Find the \$\left(n^2\right)^\text{th }n\$-gonal number

7

Given a non-negative integer, \$n\$, yield the \$(n^2)^\text{th } n\$-gonal number.

Further Detail:

The \$x\$-gonal numbers, or polygonal numbers, are also known as the two-dimensional figurate numbers.

Many people will be familiar with the triangular numbers, these are the \$3\$-gonal numbers:

$$F(3,n)=\sum_{i=1}^{n}(i)=\frac{n(n+1)}{2}$$

The triangular numbers are OEIS A000217:

0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, ...

Probably even more will be familiar with the square numbers, these are the \$4\$-gonal numbers:

$$F(4,n)=\sum_{i=1}^{n}(2i-1)=n^{2}$$

The square numbers are OEIS A000290:

0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, ...

In general the \$k^\text{th }x\$-gonal number is the number of pebbles required to iteratively build up an \$x\$-sided polygon by adding pebbles to \$x-2\$ of the sides, here are a few formulae:

\begin{align} F(x,k)&=\sum_{i=1}^{k}\left(1+\sum_{j=2}^{i}(x-2)\right) \\ &=\frac{k^2(x-2)-k(x-4)}{2} \\ &=\frac{k(k-1)}{2}(x-2)+k\\ &=(x-3)\sum_{i=1}^{k-1}(i)+\sum_{i=1}^{k}(i)\\ &=(x-2)(k-1)+1+F(x,k-1) \end{align}

I'm sure there are plenty more.
Note that the last one is recursive but that \$F(x,0)=0\$.

The challenge

...is to golf code for \$G(n)=F(n,n^2)\$ for non-negative \$n\$.
i.e. Given a non-negative integer, \$n\$, yield the \$(n^2)^\text{th } n\$-gonal number.
This is not (currently) in the OEIS:

0, 1, 4, 45, 256, 925, 2556, 5929, 12160, 22761, 39700, 65461, 103104, 156325, 229516, 327825, 457216, 624529, 837540, 1105021, ...

Notes

You may yield a list of these numbers up to and including the required one if preferred.
For example, given an input of 5 you may yield either:

  • the integer 925, or
  • the list [0, 1, 4, 45, 256, 925]
  • ...but not [0, 1, 4, 45, 256] or [1, 4, 45, 256, 925]

Results may also be results of floating point calculation and may deviate as such, so long as infinite precision floating point arithmetic would yield correct results.

Win by creating the shortest code in bytes in a language. The overall winner will be the shortest across all languages, but please don't let golfing languages dissuade you from entering in your favourite language - the primary goals are to challenge yourself and have fun!

Jonathan Allan

Posted 2018-08-26T20:30:26.683

Reputation: 67 804

Results may also be results of floating point calculation and may deviate as such -- are possible integer overflows also acceptable? – Jonathan Frech – 2018-08-27T00:07:39.793

@JonathanFrech indeed, that would certainly be a default (the floating point stuff probably is too, but thought it best to say since it's an integer based challenge) – Jonathan Allan – 2018-08-27T00:17:03.593

Answers

6

Wolfram Language (Mathematica), 19 bytes

#(4-#-2#^2+#^3)#/2&

Try it online!

Just a golfy version of the second formula.

With PolygonalNumber built-in, 23 bytes

PolygonalNumber[#,#^2]&

Try it online!

As expected, the Mathematica built-in is longer than the golfed version.

JungHwan Min

Posted 2018-08-26T20:30:26.683

Reputation: 13 290

5

Neim, 2 bytes

ᛦℙ

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¯\_(ツ)_/¯ First Neim answer, right tool for the job. By the way, the advice I give on my About me page still holds for my own posts: Don't upvote trivial solutions, please.

Mr. Xcoder

Posted 2018-08-26T20:30:26.683

Reputation: 39 774

4Finding the right language is a +1 from me :p – Jonathan Allan – 2018-08-26T21:11:38.150

Upvoted because when I saw the question I immediately thought "oh, Neim can do this in 2 bytes". But someone beat me to it! – Okx – 2018-08-27T13:54:00.637

Upvoted because I am oppositional by nature. – ngm – 2018-08-27T14:53:13.780

3

Jelly, 8 bytes

²Ḷ×_2$‘S

Try it online!

In short, it generates the range \$[0,1,2,\dots,(n-1)^2]\$, multiplies every integer in that range with \$n-2\$, increments each of them (to avoid adding \$n^2\$ at the end, saving 1 byte), then sums the resulting list.

Mr. Xcoder

Posted 2018-08-26T20:30:26.683

Reputation: 39 774

Ah ha you found the 1-byte save I knew must have been there (had _2ײṖ$S+² , ²ṖS×’’$+² and many other 9 variants) – Jonathan Allan – 2018-08-27T00:05:18.030

1@LuisMendo >_< I was trying to make it clearer - thanks for the heads-up! – Jonathan Allan – 2018-08-27T00:44:17.800

@JonathanAllan Thank you for the edits. – Mr. Xcoder – 2018-08-27T06:43:27.790

2

JavaScript (ES7), 26 bytes

n=>n**4*(n-2)-n*n*(n-4)>>1

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Recursive (ES6), 33 bytes

A quick attempt from mobile. Not really optimized for \$(n,n^2)\$.

x=>(g=k=>k&&(x-2)*--k-~g(k))(x*x)

Try it online!

Arnauld

Posted 2018-08-26T20:30:26.683

Reputation: 111 334

Golfing from mobile and still FGITW :o (helps having no weird code-page I guess) – Jonathan Allan – 2018-08-26T20:49:37.767

It definitely helps. :) (Or I'd need another phone, I guess...) – Arnauld – 2018-08-26T20:51:22.503

2

05AB1E, 4 bytes

nsÅU

Try it online!

Mr. Xcoder

Posted 2018-08-26T20:30:26.683

Reputation: 39 774

I thought you'd been really clever for a second there. – Jonathan Allan – 2018-08-26T21:07:14.657

@JonathanAllan ¯\(ツ)/¯... Right tool for the job. Actually, give me a second to try Neim :P – Mr. Xcoder – 2018-08-26T21:07:51.577

2

Pyth, 14 9 bytes

smh*d-Q2*

Try it online!

hakr14

Posted 2018-08-26T20:30:26.683

Reputation: 1 295

shM*R-Q2* saves 5 whole bytes. – Mr. Xcoder – 2018-08-27T00:02:21.067

@Mr.Xcoder yeah, found a similar 9 byte solution right before you posted that... – hakr14 – 2018-08-27T00:04:33.050

2

Python 2, 31 30 bytes

lambda k:(k*k*(k-2)-k+4)*k*k/2

Try it online!

Saved:

  • -1 byte, thanks to joH1

TFeld

Posted 2018-08-26T20:30:26.683

Reputation: 19 246

30 bytes: lambda k:(k*k*(k-2)-k+4)*k*k/2

– joH1 – 2018-08-30T11:19:29.427

1

cQuents, 20 bytes

$0:$$+$$/2($$-1)($-2

Try it online!

Explanation

$0                      Zero indexing
  :                     Mode: sequence (output nth term)
                        Each term equals:
   $$+                    index * index +
      $$/2                index * index / 2 * 
          ($$-1)($-2      (index * index - 1) * (index - 2
                    )    ) implicit

Stephen

Posted 2018-08-26T20:30:26.683

Reputation: 12 293

1

Charcoal, 17 bytes

Ｉ⊘↨⟦¹±²±¹¦⁴¦⁰¦⁰⟧Ｎ

Try it online! Link is to verbose version of code. Explanation: Uses the second formula expanded into the polynomial \$ \frac{1}{2} ( x^5 - 2x^4 - x^3 + 4x^2 ) \$ which is calculated by treating it as an arbitrary base conversion before halving and casting to string for implicit output.

Neil

Posted 2018-08-26T20:30:26.683

Reputation: 95 035

1

Physica, 25 bytes

->n:(n^4-n^2)/2*(n-2)+n^2

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->n:n^2*(4-n-2*n^2+n^3)/2

Try it online!

->n:n^5/2-n^4-n^3/2+2*n*n

Try it online! | Try all of them at once as a test suite!

Mr. Xcoder

Posted 2018-08-26T20:30:26.683

Reputation: 39 774

1

Jelly, 11 10 bytes

²µ½_2×’H×+

Try it online!

Uses the formula \$F(x,k)=\frac{k(k-1)}{2}(x-2)+k\$ as given in the challenge.

dylnan

Posted 2018-08-26T20:30:26.683

Reputation: 4 993

I have less but don't think I have optimal yet. – Jonathan Allan – 2018-08-26T22:44:29.887

1

Clean, 38 bytes

import StdEnv
$x=x^2*(x^3/2-x^2-x/2+2)

Try it online!

Οurous

Posted 2018-08-26T20:30:26.683

Reputation: 7 916

1

APL (Dyalog), 17 bytes

×⍨×2+×⍨-⍨.5×*∘3-⊢

Try it online!

Uriel

Posted 2018-08-26T20:30:26.683

Reputation: 11 708

You may want to include 0 in your test-suite – Jonathan Allan – 2018-08-27T00:07:10.543

2+×⍨-⍨ → 2-×⍨- – Adám – 2018-08-27T12:50:31.063

1

MATL, 9 bytes

U:qG2-*Qs

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This uses Mr. Xcoder's Jelly approach.

Luis Mendo

Posted 2018-08-26T20:30:26.683

Reputation: 87 464

FYI the approach is just a small step from the 4th formula (with the two sums), or, indeed, the 3rd (replace the triangle-number-esque formula with a sum) :) – Jonathan Allan – 2018-08-27T00:20:18.140

@JonathanAllan Yes, I had noticed about the third, but not about the fourth. Thanks! – Luis Mendo – 2018-08-27T00:40:56.763

1

K (oK), 23 bytes

{((x-2)*k%2%k-1)+k:x*x}

Try it online!

Prefix function; implementation of the second formula: \$\frac{k(k-1)}{2}(x-2)+k\$

How:

{((x-2)*k%2%k-1)+k:x*x} # Main function, argument x.
                 k:x*x  # def k = n²
 ((x-2)*k%2%k-1)+       # calculate k + (((k/2)/(k-1))*(x-2))

J. Sallé

Posted 2018-08-26T20:30:26.683

Reputation: 3 233

1

Haskell (31 bytes)

g k=(k*k*k*k*(k-2)-k*k*(k-4))/2

Davide Spataro

Posted 2018-08-26T20:30:26.683

Reputation: 211

k**4 instead of k*k*k*k. – nimi – 2018-08-28T15:50:43.023

2

or k^4. But the equation may be rearranged a little: (k^4*(k-2)-k*k*(k-4))/2 = ((k*k*(k-2)-k+4)*k*k)/2 = ((k^3-2*k-k+4)*k*k)/2 - so this gives the 25 byte version: g k=(k^3-2*k*k-k+4)*k*k/2 Here is an online IDE link - Try It Online!

– Jonathan Allan – 2018-08-28T17:39:50.987

1

Haskell, 27 bytes

f k=(k^4*(k-2)-k^2*(k-4))/2

Try it online!

My first attempt at golfing in Haskell, basically same as almost any answer :

Nothing worth explaining here.

Any3nymous user

Posted 2018-08-26T20:30:26.683

Reputation: 539

Nice. I gave a golfed version to Davide Spataro, since they posted earlier, (my first attempt at golfing Haskell too :)) – Jonathan Allan – 2018-08-28T17:43:08.317

@JonathanAllan you did better than me, (your formula was better). Seeing there is already an existing Haskell answer, do you want me to delete mine ? – Any3nymous user – 2018-08-29T06:59:20.303

Some people delete if they find the same (or effectively the same) solution, but I think if one finds it independently there is nothing wrong with keeping it. – Jonathan Allan – 2018-08-29T07:28:44.023

1@JonathanAllan then for now, I'll let it stay :)\ – Any3nymous user – 2018-08-29T08:23:09.547

1

C (gcc), 60 36 34 bytes

-24 bytes thanks to Jonathan Allan

-2 bytes by factorizing again an n*n expression.

g(n){return(n*n*(n-2)-n+4)*n*n/2;}

Try it online!

Factored version of the second formula, uses a single function.

Entry point is function g(n), value is returned as integer.

joH1

Posted 2018-08-26T20:30:26.683

Reputation: 391

1g(n){return(n*n*n-2*n*n-n+4)*n*n/2;} is 36 bytes (is it golfable?) – Jonathan Allan – 2018-08-30T09:40:54.537

@JonathanAllan thanks! I factorized as much as I could, I don't think it can't be golfed down again, at least not using this approach. – joH1 – 2018-08-30T11:14:42.880

1

Ruby, 25 bytes

->n{(4-n+(n-2)*n*=n)*n/2}

Try it online!

G B

Posted 2018-08-26T20:30:26.683

Reputation: 11 099

0

Pyt, 4 bytes

Đ²⇹ᑭ

Try it online!

Explanation:

        Implicit input
Đ       Duplicate
 ²      Square
  ⇹     Swap top two elements on the stack
   ᑭ    Get the (n^2)th n-gonal number
        Implicit output

mudkip201

Posted 2018-08-26T20:30:26.683

Reputation: 833

Asked: 2018-08-26T20:30:26.683

Viewed: 469 times

Active: 2018-08-30T11:31:53.547