Is this number tetrahedral?

Question 1

Happy New Year 2024!

2024 is a tetrahedral number. A tetrahedral number is a number that can be represented in the form \$n(n+1)(n+2)/6\$ for some positive integer \$n\$. Or, equivalently, they are the sum of the first \$n\$ triangular numbers. They are also the number of objects in a triangular pyramid which has \$n\$ objects on each of its edges.

For example, \10ドル\$ is a tetrahedral number because \10ドル = \frac{3 \times 4 \times 5}{6}\$.

Here are the first few tetrahedral numbers:

1, 4, 10, 20, 35, 56, 84, 120, 165, 220, 286, 364, 455, 560, 680, 816, 969, 1140, 1330, 1540, 1771, 2024, ...

This is sequence A000292 in the OEIS.

Task

Given a positive integer \$n\$, determine whether \$n\$ is a tetrahedral number.

This is code-golf, so the shortest code in bytes in each language wins.

This is also a decision-problem, so you may use your language's convention for truthy/falsy (swapping truthy and falsy is allowed), or use two distinct, fixed values to represent true or false.

Question 2

R, (削除) 32 (削除ここまで) (削除) 31 (削除ここまで) 29 bytes

function(n)n%in%choose(3:n,3)

Try it online!

Based on the observation on wiki or OEIS that \$Te_n={{n+2}\choose{3}}\$; : ranges downwards for the case of \$n=1\$.

Question 3

Vyxal 3 `R`, 3 bytes

@@c

Try it Online! (link is to literate version)

Simply cumulative sum twice, and check if the input is in that.

R flag casts num to range for cumsum

Question 4

Nice. Should the R flag be part of the language name for your answer?

Question 5

@NickKennedy it's the result of a new website for the new language - there'll be things I've forgotten like the part that includes flags in the submission :p

Question 6

On closer inspection, it was a bug where for some reason, using literate mode completely wiped all flags in the template instead of just removing the literate mode flag. A very silly bug

Question 7

Retina 0.8.2, 26 bytes

.+
$*
(^(1)|(?<2>12円)1円)+$

Try it online! Link includes test cases. Explanation: The first stage just converts to unary; the second stage adds successive triangular numbers to see if this will exactly consume the unary number. The ?<2> syntax reuses capturing group 2; the equivalent in Perl is to use alternative capture group numbering:

Perl 5 `-p`, 33 bytes

$_=(1x$_)=~/((?|^(1)|(12円)1円))+$/

Try it online! Link includes test cases.

Question 8

MATL, 6 bytes

t:tY+m

Try at MATL online!

How it works

This uses the fact (see OEIS) that tetrahedral numbers are

the convolution of the natural numbers with themselves

It suffices to compute the convolution of the finite sequence 1, 2, ..., n with itself. Also, it is not necessary to discard the "transient" (non-valid part) at the end of the convolution result, because all the values there exceed n and thus cannot cause false positives.

t % Implicit input: n. Duplicate
: % Range: gives [1 2 ... n]
t % Duplicate
Y+ % Convolution
m % Ismember. Implicit output

Question 9

Vyxal `R`, 23 bits^v2 , 2.875 bytes

¦¦c

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Bitstring:

10110100000101111100100

Same as my vyxal 3 answer but with the funny haha fracbyte encoding.

Question 10

+1 for "funny haha fracbyte encoding"

Question 11

Ruby, 41 bytes

->n{n*=6;a=(n**3**-1).to_i;n==a*-~a*a+=2}

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Finds the cube root of n*6 rounded down,then applies the formula a*(a+1)*(a+2) to see if this equals n*6

Question 12

JavaScript (ES7), 28 bytes

-2 thanks to @tsh

Returns a falsy value for tetrahedral.

n=>(n*=6)+(n**=1/3,~n)**3-~n

Try it online!

Method

Computes:

$$k=\left\lfloor (6n)^{1/3}\right\rfloor$$

And tests whether:

$$(k+1)^3-k-1=6n$$

Question 13

Since swap truthy falsy is allowed, you can change == into -. Also, this is 28 bytes n=>(n*=6)+(n**=1/3,~n)**3-~n which use \$k(k+1)(k+2)=(k+1)^3-(k+1)\$

Question 14

Jelly, 4 bytes

RÄÄċ

Try it online!

A monadic link taking an integer and returning 1 for tetrahedral and 0 for not. Similar to many of the other answers: range, cumsum, cumsum, count.

Question 15

Uiua ^SBCS, 10 bytes

∊:\+\++1⇡.

Try it!

Port of lyxal's Vyxal 3 answer.

∊:\+\++1⇡.
 . # duplicate
 ⇡ # range
 +1 # increment
 \+\+ # cumulative sum twice
∊: # is the input in this?

Question 16

Python, 42 bytes

lambda n:n*6in(y**3-y for y in range(n+2))

Attempt This Online!

Returns True/False. Saves a few bytes by shifting the closed formula by 1.

Question 17

I was about to suggest the improvement of shifting the closed formula on Joao's answer (which I somehow saw first), but I somehow missed the exponentiation trick.

Question 18

TI-BASIC, 14 bytes

max(Ans=cumSum(cumSum(cumSum(1 or rand(Ans

Takes input in Ans. Works on numbers that are within the list size limit.

20 bytes

Input N
int(3√(6N
Ans3+3Ans2+2Ans=6N

Based on Arnauld's answer.

Question 19

C++, (削除) 63 (削除ここまで) 57 bytes

(-6 bytes thanks to AZTECCO)

[](int n){for(int i=n+3;i--;)n*=n!=-~i*i*~-i/6;return n;}

Try it online!

This swaps truthy/falsy, so it returns an integer greater than 0 for non-tetrahedral numbers, and 0 for tetrahedral numbers.

63 byte version:

[](int n){for(int i=n+2;i-->0;)if(n==-~i*i*~-i/6)n=0;return n;}

Try it online!

Question 20

Welcome! And this can be 57

Question 21

This can easily be a C function that returns through arguments.

Question 22

@qwr like this? i wasn't planning on switching languages, and i'm not sure if returning through the arguments is allowed

Question 23

Yes there's some meta post about it. And you can define C functions instead of full programs. In C++ you have anonymous functions as you did too.

Question 24

Python 3, 49 bytes

lambda x:any(x==n*-~n*(n+2)/6for n in range(x+1))

Try it online!

No extra work, just the original formula

Python 3, 47 bytes swapping T/F

lambda x:all(x+n*~n*(n+2)/6for n in range(x+1))

Try it online!

Question 25

You can save two bytes from the first version with lambda n:n in[j*-~j*(j+2)/6for j in range(n+1)]

Question 26

-2 on top of that with j*~j*~-~j/6

Question 27

@Jakav Would finally result in Albert.Lang's solution

Question 28

Desmos, 31 bytes

f(k)=∏_{n=1}^k(6k-nnn-3nn-2n)

Try It On Desmos!

Try It On Desmos! - Prettified

Returns 0 if it is a tetrahedral number, otherwise it returns a non-zero integer. For 5 more bytes, you can make it return 0 if it is a tetrahedral number, and 1 otherwise:

36 bytes

f(k)=∏_{n=1}^ksgn(6k-nnn-3nn-2n)^2

Try It On Desmos!

Try It On Desmos! - Prettified

Question 29

Haskell + hgl, 12 bytes

e*^xc scp<e0

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Explanation

e0 integers from 0 to the input
xc scp perform cumulative sums twice
e determine if the input is an element.

Reflection

It is infuriating I don't have a way to check for membership in an infinite monotonic list. I would have sworn I had implemented it, but I've spent a long time looking and I can't find it.

It wouldn't save bytes here, in fact if it were 3 bytes it would make this longer by a byte, but it should exist because it will be useful in the future.

Question 30

Perl 5 `-p`, 44 bytes

$i++while($;=$i*($i+1)*($i+2)/6)<$_;$_=$_==$

Try it online!

Question 31

Ah, -l, that's how it's done (updates his 33 byte Perl 5 -p answer).

Question 32

38 bytes Try it online!

Question 33

Haskell, 32 bytes

f n=elem(6*n)[k^3-k|k<-[1..n+1]]

Try it online!

Check whether \6ドルn\$ is of the form \$k^3-k\$.

32 bytes

q=scanl1(+)
f n=elem n$q$q[1..n]

Try it online!

Using the cumulative-sum-twice method from other answers.

Question 34

Mathematica, (削除) 34 (削除ここまで) (削除) 33 (削除ここまで) 32 bytes

Outputs truthy/falsy reversed.

k^3-k-6#/.k->⌊(6#)^(1/3)⌋+1&

Try this online!

Question 35

Welcome to Code Golf! Nice first answer!

Question 36

27 bytes

Question 37

Scala 3, 41 bytes

x=>(0 to x).exists(n=>x==n*(n+1)*(n+2)/6)

Attempt This Online!

Question 38

Charcoal, 20 bytes

NθW‹↨υ1θ⊞υ+Lυ↨υ0=Συθ

Try it online! Link is to verbose version of code. Outputs a Charcoal boolean, i.e. - for tetrahedral, nothing if not. Explanation:

Nθ

Input n.

W‹↨υ1θ

Until the sum reaches n, ...

⊞υ+Lυ↨υ0

... push successive triangular numbers to the predefined empty list.

=Συθ

See whether n was reached exactly.

Question 39

x86 32-bit machine code, 11 bytes

99 31 C9 01 CA 41 29 D0 77 F9 C3

Try it online!

Following the regparm(1) calling convention, this function takes a number in EAX and returns a number in EAX, which uses reversed truthy/falsy – it is 0 for tetrahedral numbers and nonzero otherwise.

In assembly:

f: cdq # Set EDX to 0 by sign-extending EAX.
 xor ecx, ecx # Set ECX to 0.
r: add edx, ecx # Add ECX to EDX. EDX will run through the triangular numbers.
 inc ecx # Add 1 to ECX.
 sub eax, edx # Subtract EDX from EAX.
 ja r # Jump back, to repeat, if EAX remains positive.
 ret # Return.

Question 40

Raku, 48 bytes

->\a{^∞ .map(->\n{(n**3-n)/6}).first(*>=a)-a};

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Generate the sequence itself until it reaches a value greater than or equal to the input. Then return the difference, i.e., return 0 if it hits the input, and a positive integer otherwise; True/False is therefore swapped.

Question 41

Python, (削除) 80 64 55 51 47 (削除ここまで) 45 bytes

lambda x:x*6in[k*~k*~-~k for k in range(x+1)]

Attempt This Online!

Based off of Aiden Chow's Desmos answer.

Question 42

-3 bytes

Question 43

-9 bytes

Question 44

Go, (削除) 80 (削除ここまで) 75 bytes

func(x int)int{for i:=1;i<=x;i++{if 6*x==i*(i+1)*(i+2){return 1}}
return 0}

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Loops over all numbers from \1ドル\$ to \$x\$ inclusive and sees if \6ドルx = i(i+1)(i+2)\$.

-5 from Joao-3 by using int as the return type instead.

Question 45

-5 bytes, uses int instead of bool to save bytes

Question 46

05AB1E, 7 bytes

LηOηOIå

Try it online or verify the first 100 test cases.

Explanation:

PowerShell, 76 bytes

$a=$b=1;while($b-ge1){$b=$_*6/$a/($a+1)/($a+2);if($b-eq1){return !0};$a++}!1

Try it online!

Question 48

APL(Dyalog Unicode), ^{(削除) 10 9 (削除ここまで)}8 bytes ^SBCS

⊢∊+\⍣2⍤⍳

Try it on APLgolf!

-1 byte thanks to att.

A tacit function which takes an integer on the right and returns 1 if tetrahedral and 0 if not. Same method as the Vyxal and Uiua answers.

⊢∊+\⍣2⍤⍳⁡‎‎⁪⁡⁪⁠⁪⁢⁤⁪‏‏⁡⁠⁡‌⁢‎‎⁪⁡⁪⁠⁪⁤⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁡⁪‏‏⁡⁠⁡‌⁣‎‎⁪⁡⁪⁠⁪⁢⁢⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁣⁪‏‏⁡⁠⁡‌⁤‎‎⁪⁡⁪⁠⁪⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁪‏‏⁡⁠⁡‌
 ⍤⍳ # ‎⁡range, then
 +\ # ‎⁢cumulative sums,
 ⍣2 # ‎⁣twice
⊢∊ # ‎⁤input is member
💎

Created with the help of Luminespire.

Question 49

+\⍣2⍤⍳ saves parens

Question 50

APL(NARS), 64 chars

r←B ×ばつ⍳w&l×ばつ(y+1)×ばつ⍳w>z⋄r←1

9+たす17+たす4+たす19+たす11+たす4=わ64

f(y)=×ばつ(y+1)×ばつ(y+2)=y^3+3y^2+2y is a crescent function and from y0=⌈(6w)^(1/3) => (y0)^3≥6w =>(y0)^3+3(y0)^2+2(y0)>6w so it is possible begin the loop in that point y0 and decrease y0 until (y0)^3+3(y0)^2+2(y0)≤6w

 a⊂⍨B ̈a←⍳123
 1 4 10 20 35 56 84 120

it seems is possible to use big rational and big float too

 B 45487864677774111x
0
 B ×ばつ(45487864677774111x+1)×ばつ(45487864677774111x+2)÷6
1

Question 51

dc, 50 bytes

[1pq]st[0pq]sf[1+dd1+d1+**6/dln=tln<flxx]sx?sn0lxx

Try it online!

This solution outputs 1 for truthy, and 0 for falsy.

Question 52

Rattle, (削除) ^{27 (削除ここまで)} 26 bytes

|r`[=#-s+2*~*#/6[\=1q]]3`=

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Outputs 1 if the number is tetrahedral, 0 otherwise.

Explanation

| take input
 r` set input to "\" (variable)
 [ ... ]3` loop int("3"+str(N)) times where N is the input. The 3 catches the edge case of 1.
=# set the top of the stack to i, where i is the loop count
 - decrement
 s save to memory
 +2 increment by 2
 *~ multiply by the value in memory (performing (i-1)(i+1))
 *# multiply by i to obtain (i-1)(i)(i+1), effectively the same 
 as n(n+1)(n+2) but with shifted indices
 /6 divide by 6
 [\...] if equal to "\" (the original input)
 =1 set the top of the stack to 1
 q quit and implicitly print the top of the stack
= if the loop exits without quitting, the number is not tetrahedral
 (set top of stack to 0 then print implicitly)

There may be a way to get this shorter. I might come back later to try to get the byte count down (someone else should try too!).

Giuseppe 29.4k3 gold badges33 silver badges106 bronze badges · Accepted Answer · 2024-01-01 01:40:56Z

R, (削除) 32 (削除ここまで) (削除) 31 (削除ここまで) 29 bytes

function(n)n%in%choose(3:n,3)

Try it online!

Based on the observation on wiki or OEIS that \$Te_n={{n+2}\choose{3}}\$; : ranges downwards for the case of \$n=1\$.

Is this number tetrahedral?

Task

31 Answers 31

R, (削除) 32 (削除ここまで) (削除) 31 (削除ここまで) 29 bytes

Vyxal 3 R, 3 bytes

Retina 0.8.2, 26 bytes

Perl 5 -p, 33 bytes

MATL, 6 bytes

How it works

Vyxal R, 23 bitsv2 , 2.875 bytes

Ruby, 41 bytes

JavaScript (ES7), 28 bytes

Method

Jelly, 4 bytes

Uiua SBCS , 10 bytes

Python, 42 bytes

TI-BASIC, 14 bytes

20 bytes

C++, (削除) 63 (削除ここまで) 57 bytes

Python 3, 49 bytes

Python 3, 47 bytes swapping T/F

Desmos, 31 bytes

36 bytes

Haskell + hgl, 12 bytes

Explanation

Reflection

Perl 5 -p, 44 bytes

Haskell, 32 bytes

Mathematica, (削除) 34 (削除ここまで) (削除) 33 (削除ここまで) 32 bytes

Scala 3, 41 bytes

Charcoal, 20 bytes

x86 32-bit machine code, 11 bytes

Raku, 48 bytes

Python, (削除) 80 64 55 51 47 (削除ここまで) 45 bytes

Go, (削除) 80 (削除ここまで) 75 bytes

05AB1E, 7 bytes

PowerShell, 76 bytes

APL(Dyalog Unicode), (削除) 10 9 (削除ここまで)8 bytes SBCS

APL(NARS), 64 chars

dc, 50 bytes

Rattle, (削除) 27 (削除ここまで) 26 bytes

Explanation

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