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

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!

• Results may also be results of floating point calculation and may deviate as such -- are possible integer overflows also acceptable? Aug 27, 2018 at 0:07
• @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) Aug 27, 2018 at 0:17

# 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.

# 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.

• Finding the right language is a +1 from me :p Aug 26, 2018 at 21:11
• 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
Aug 27, 2018 at 13:54
• Upvoted because I am oppositional by nature.
– ngm
Aug 27, 2018 at 14:53

$$/2 index * index / 2 * ($$-1)($-2 (index * index - 1) * (index - 2 ) ) implicit  # Physica, 25 bytes ->n:(n^4-n^2)/2*(n-2)+n^2  Try it online! ->n:n^2*(4-n-2*n^2+n^3)/2  Try it online! ->n:n^5/2-n^4-n^3/2+2*n*n  # Clean, 38 bytes import StdEnv$x=x^2*(x^3/2-x^2-x/2+2)


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# APL (Dyalog), 17 bytes

×⍨×2+×⍨-⍨.5×*∘3-⊢


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• You may want to include 0 in your test-suite Aug 27, 2018 at 0:07
• 2+×⍨-⍨2-×⍨-
Aug 27, 2018 at 12:50

# 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.

• I have less but don't think I have optimal yet. Aug 26, 2018 at 22:44

# MATL, 9 bytes

U:qG2-*Qs


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This uses Mr. Xcoder's Jelly approach.

• 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) :) Aug 27, 2018 at 0:20
• @JonathanAllan Yes, I had noticed about the third, but not about the fourth. Thanks! Aug 27, 2018 at 0:40

# 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))


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

• k**4 instead of k*k*k*k.
– nimi
Aug 28, 2018 at 15:50
• 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! Aug 28, 2018 at 17:39

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.

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

# C (gcc), 6036 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.

• g(n){return(n*n*n-2*n*n-n+4)*n*n/2;} is 36 bytes (is it golfable?) Aug 30, 2018 at 9:40
• @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
Aug 30, 2018 at 11:14

# Ruby, 25 bytes

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


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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