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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!

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2
  • \$\begingroup\$ Results may also be results of floating point calculation and may deviate as such -- are possible integer overflows also acceptable? \$\endgroup\$
    – Jonathan Frech
    Aug 27, 2018 at 0:07
  • \$\begingroup\$ @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) \$\endgroup\$
    – Jonathan Allan
    Aug 27, 2018 at 0:17

20 Answers 20

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6
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Wolfram Language (Mathematica), 19 bytes

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

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Just a golfy version of the second formula.

With PolygonalNumber built-in, 23 bytes

PolygonalNumber[#,#^2]&

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

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5
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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.

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3
  • 4
    \$\begingroup\$ Finding the right language is a +1 from me :p \$\endgroup\$
    – Jonathan Allan
    Aug 26, 2018 at 21:11
  • \$\begingroup\$ Upvoted because when I saw the question I immediately thought "oh, Neim can do this in 2 bytes". But someone beat me to it! \$\endgroup\$
    – Okx
    Aug 27, 2018 at 13:54
  • \$\begingroup\$ Upvoted because I am oppositional by nature. \$\endgroup\$
    – ngm
    Aug 27, 2018 at 14:53
3
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Jelly, 8 bytes

²Ḷ×_2$‘S

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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.

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2
  • \$\begingroup\$ 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) \$\endgroup\$
    – Jonathan Allan
    Aug 27, 2018 at 0:05
  • \$\begingroup\$ @JonathanAllan Thank you for the edits. \$\endgroup\$
    – Mr. Xcoder
    Aug 27, 2018 at 6:43
2
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05AB1E, 4 bytes

nsÅU

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2
  • \$\begingroup\$ I thought you'd been really clever for a second there. \$\endgroup\$
    – Jonathan Allan
    Aug 26, 2018 at 21:07
  • \$\begingroup\$ @JonathanAllan ¯\_(ツ)_/¯... Right tool for the job. Actually, give me a second to try Neim :P \$\endgroup\$
    – Mr. Xcoder
    Aug 26, 2018 at 21:07
2
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Pyth, 14 9 bytes

smh*d-Q2*

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2
  • 1
    \$\begingroup\$ shM*R-Q2* saves 5 whole bytes. \$\endgroup\$
    – Mr. Xcoder
    Aug 27, 2018 at 0:02
  • \$\begingroup\$ @Mr.Xcoder yeah, found a similar 9 byte solution right before you posted that... \$\endgroup\$
    – hakr14
    Aug 27, 2018 at 0:04
2
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Python 2, 31 30 bytes

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

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Saved:

  • -1 byte, thanks to joH1
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1
1
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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)

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2
  • \$\begingroup\$ Golfing from mobile and still FGITW :o (helps having no weird code-page I guess) \$\endgroup\$
    – Jonathan Allan
    Aug 26, 2018 at 20:49
  • \$\begingroup\$ It definitely helps. :) (Or I'd need another phone, I guess...) \$\endgroup\$
    – Arnauld
    Aug 26, 2018 at 20:51
1
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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.

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1
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cQuents, 20 bytes

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

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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
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0
1
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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

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->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!

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1
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Clean, 38 bytes

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

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1
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APL (Dyalog), 17 bytes

×⍨×2+×⍨-⍨.5×*∘3-⊢

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2
  • \$\begingroup\$ You may want to include 0 in your test-suite \$\endgroup\$
    – Jonathan Allan
    Aug 27, 2018 at 0:07
  • \$\begingroup\$ 2+×⍨-⍨ → 2-×⍨- \$\endgroup\$
    – Adám
    Aug 27, 2018 at 12:50
1
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Jelly, 11 10 bytes

²µ½_2×’H×+

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

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1
  • \$\begingroup\$ I have less but don't think I have optimal yet. \$\endgroup\$
    – Jonathan Allan
    Aug 26, 2018 at 22:44
1
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MATL, 9 bytes

U:qG2-*Qs

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This uses Mr. Xcoder's Jelly approach.

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2
  • \$\begingroup\$ 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) :) \$\endgroup\$
    – Jonathan Allan
    Aug 27, 2018 at 0:20
  • \$\begingroup\$ @JonathanAllan Yes, I had noticed about the third, but not about the fourth. Thanks! \$\endgroup\$
    – Luis Mendo
    Aug 27, 2018 at 0:40
1
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K (oK), 23 bytes

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

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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))
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1
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Haskell (31 bytes)

g k=(k*k*k*k*(k-2)-k*k*(k-4))/2
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2
  • \$\begingroup\$ k**4 instead of k*k*k*k. \$\endgroup\$
    – nimi
    Aug 28, 2018 at 15:50
  • 2
    \$\begingroup\$ 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! \$\endgroup\$
    – Jonathan Allan
    Aug 28, 2018 at 17:39
1
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Haskell, 27 bytes

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

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

Nothing worth explaining here.

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4
  • \$\begingroup\$ Nice. I gave a golfed version to Davide Spataro, since they posted earlier, (my first attempt at golfing Haskell too :)) \$\endgroup\$
    – Jonathan Allan
    Aug 28, 2018 at 17:43
  • \$\begingroup\$ @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 ? \$\endgroup\$
    – user82328
    Aug 29, 2018 at 6:59
  • \$\begingroup\$ 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. \$\endgroup\$
    – Jonathan Allan
    Aug 29, 2018 at 7:28
  • 1
    \$\begingroup\$ @JonathanAllan then for now, I'll let it stay :)\ \$\endgroup\$
    – user82328
    Aug 29, 2018 at 8:23
1
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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;}

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

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

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2
  • 1
    \$\begingroup\$ g(n){return(n*n*n-2*n*n-n+4)*n*n/2;} is 36 bytes (is it golfable?) \$\endgroup\$
    – Jonathan Allan
    Aug 30, 2018 at 9:40
  • \$\begingroup\$ @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. \$\endgroup\$
    – joH1
    Aug 30, 2018 at 11:14
1
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Ruby, 25 bytes

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

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0
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Pyt, 4 bytes

Đ²⇹ᑭ

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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
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