Julia, 16 14 bytes

2 bytes saved thanks to @MartinEnder

!n=(x=1:n)⋅x

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How?

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

Martin Ender · Accepted Answer · 2017-10-17 15:05:09Z

Labyrinth, 11 bytes

:!\ + : *:#

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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.

Giuseppe · Accepted Answer · 2017-10-17 16:02:58Z

Cubix, 15 bytes

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

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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

Cinaski · Accepted Answer · 2017-10-17 16:18:58Z

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!

nimi · Accepted Answer · 2017-10-17 16:29:27Z

2

Haskell, 20 bytes

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

Try it online!

answered Oct 17, 2017 at 16:29

nimi

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Wernisch · Accepted Answer · 2017-10-18 11:15:04Z

2

Excel, 19 bytes

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

answered Oct 18, 2017 at 11:15

Wernisch

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Emigna · Accepted Answer · 2017-10-18 15:08:10Z

2

><>, 15 13 11 bytes

Saved 2 bytes thanks to Not a tree

0:n:l1-:*+!

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Outputs the sequence indefinitely.

edited Oct 18, 2017 at 15:08

answered Oct 17, 2017 at 14:18

Emigna

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1

\$\begingroup\$ 14 bytes (12+2 for -v flag): ::1+:}+**6,n (Try it online!) \$\endgroup\$

Not a tree
– Not a tree

2017-10-18 01:43:30 +00:00
Commented Oct 18, 2017 at 1:43
1

\$\begingroup\$ Or 11 bytes (prints forever, starting at N=1): Try it online! \$\endgroup\$

Not a tree
– Not a tree

2017-10-18 02:00:53 +00:00
Commented Oct 18, 2017 at 2:00
\$\begingroup\$ @Notatree: Very nice idea using l. Checking with OP if it is okay to start at 1. \$\endgroup\$

Emigna
– Emigna

2017-10-18 06:28:49 +00:00
Commented Oct 18, 2017 at 6:28
\$\begingroup\$ @Notatree: Unfortunately we aren't allowed to start at 1, but it still saves 2 bytes. Thanks! \$\endgroup\$

Emigna
– Emigna

2017-10-18 09:52:08 +00:00
Commented Oct 18, 2017 at 9:52
1

\$\begingroup\$ (I should mention that I got the l idea from Martin Ender's Labyrinth answer.) \$\endgroup\$

Not a tree
– Not a tree

2017-10-18 14:59:52 +00:00
Commented Oct 18, 2017 at 14:59

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

2

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

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 17, 2017 at 15:26

Dave

4822 silver badges10 bronze badges

\$\begingroup\$ Is there no square built in in Pyth? \$\endgroup\$

caird coinheringaahing
– caird coinheringaahing ♦

2017-10-17 15:54:32 +00:00
Commented Oct 17, 2017 at 15:54
\$\begingroup\$ Not as far as I know... \$\endgroup\$

Dave
– Dave

2017-10-17 16:01:34 +00:00
Commented Oct 17, 2017 at 16:01
\$\begingroup\$ No there is no square built-in Pyth. Also 6 bytes \$\endgroup\$

Mr. Xcoder
– Mr. Xcoder

2017-10-17 16:15:48 +00:00
Commented Oct 17, 2017 at 16:15
1

\$\begingroup\$ 5 bytes. \$\endgroup\$

Steven H.
– Steven H.

2017-10-17 18:48:35 +00:00
Commented Oct 17, 2017 at 18:48
\$\begingroup\$ Fixable with +1 byte to each answer, I'll edit once I get off of mobile. \$\endgroup\$

Dave
– Dave

2017-10-18 10:55:44 +00:00
Commented Oct 18, 2017 at 10:55

| Show 2 more comments

Conor O'Brien · Accepted Answer · 2017-10-19 13:53:18Z

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.

\$\begingroup\$ 4%~3!2*>: saves a byte \$\endgroup\$

miles
– miles

2017-10-19 11:45:38 +00:00
Commented Oct 19, 2017 at 11:45 — miles
– miles, Commented Oct 19, 2017 at 11:45

The Thonnu · Accepted Answer · 2023-06-10 16:53:18Z

Thunno 2 `S`, 2 bytes

R²

Attempt This Online!

Explanation

R² # Implicit input R # Range of input ² # Square each # Sum the list # Implicit output

alephalpha · Accepted Answer · 2023-06-12 03:56:50Z

Nekomata, 3 bytes

R:∙

Attempt This Online!

R:∙ R Range : Duplicate ∙ Dot product

Because there isn't a square built-in yet.

Square pyramidal numbers

Task

Example:

Specification:

78 Answers 78

Python 2, 22 bytes

Python 2, 27 bytes

JavaScript (ES6), 16 bytes

Demo

How?

MATL, 3 bytes

Explanation

05AB1E, 3 bytes

Brain-Flak, 36 bytes

Brain-Flak, 34 bytes

How does it work?

Brain-Flak, 36 bytes

Jelly, 3 bytes

Explanation

Husk, 3 bytes

Hexagony, 23 bytes

Explanation

Ohm v2, 3 bytes

Japt, 3 bytes

Brain-Flak, 46 bytes

CJam, 10 bytes

Wolfram Language (Mathematica), 14 bytes

Brachylog, 5 bytes

Explanation

Actually, 3 bytes

APL (Dyalog), 7 5 bytes

R, 17 bytes

CJam, 9 bytes

Explanation

Julia, 16 14 bytes

Labyrinth, 11 bytes

Explanation

Cubix, 15 bytes

Befunge, 16 bytes

Befunge, 38 bytes

Haskell, 20 bytes

Excel, 19 bytes

><>, 15 13 11 bytes

Pyth, 7 5 bytes thanks to Steven H

J, 10 9 bytes

J, 10 bytes

Other contenders

Thunno 2 S, 2 bytes

Explanation

Nekomata, 3 bytes

Related

Hot Network Questions

Thunno 2 `S`, 2 bytes