Is this number tetrahedral?

Question

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\$ is a tetrahedral number because \$10 = \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.

Answer 1

R, 32 31 29 bytes

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

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

Answer 2

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

Answer 3

Retina 0.8.2, 26 bytes

.+
$*
(^(1)|(?<2>1\2)\1)+$

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)|(1\2)\1))+$/

Try it online! Link includes test cases.

Answer 4

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

Answer 5

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.

Answer 6

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

Answer 7

JavaScript (ES7), 28 bytes

-2 thanks to @tsh

Returns a falsy value for tetrahedral.

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

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Method

Computes:

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

And tests whether:

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

Answer 8

Jelly, 4 bytes

RÄÄċ

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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.

Answer 9

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?

Answer 10

Python, 42 bytes

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

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

Answer 11

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.

Answer 12

C++, 63 57 bytes

(-6 bytes thanks to AZTECCO)

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

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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;}

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Answer 13

Python 3, 49 bytes

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

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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))

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Answer 14

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

Answer 15

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.

Answer 16

Perl 5 -p, 44 bytes

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

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Answer 17

Haskell, 32 bytes

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

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

32 bytes

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

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

Answer 18

Mathematica, 34 33 32 bytes

Outputs truthy/falsy reversed.

k^3-k-6#/.k->⌊(6#)^(1/3)⌋+1&

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Answer 19

Scala 3, 41 bytes

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

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Answer 20

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.

Answer 21

x86 32-bit machine code, 11 bytes

99 31 C9 01 CA 41 29 D0 77 F9 C3

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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.

Answer 22

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.

Answer 23

Python, 80 64 55 51 47 45 bytes

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

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Based off of Aiden Chow's Desmos answer.

Answer 24

Go, 80 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\$ to \$x\$ inclusive and sees if \$6x = i(i+1)(i+2)\$.

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

Answer 25

05AB1E, 7 bytes

LηOηOIå

Try it online or verify the first 100 test cases.

Explanation:

Similar as other answers.

L        # Push a list in the range [1, (implicit) input]
 ηOηO    # Cumulative sum twice:
 η       #  Get the prefixes of this list
  O      #  Sum each prefix
   ηO    #  And again
     Iå  # Check if the input is in this list
         # (which is output implicitly as result)

Answer 26

PowerShell, 76 bytes

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

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Answer 27

APL(Dyalog Unicode), ^{10 9}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.

Answer 28

APL(NARS), 64 chars

r←B w;y;z
y←⌈3√w←w×6⋄r←0⋄→3
y-←1
→2×⍳w<z←y×(y+1)×y+2
→0×⍳w>z⋄r←1

9+17+4+19+11+4=64

f(y)=y×(y+1)×(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 45487864677774111x×(45487864677774111x+1)×(45487864677774111x+2)÷6
1

Answer 29

dc, 50 bytes

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

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

Answer 30

Rattle, ²⁷ 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!).