Nth Ngonal Numbers

Question

Most of us are probably familiar with the concept of triangular and square numbers. However, there are also pentagonal numbers, hexagonal numbers, septagonal numbers, octagonal numbers, etc. The Nth Nagonal number is defined as the Nth number of the sequence formed with a polygon of N sides. Obviously, N >= 3, as there are no 2 or 1 sided closed shapes. The first few Nth Ngonal numbers are 0, 1, 2, 6, 16, 35, 66, 112, 176, 261, 370, 506, 672, 871.... This is sequence A060354 in the OEIS.

Your Task:

Write a program or function that, when given an integer n as input, outputs/returns the Nth Nagonal number.

Input:

An integer N between 3 and 10^6.

Output:

The Nth Nagonal number where N is the input.

Test Case:

25 -> 6925
35 -> 19670
40 -> 29680

Scoring:

This is code-golf, lowest score in bytes wins!

\$\begingroup\$ Related \$\endgroup\$
– Digital Trauma
Commented Oct 11, 2017 at 17:55 — Digital Trauma, Commented Oct 11, 2017 at 17:55
\$\begingroup\$ It would not hurt to add a direct precise definition. \$\endgroup\$
– Wlod AA
Commented Oct 12, 2017 at 19:11 — Wlod AA, Commented Oct 12, 2017 at 19:11

Okx · Accepted Answer · 2017-10-11 15:28:07Z

13

Neim, 1 byte

ℙ

¯\_(ツ)_/¯

Try it online!

answered Oct 11, 2017 at 15:28

Okx

16.4k5 gold badges43 silver badges113 bronze badges

1

\$\begingroup\$ A built-in for this? Really? Obviously less obscure than I'd thought. \$\endgroup\$
– Gryphon
Commented Oct 11, 2017 at 15:29
1

\$\begingroup\$ @Gryphon It's existed for ages. \$\endgroup\$
– Okx
Commented Oct 11, 2017 at 15:29
2

\$\begingroup\$ @Gryphon Yeah. It's existed since May. \$\endgroup\$
– Okx
Commented Oct 11, 2017 at 15:30
1

\$\begingroup\$ @Gryphon At that point, there were a lot of challenges for polygonal numbers, so I just added a bunch to Neim :P \$\endgroup\$
– Okx
Commented Oct 11, 2017 at 15:35
3

\$\begingroup\$ This seems to be a built-in for calculating the ath b-gonal number, which gets both its parameters autofilled with the only argument given. \$\endgroup\$
– lynn
Commented Oct 11, 2017 at 18:08

| Show 3 more comments

Emigna · Accepted Answer · 2017-10-11 16:04:34Z

7

05AB1E, 7 6 bytes

Saved 1 byte thanks to Neil

<ÐP+>;

Try it online!

Explanation

<        # push input-1
 Ð       # triplicate
  P      # product of stack
   +     # add input
    >    # increment
     ;   # divide by 2

edited Oct 11, 2017 at 16:04

answered Oct 11, 2017 at 15:33

Emigna

53k5 gold badges43 silver badges162 bronze badges

\$\begingroup\$ <Dn*+>; also works for 7 bytes. \$\endgroup\$
– Neil
Commented Oct 11, 2017 at 15:57
\$\begingroup\$ @Neil: Thanks! A modified version of that saved me a byte :) \$\endgroup\$
– Emigna
Commented Oct 11, 2017 at 16:05
\$\begingroup\$ Can't you do <3m instead of <ÐP? \$\endgroup\$
– Erik the Outgolfer
Commented Oct 11, 2017 at 16:34
\$\begingroup\$ @EriktheOutgolfer: Sure, but it's the same amount of bytes. \$\endgroup\$
– Emigna
Commented Oct 11, 2017 at 17:24
\$\begingroup\$ @Emigna Just asked to be sure. \$\endgroup\$
– Erik the Outgolfer
Commented Oct 11, 2017 at 17:25

Add a comment |

lynn · Accepted Answer · 2017-10-11 18:43:53Z

5

Jelly, 5 bytes

c3×3+

Try it online!

Computes choose(n, 3) × 3 + n.

This translates quite readily to 05AB1E:

05AB1E, 5 bytes

3c3*+

Try it online!

edited Oct 11, 2017 at 18:43

answered Oct 11, 2017 at 18:31

lynn

69.2k11 gold badges133 silver badges283 bronze badges

1

\$\begingroup\$ Umm, why do you have a f=\ in a Jelly submission header? :P \$\endgroup\$
– Mr. Xcoder
Commented Oct 11, 2017 at 18:41

Add a comment |

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

4

Pyke, 6 bytes

t3^+he

Try it here!

t      - Decrement.
 3^    - Raise to the power of 3.
   +   - Add the input.
    h  - Increment.
     e - Floor Halve.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 11, 2017 at 16:28

Mr. Xcoder

42.5k9 gold badges81 silver badges214 bronze badges

Add a comment |

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

4

Japt, 9 8 bytes

´U+³ z Ä

Try it

1 byte saved thanks to ETH

Explanation

Decrement (´) the input (U), add the input cubed (³) to that, floor divide by 2 (z) and add 1 (Ä).

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 11, 2017 at 16:18

Shaggy

43.5k4 gold badges36 silver badges94 bronze badges

\$\begingroup\$ There are about a bazillion ways to rewrite this in 9 bytes, but it's tough to find an 8-byte solution... oh, here's one I think should work: ´U+³ z Ä \$\endgroup\$
– ETHproductions
Commented Oct 11, 2017 at 16:35
\$\begingroup\$ Nice one, @ETHproductions; got distracted by work while trying to crunch it down. \$\endgroup\$
– Shaggy
Commented Oct 11, 2017 at 16:45
\$\begingroup\$ @ETHproductions Nice one, that's tricky...another one would be ´U+³+2 z. \$\endgroup\$
– Erik the Outgolfer
Commented Oct 11, 2017 at 16:46
\$\begingroup\$ @EriktheOutgolfer: or ÄÄ instead of +2. \$\endgroup\$
– Shaggy
Commented Oct 11, 2017 at 16:48
1

\$\begingroup\$ 7 bytes: à3 *3+U or U+3*Uà3 (credit goes partly to Lynn)... Quite sure it can be rearranged to save more bytes. \$\endgroup\$
– Mr. Xcoder
Commented Oct 11, 2017 at 18:42

| Show 1 more comment

AdmBorkBork · Accepted Answer · 2017-10-11 17:57:21Z

4

PowerShell, 34 28 bytes

param($n)$n*($n*$n-3*$n+4)/2

Try it online!

Closed-form solution golfed from the OEIS page. Used FOIL for another 6 byte savings.

edited Oct 11, 2017 at 17:57

answered Oct 11, 2017 at 15:32

AdmBorkBork

43.5k5 gold badges103 silver badges284 bronze badges

Add a comment |

Mr. Xcoder · Accepted Answer · 2017-10-11 16:29:46Z

3

Python 2, 23 bytes

lambda n:(~-n)**3-~n>>1

Try it online!

edited Oct 11, 2017 at 16:29

answered Oct 11, 2017 at 16:24

Mr. Xcoder

42.5k9 gold badges81 silver badges214 bronze badges

Add a comment |

Giuseppe · Accepted Answer · 2017-10-11 19:16:31Z

3

MATL, 7 bytes

t3Xn3*+

Luis Mendo's suggestion, which is a little clearer.

    (implicit input)
t                         duplicate
 3Xn                      n choose 3
    3*                    multiply by 3
      +                   add
(implicit output)

Try it online!

t:3XNn+

Try it online!

Both solutions port Lynn's algorithm

(implicit input)
t                         duplicate
 :                        range (1...n)
  3XN                     push 3, compute all 3-combinations of the range
     n                    number ( equal to 3*choose(n,3) )
      +                   add
(implicit output)

edited Oct 11, 2017 at 19:16

answered Oct 11, 2017 at 18:38

Giuseppe

28.8k3 gold badges31 silver badges105 bronze badges

\$\begingroup\$ Perhaps a little simpler: t3Xn3*+ \$\endgroup\$
– Luis Mendo
Commented Oct 11, 2017 at 18:54
\$\begingroup\$ huh, I swear I looked for Xn and didn't see it. I'm not convinced I know how to read documentation... \$\endgroup\$
– Giuseppe
Commented Oct 11, 2017 at 19:12
\$\begingroup\$ Documentation could be clearer :-) It assumes you more or less know the equivalent Matlab functions. In this case, if you search for combinations of nchoosek in Suever's server it gives both functions as results \$\endgroup\$
– Luis Mendo
Commented Oct 11, 2017 at 19:27

Add a comment |

Jonathan Frech · Accepted Answer · 2017-10-11 15:43:44Z

2

Python 2, 25 24 bytes

Saved one byte thanks to Neil; golfed >>1 to /2.

lambda n:n*(n*n-3*n+4)/2

Try it online!

edited Oct 11, 2017 at 15:43

answered Oct 11, 2017 at 15:27

Jonathan Frech

7,3411 gold badge22 silver badges44 bronze badges

\$\begingroup\$ Can you not use /2 instead of >>1? \$\endgroup\$
– Neil
Commented Oct 11, 2017 at 15:40
\$\begingroup\$ @Neil True, true. \$\endgroup\$
– Jonathan Frech
Commented Oct 11, 2017 at 15:40
\$\begingroup\$ @Neil It was an artifact from back when there had been a sum that needed division. \$\endgroup\$
– Jonathan Frech
Commented Oct 11, 2017 at 15:44

Add a comment |

0xffcourse · Accepted Answer · 2017-10-11 15:51:36Z

2

Recursiva, 11 bytes

*Ha+-Sa*3a4

Try it online!

answered Oct 11, 2017 at 15:51

0xffcourse

2,7793 gold badges14 silver badges27 bronze badges

Add a comment |

ETHproductions · Accepted Answer · 2017-10-11 15:51:59Z

2

JavaScript (ES6), 38 bytes

f=(n,k=n)=>k<2|n<3?k:f(n-1,k)+f(3,k-1)

Recursion FTW (or maybe only for seventh...)

answered Oct 11, 2017 at 15:51

ETHproductions

50.1k6 gold badges94 silver badges240 bronze badges

\$\begingroup\$ Or you could port the closed form solution for 17 bytes... \$\endgroup\$
– Neil
Commented Oct 11, 2017 at 18:29

Add a comment |

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

2

Mathematica, 14 bytes

shorter than built-in!!!

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

Try it online!

and 3 bytes shorter with Martin Ender's help

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 11, 2017 at 15:55

ZaMoC

24.9k2 gold badges30 silver badges81 bronze badges

Add a comment |

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

2

Cubix, 20 17 bytes

Saved 3 bytes porting Emigna's answer.

Iu(:^\:**p+u@O,2)

Try it online!

    I u
    ( :
^ \ : * * p + u
@ O , 2 ) . . .
    . .
    . .

original answer:

Iu-2^\:*p*qu@O,2+p*:

Try it online!

Expands to the cube

    I u
    - 2
^ \ : * p * q u
@ O , 2 + p * :
    . .
    . .

which implements the (n*(n-2)^2+n^2)/2 approach.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 11, 2017 at 16:51

Giuseppe

28.8k3 gold badges31 silver badges105 bronze badges

Add a comment |

Cinaski · Accepted Answer · 2017-10-11 15:30:33Z

1

Ohm v2, 3 bytes

DÆ¤

Try it online!

answered Oct 11, 2017 at 15:30

Cinaski

2,2581 gold badge8 silver badges17 bronze badges

Add a comment |

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

1

Mathematica, 20 bytes

#~PolygonalNumber~#&

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 11, 2017 at 15:32

Jonathan Frech

7,3411 gold badge22 silver badges44 bronze badges

Add a comment |

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

1

Pyth, 7 bytes

+*3.cQ3

Try it here!

Uses Lynn's algorithm.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 11, 2017 at 18:33

Mr. Xcoder

42.5k9 gold badges81 silver badges214 bronze badges

Add a comment |

juh · Accepted Answer · 2017-10-11 20:48:58Z

dc, 13 bytes

dd2-2^*r2^+2/

A fairly straightforward implementation of the first formula listed on the OEIS page.

# Commands           # Stack Tracker (tm)
# Begin with input   # n
d                    # n n
d                    # n n n
2-                   # n-2 n n
2^                   # (n-2)^2 n n
*                    # n*(n-2)^2 n
r                    # n n*(n-2)^2
2^                   # n^2 n*(n-2)^2
+                    # n*(n-2)^2+n^2
2/                   # (n*(n-2)^2+n^2)/2 # matches first formula
# End with output on stack

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

1

Japt, 7 bytes

à3 *3+U

Try it here!

First, it was a comment on Shaggy's answer, but they told me I should post it myself.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 12, 2017 at 12:47

Mr. Xcoder

42.5k9 gold badges81 silver badges214 bronze badges

\$\begingroup\$ Only noticing this now - +1. \$\endgroup\$
– Shaggy
Commented Jan 19, 2018 at 17:08

Add a comment |

Mr. Xcoder · Accepted Answer · 2017-12-23 21:53:33Z

1

05AB1E, 2 bytes

ÅU

Try it online!

How?

¯\_(ツ)_/¯

answered Dec 23, 2017 at 21:53

Mr. Xcoder

42.5k9 gold badges81 silver badges214 bronze badges

Add a comment |

oeuf · Accepted Answer · 2022-07-21 05:37:19Z

1

K (ngn/k), 23 bytes

{x*(((x*x)-(3*x))+4)%2}

Try it online!

Having to add parenthesis because K parses from right-to-left.

answered Jul 21, 2022 at 5:37

oeuf

1,8434 silver badges18 bronze badges

1

\$\begingroup\$ "Having to add parenthesis because K parses from right-to-left." - you don't have to add all of them \$\endgroup\$
– ngn
Commented Jul 21, 2022 at 16:41
1

\$\begingroup\$ for instance the ( ) around 3*x are unnecessary, and you can avoid more pairs by rewriting things in equivalent ways \$\endgroup\$
– ngn
Commented Jul 21, 2022 at 17:17

Add a comment |

naffetS · Accepted Answer · 2022-07-21 16:42:16Z

1

HOPS, 17 bytes

{n*(n*n-3*n+4)/2}

Attempt This Online!

Using the generating function given on OEIS by R. J. Mathar:

HOPS, 23 bytes

x*(1-2*x+4*x^2)/(1-x)^4

Attempt This Online!

Or using the exponential generating function given on OEIS by Paul Barry:

HOPS, 24 bytes

(exp(x)*(x+x^3/2)).*{n!}

Attempt This Online!

-3 bytes thanks to alephalpha

edited Jul 21, 2022 at 16:42

answered Jul 21, 2022 at 1:28

naffetS

12.8k2 gold badges20 silver badges53 bronze badges

\$\begingroup\$ -3 bytes for the last one: (exp(x)*(x+x^3/2)).*{n!}. \$\endgroup\$
– alephalpha
Commented Jul 21, 2022 at 7:01

Add a comment |

Stephen · Accepted Answer · 2017-10-11 16:57:16Z

0

cQuents 0, 16 bytes

#|1:A(AA-3A+4)/2

Try it online!

answered Oct 11, 2017 at 16:57

Stephen

14k3 gold badges52 silver badges117 bronze badges

Add a comment |

Erik the Outgolfer · Accepted Answer · 2017-10-11 18:24:39Z

0

Jelly, 6 bytes

’*3+‘H

Try it online!

Uses Emigna's algorithm inspired by Neil.

edited Oct 11, 2017 at 18:24

answered Oct 11, 2017 at 16:35

Erik the Outgolfer

40.6k5 gold badges45 silver badges122 bronze badges

\$\begingroup\$ I think you'll find that the <Dn*+>; comment was mine... \$\endgroup\$
– Neil
Commented Oct 11, 2017 at 18:23
\$\begingroup\$ I'm using his modification, but I'll add you as well. \$\endgroup\$
– Erik the Outgolfer
Commented Oct 11, 2017 at 18:24

Add a comment |

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

0

Java 8, 18 bytes

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

Try it here.

The approach used by most other answers is the shortest in Java. For funsies I also ported two other answers:

Port of Mr. Xcoder's Python 2 answer (29 bytes):

n->(int)Math.pow(n-1,3)-~n>>1

Try it here.

Port of Lynn's Jelly answer (with manual calculation of a choose b) (76 bytes):

n->c(n,3)*3+nint c(int t,int c){return t<c?0:c==t|c==0?1:c(--t,c-1)+c(t,c);}

Try it here.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Oct 12, 2017 at 7:38

Kevin Cruijssen

131k13 gold badges144 silver badges384 bronze badges

Add a comment |

Nth Ngonal Numbers

Your Task:

Input:

Output:

Test Case:

Scoring:

24 Answers 24

Neim, 1 byte

05AB1E, 7 6 bytes

Jelly, 5 bytes

05AB1E, 5 bytes

Pyke, 6 bytes

Japt, 9 8 bytes

Explanation

PowerShell, 34 28 bytes

Python 2, 23 bytes

MATL, 7 bytes

Python 2, 25 24 bytes

Recursiva, 11 bytes

JavaScript (ES6), 38 bytes

Mathematica, 14 bytes

Cubix, 20 17 bytes

original answer:

Ohm v2, 3 bytes

Mathematica, 20 bytes

Pyth, 7 bytes

dc, 13 bytes

Japt, 7 bytes

05AB1E, 2 bytes

How?

K (ngn/k), 23 bytes

HOPS, 17 bytes

HOPS, 23 bytes

HOPS, 24 bytes

cQuents 0, 16 bytes

Jelly, 6 bytes

Java 8, 18 bytes

Your Answer

Sign up or log in

Post as a guest

Not the answer you're looking for? Browse other questions tagged code-golfsequenceinteger or ask your own question.

Linked

Related

Hot Network Questions

Not the answer you're looking for? Browse other questions tagged
code-golf
sequence
integer
or ask your own question.