# Is this number triangular?

## Challenge

Given a positive integer, determine whether it is a triangular number, and accordingly output one of any two constant, distinct values.

### Definition

A triangular number is a number that can be expressed as the sum of consecutive positive integers, starting at 1. They can also be expressed with the formula $$\\frac {n(n + 1)} 2\$$, where $$\n\$$ is some positive integer.

## Test cases

Truthy:

1
3
6
10
15
21
55
276
1540
2701
5050
7626
18915
71253
173166
222111
303031
307720
500500
998991


Falsy:

2
4
5
7
8
9
11
16
32
50
290
555
4576
31988
187394
501500
999999


## Rules

• Your entry may be a function or a program.
• You may assume that the input is a positive integer under $$\10^6\$$.
• You must pick two constant, distinct outputs to distinguish the two categories.

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

• May 22, 2017 at 20:27
• Related OEIS sequence
– ovs
May 22, 2017 at 21:14
• Related May 22, 2017 at 21:28
• Why didn't you include zero?
– Neil
May 22, 2017 at 21:40
• @Neil I wanted to minimize the number of possible edge cases, and handling zero is one of them that I felt wasn't too important. Do you think it would have been better if zero needed to be handled? (The Jelly answer currently fails on zero, for instance) May 22, 2017 at 21:44

EDIT:

• -1 byte: @xnor got rid of parentheses with a $. An anonymous function taking an Int and returning a Char. Output is '1' for triangular numbers and '0' for others. (!!)$show.(10^)=<<[0..]


Try it online!

• Use as ((!!)$show.(10^)=<<[0..]) 998991. • Generates the numbers 1, 10, 100, 1000, ..., converts those to strings, and concatenates them. Then indexes into the resulting infinite string "1101001000100001000001000000...  • An imaginative method! You can save a byte with (!!)$show.(10^)=<<[0..].
– xnor
May 23, 2017 at 6:07

# Python, 24 bytes

lambda n:(8*n+1)**.5%1>0


Try it online!

Outputs False for triangular numbers, True for the rest. Checks if 8*n+1 is a perfect square. Python's float precision for square roots easily suffices for the challenge limits of n up to a million, first giving a false positive on n=6896076976160002 as found by Deadcode.

• (1<<10000)**.5: OverflowError: int too large to convert to float May 23, 2017 at 3:42
• @trichoplax I think I was disputing xnor's claim in the text. The submission is fine, I agree. Jul 29, 2017 at 19:03
• Python does not do arbitrary-precision exponentiation with ** unless both operands are integer. Otherwise it uses IEEE 374. This function reports both 6896076976160001 and its successor, 6896076976160002, as being triangular (that is the smallest number with which it fails). Jan 20, 2020 at 20:23
• @Deadcode Thanks, I stand corrected. I edited the post.
– xnor
Jan 20, 2020 at 20:27
• @Deadcode You mean 754? I don't see a 374. Mar 20, 2020 at 20:44

# Jelly, 4 bytes

R+\ċ


Try it online!

### How?

R+\ċ - Main link: n
R    - range(n)   -> [1,2,3,...,N]
\  - cumulative reduce by:
ċ - count occurrences of right (n) in left -> 1 if triangular, 0 otherwise

• I'm surprised cumulative reduce doesn't automatically make a range. Is there a design choice behind this? May 22, 2017 at 20:45
• I am not 100% sure, but I think it would (at least currently) need to by the dyadic operation being reduced over that would cause a range to be made. May 22, 2017 at 20:52
• ...actually even that does not seem to apply (e.g. this vs this. It seems the quicklink implementation overrides such that the iterable does not make a range even if the dyadic operation defines it to do so for an argument. Pinged Dennis to field this one :) May 22, 2017 at 20:59
• @ETHproductions / and \ probably were among the first five quicks to be implemented, predating the idea to cast integer arguments to range. May 22, 2017 at 21:43

## Retina, 10 bytes

(^1|1\1)+$ Input is in unary. Output is 0 or 1. Try it online! (As a test suite that does decimal-to-unary conversion for convenience.) ### Explanation This is the most basic exercise in forward-references. Most people are familiar with backreferences in regex, e.g. (.)\1 to match a repeated character. However, some of the more advanced flavours allow you to use a backreference before or inside the group it's referring to. In that case, it's usually called a forward-reference. This can make sense if the reference is repeated. It might not be well defined on the first iteration, but on subsequent iterations, the later or surrounding group has captured something and can be reused. This is most commonly used to implement recurrent patterns on unary strings. In this case, we try to match the input as the sum of consecutive integers: ( # This is group 1, which we'll repeat 1 or more times. ^1 # Group 1 either matches a single 1 at the beginning of the string. | # or 1\1 # It matches whatever the previous iteration matched, plus another # 1, thereby incrementing our counter. # Note that the first alternative only works on the first iteration # due to the anchor, and the second alternative only works *after* # the first iteration, because only then the reference is valid. )+$        # Finally, we make sure that we can exactly hit the end of the
# string with this process.

• Why doesn't (^|1\1)+$ work? May 23, 2017 at 2:39 • @LeakyNun regex engines have an optimisation that they stop repeating a group if it was empty n times where n is the minimum of the quantifier you're using (in your case 1; if the minimum was 0, it would be tried once anyway). If you change the + to {2,}, it should work. This optimisation prevents infinite loops but it's also the only thing that keeps .NET regex from being Turing-complete on its own. May 23, 2017 at 4:53 • This just saved me 70 bytes: codegolf.stackexchange.com/a/118387 – Neil May 23, 2017 at 13:59 • Make that 74 bytes, thanks to \G! – Neil Jun 8, 2017 at 11:39 # Python 2, 25 bytes Checks if (8x+1) is a square number. lambda x:(8*x+1)**.5%1==0  Try it online! ## Mathematica, 16 bytes OddQ@Sqrt[1+8#]&  Essentially a port of xnor's Python solution. Outputs True for triangular numbers, False otherwise. ## JavaScript (ES6), 30 27 bytes Saved 2 bytes thanks to kamoroso94 f=(n,k)=>n>0?f(n+~k,-~k):!n  ### Test cases f=(n,k)=>n>0?f(n+~k,-~k):!n console.log('Testing truthy test cases'); console.log(f(1)) console.log(f(3)) console.log(f(6)) console.log(f(10)) console.log(f(15)) console.log(f(21)) console.log(f(55)) console.log(f(276)) console.log(f(1540)) console.log(f(2701)) console.log(f(5050)) console.log(f(7626)) console.log(f(18915)) console.log(f(71253)) console.log(f(173166)) console.log(f(222111)) console.log(f(303031)) console.log(f(307720)) console.log(f(500500)) console.log(f(998991)) console.log('Testing falsy test cases'); console.log(f(2)) console.log(f(4)) console.log(f(5)) console.log(f(7)) console.log(f(8)) console.log(f(9)) console.log(f(11)) console.log(f(16)) console.log(f(32)) console.log(f(50)) console.log(f(290)) console.log(f(555)) console.log(f(4576)) console.log(f(31988)) console.log(f(187394)) console.log(f(501500)) console.log(f(999999)) ### Non-recursive version (ES7), 19 bytes Port of Adnan's answer. x=>(8*x+1)**.5%1==0  • Only seeing now that you edited the 19 byte solution into your answer a few minutes before I posted mine. Should I delete mine? What's the generally accepted etiquette on that? May 23, 2017 at 10:34 • @Shaggy I don't think it's a real problem here. My 'main' answer really is the recursive one. May 23, 2017 at 10:43 • Reduce to 28 bytes with f=(n,k=1)=>n>0?f(n-k,k+1):!n? May 23, 2017 at 12:20 • @kamoroso94 Thanks! Updated. And a third byte was saved by omitting the initialization of k. May 23, 2017 at 12:35 • Elegant use of bitwise NOT as an incrementor for an initially-undefined value; your edit was a pleasure to read after I independently arrived at your prior solution. May 23, 2017 at 16:15 # Brain-Flak, 40 bytes (([{}](((()))<>))<>){<>({}({}({})))}{}{}  Wheat Wizard and I had a duel over this question. When we decided to post our solutions we were tied at 42 bytes, but I found a 2 byte golf of his solution. We decided that would count as the tie breaker (my solution is below). Try it online! ### Explanation: # Set up the stacks like this: -input 1 -input 1 1 (([{}](((()))<>))<>) ^ # Output 1 for triangular and 0 for non-triangular {<>({}({}({})))}{}{}  For a full explanation please see Wheat Wizard's answer. # Brain-Flak, 42 bytes (([({})])<>){(({}())<>{}({})){((<>))}{}{}}  Outputs 0\n (literal newline) for truthy, and the empty string for falsy. The idea is to subtract 1 then 2 then 3 all the way up to the input. If you hit 0, then you know this is a triangular number, so you can stop there. # Push -input on both stacks. One is a counter and the other is a running total (([({})])<>) # Count up from -input to 0 { # Push the new total which is: (counter += 1) + total (popped) + input (not popped) # This effectively adds 1, then 2, then 3 and so on to the running total (({}())<>{}({})) # If not 0 { # Push to 0s and switch stacks to "protect" the other values ((<>)) # End if } # Pop the two 0s, or empty the stack if we hit 0 {}{} # End loop }  ### Here's a 46 byte solution that I found interesting. {<>(({}())){({}[()]<>{(<({}[()])>)}{}<>)}{}<>}  Outputs 0\n (literal newline) for truthy, the empty string for falsy. The idea is to count down from input by consecutive numbers, 1 at a time. E.g. input - (1) - (1,1) - (1,1,1). Each time we subtract, if we aren't at 0 yet, we leave an extra value on the stack. That way, if we are at 0 and are still subtracting when we pop we remove the last value on the stack. If the input was a triangular number, we will end exactly at 0, and wont pop the 0. # Implicit input (call it I) # Until we reach 0, or the stack is empty { # Add 1 to the other stack and push it twice. This is our counter. <>(({}())) # While counter != 0 { # counter -= 1 ({}[()] # if I != 0 <>{ # I -= 1, and push 0 to escape the if (<({}[()])>) # End if } # Pop from the stack with I. This is either the 0 from the if, or I {} # Get ready for next loop End while <>) # End While } # Pop the counter that we were subtracting from {}<> # End Until we reach 0, or the stack is empty. }  # CJam, 11 bytes ri2*_mQ_)*=  Outputs 1 for triangular, 0 otherwise. Try it online! ### Explanation Consider input 21. ri e# Input integer. STACK: 21 2* e# Multiply by 2. STACK: 42 _ e# Duplicate. STACK: 42, 42 mQ e# Integer square root. STACK: 42, 6 _) e# Duplicate, increment. STACK: 42, 6, 7 * e# Multiply. STACK: 42, 42 = e# Equal? STACK: 1  # Jelly, 5 bytes ×8‘Æ²  Try it online! ### Background Let n be the input. If n is the kth triangular number, we have $$n = \frac{k(k+1)}{2} \iff k^2+k-2n = 0 \iff k = \frac12 (-1 \pm \sqrt{1+8n}),$$ which means there will be a natural solution if and only if 1 + 8n is an odd, perfect square. Clearly, checking the parity of 1 + 8n is not required. ### How it works ×8‘Æ² Main link. Argument: n ×8 Yield 8n. ‘ Increment, yielding 8n + 1. Æ² Test if the result is a perfect square.  # PowerShell, 31 30 bytes "$args"-in(1..1e6|%{($s+=$_)})


Try it online!

Nice and slow brute force method. Make an array of every sum of 1 through 106, and see if the argument is in there.

# Brain-Flak, 42 bytes

(([{}](<((())<>)>))<>){<>({}({}({})))}{}{}


Try it online!

## Explanation

The goal of this program is to create a state on two stacks and perform constant operation on both stacks until one of them zeros, we can then output depending on which stack we are on. This is similar to programs that determine the sign of a number. These programs put n on one stack and -n on the other and add one and switch stacks until one of the stacks is zero. If the number was negative in the first place the first stack will hit zero, if the number was positive the other stack will hit zero.

Here we create two stacks one that subtracts consecutive numbers from the input and one that just subtracts one. The one that subtracts consecutive numbers will only terminate if the number is triangular, (other wise it will just pass zero and keep going into the negatives). The other one will always terminate for any positive number, but will always do so slower than the first, thus non-triangular numbers will terminate on that stack.

So how do we set up stacks so that the same operation subtracts consecutive numbers on one and subtracts one on the other? On each stack we have the input on top so that in can be checked, below that we have the difference and below that we have the difference of the difference. Each time we run we add the "difference of the difference" to the regular "difference" and subtract that from the input. For the stack that checks for triangularity we set our double difference to be 1 so that we get consecutive integers each time we run, for the other stack we set it to 0 so that we never change the difference, that is it always stays 1. Here is how the stack is set up at the beginning, where n is the input:

-n  -n
0   1
1   0


When we finally do terminate we can use these differences to check which stack we are on we pop the top two values and we get 1 for a triangular number and 0 for a non-triangular number.

### Annotated code

(([{}](<((())<>)>))<>) Set up the stack
{                      While
<>                    Switch stacks
({}({}({})))          Add bottom to second to bottom, add second to bottom to top
}                      End while
{}{}                   Pop the top two values


Here's a 50 byte solution I like as well.

{(({}[()]))}(([[]])<>){({}{}())<>}({}{()<><{}>}{})


Try it online!

# Cubix, 23 2425 bytes

I1Wq/)s.;0..s;p-?\.+O@u


0 for truthy and nothing 0 for falsey. Brutes forces by incrementing counter, adding to cumulative sum and comparing to input. Now to try and fit it on a 2x2x2 cube. Did it!

    I 1
W q
/ ) s . ; 0 . .
s ; p - ? \ . +
O @
u .


Try it online!

• / Reflect to to face.
• I10\ get integer input, push 1 (counter), push 0 (sum) and reflect
• +s;p- loop body. Add sum and counter, drop previous sum, raise input and subtract
• ? Test the result of the subtraction
• For 0 result carrying on straight ahead \.uO@ reflect to bottom face, no-op, U-turn, output and halt.
• For positive result turn right onto bottom face and @ halt
• For negative result turn left ;qWs)/su drop subtraction, put input to bottom, shift left, swap counter and sum, increment counter, reflect, swap sum and counter, U-turn onto main loop body.
• So tantalizingly close... that last byte is going to take a lot of effort and cleverness though. May 22, 2017 at 23:44
• Yep, thought I had it but is being elusive May 22, 2017 at 23:48
• @ETHproductions found the byte May 23, 2017 at 0:15
• Your code and your unfolded cube appear to be different, the lower right corner is a . on the cube but a 1 in your code. Jun 13, 2017 at 13:39
• @WheatWizard Thanks for that, bad editing on my part Jun 14, 2017 at 18:31

# R, 23 19 bytes

Similar approach as other answers. Checks to see if 8x+1 is a perfect square.
-4 bytes thanks Giuseppe and MickyT.

!(8*scan()+1)^.5%%1


Try it online!

• you can use ! instead of ==0 Dec 6, 2018 at 19:18
• This is extra nice since it's vectorized, too! Dec 6, 2018 at 19:19
• I think you can get rid of the exterior brackets as well !(8*scan()+1)^.5%%1 Dec 6, 2018 at 19:21

# MATL, 5 bytes

t:Ysm


Try it online!

Explanation:

t       % Duplicate input
:      % Range(1, input)
Ys    % Cumulative sum. This will push the first *n* triangular numbers
m   % ismember. Pushes true if the input is contained within the array we just pushed

• I was going to post t:Ys=a. Forgot about m :-) May 22, 2017 at 20:38
• @LuisMendo I didn't know about m until I saw this answer. Funny how the two answer are almost identical :D May 23, 2017 at 17:40

# 05AB1E, 7 6 bytes

EDIT: Thanks to @Dennis: Saved a byte because I forgot about the increment operator

8*>t.ï


Try it online!

n is triangular if sqrt(8n + 1) is an integer

How it works

8* # multiply implicit input by 8
t # sqrt
.ï # is integer

• Probably wasn't available yet at the time, but t.ï can be Å² these days, which is a builtin to check if a number is a square. Feb 14, 2019 at 9:19

# Perl 6, 17 bytes

{$_∈[\+] 1..$_}


Just checks whether $_, the input to the function, is equal to any of the elements of the triangular addition reduction (1, 1+2, ..., 1+2+...+$_).

# Brachylog, 5 bytes

≥ℕ⟦+?


Try it online!

## Explanation

≥ℕ⟦+?
≥ℕ     There is a number from 0 to {the input} inclusive
⟦    such that the range from 0 to that number
+   has a sum
?  that equals the input


# Alice, 38 22 bytes

A lot of bytes saved thanks to Martin and Leo

/ i \2*.2RE.h*-n/ o @



There is a trailing newline. Outputs 1 for triangular, 0 otherwise.

Try it online!

### Explanation

This uses the same approach as my CJam answer, only clumsier. In linearized form, the program becomes

i2*.2RE.h*-no@


where the i and o are actually in ordinal mode.

Consider input 21 as an example.

i         Input integer                       STACK: 21
2*        Multiply by 2                       STACK: 42
.         Duplicate                           STACK: 42, 42
2RE       Integer square root                 STACK: 42, 6
.         Duplicate                           STACK: 42, 6, 6
h         Increment                           STACK: 42, 6, 7
*         Multiply                            STACK: 42, 42
-         Subtract                            STACK: 0
n         Logical negation                    STACK: 1
o         Output integer                      STACK:
@         End program

• My first Alice answer May 22, 2017 at 21:47
• I have a feeling this could be roughly halved with one of Martin's fancy control structures... May 22, 2017 at 21:58
• So do I ... :-) May 22, 2017 at 22:02
• My first Alice golf: Same code, 23 bytes May 23, 2017 at 3:05
• A more "standard" layout for this kind of program would be this. That said, you could get rid of the 1 on the stack, and simply output the logical negation of the subtraction (i.e. ...h*-no@)
– Leo
May 23, 2017 at 8:21

# Japt, 10 7 bytes

Saved 3 bytes thanks to @Luke and @ETHproductions

*8Ä ¬v1


Try it online!

### Explanation:

*8Ä ¬v1
¬    // Square root of:
*8       //   Input * 8
Ä      //   +1
v1  // Return 1 if divisible by 1; Else, return 0


õ å+ øU


### Explanation:

õ å+ øU
õ           // Create a range from [1...Input]
å+        // Cumulative reduce by addition
øU     // Does it contain the input?


Try it online!

• The question asks for two constant distinct ouputs.
– xnor
May 22, 2017 at 20:54
• *8Ä ¬u1 c for 9B (outputs 0 if input is triangular, 1 otherwise)
– Luke
May 22, 2017 at 21:33
• @Luke You could change u1 c to v1, I believe (switching the outputs) May 22, 2017 at 22:21
• 7 bytes? Nice! Somehow missed this while posting my own, similar solution late last might. Let me know if you'd like me to delete it. May 23, 2017 at 10:36

# 05AB1E (legacy), 4 bytes

ÅTs¢


Try it online!

Explanation:

ÅT     //push all triangle numbers <= (implicit) input
s    //push input onto stack
¢   //count occurrences of input in triangle numbers (i.e. 1 if triangle, 0 if not)


I'm using 05AB1E (legacy) since it pre-dates this challenge, but it works in current 05AB1E as well

• You can use 05AB1E rewrite for your answer as answers in languages newer than the challenge are now allowed. Mar 29, 2021 at 20:14

# TypeScript’s type system, 91 bytes

//@ts-ignore
type F<N,A=[],I=N["length"]>=N extends[...infer R,0]?F<R,[1,...R,...A],I>:A[I]


Try it at the TS playground

This is a generic type F taking a unary number N as a tuple of N zeroes. I never would have expected a challenge like this to be under 100 bytes in TS types, but in this instance it’s quite doable. This is almost definitely ungolfable.

This works by creating the string 1101001000100001…, which contains N ones separated by as many zeroes as there are already added ones, and indexing into this string with N.

Explanation of the code:

//@ts-ignore - Suppress any type errors on the following line.
type F<
N,            // N is a tuple type of 0s
A=[],         // A becomes the string, starts as an empty list
I=N["length"] // I is the original N which we use to index at the end
>=
N extends[...infer R,0] // If a 0 fits in N, take the difference R (i.e. decrement)
?F<R,            // If that succeeded, recurse with N = R,
[1,...R,...A], // A with 1 then R prepended,
I>               // and I kept as it was.
:A[I] // If it failed, N is 0, so index into A with the original number.


## Batch, 72 bytes

@set/aj=i=0
:l
@if %1% gtr %j% set/aj+=i+=1&goto l
@if %1==%j% echo 1


Outputs 1 on success, nothing on failure. Works for zero too, although not requested by the question for some reason.

# JavaScript (ES7), 19 18 bytes

From my answer to a related question.

Outputs false for triangular numbers or true for non-triangular, as permitted by the OP.

n=>(8*n+1)**.5%1>0


## Try It

f=
n=>(8*n+1)**.5%1>0
oninput=_=>o.innerText=f(+i.value)
<input id=i type=number><pre id=o>

• I think you could save a byte with n=>(8*n+1)**.5%1>0 (which would reverse the outputs) May 22, 2017 at 21:41
• @ETHproductions: OK, as long as you're allowing it. Is doing so normally permitted, though? May 22, 2017 at 22:15
• It qualifies as "two constant, distinct outputs", so yes. Other decision-problem challenges may require truthy/falsy though. May 22, 2017 at 22:20

# PHP, 30 Bytes

Prints 1 for true and nothing for false

<?=fmod(sqrt(8*$argn+1),2)==1;  Try it online! fmod # PHP, 37 Bytes Prints 1 for true and nothing for false <?=($x=sqrt($q=2*$argn)^0)*$x+$x==\$q;


Try it online!

## Mathematica, 28 bytes

!Accumulate@Range@#~FreeQ~#&

• I recommend replacing 7! by #. First, it's shorter; more importantly, the current solution is not correct, as it artificially imposes a limit on the size of the input it works on. May 23, 2017 at 0:35
• OP says: "You may assume that the input is a positive integer under 10^6".But I like your idea and I will take it ,although mine gives the right result for every case using a list of 5040 elements but yours worst case needs a list of 999999 elements.Thanks for the tip! May 23, 2017 at 1:03
• Oops sorry, didn't see the OP's comment! Yes, there are some "perverse" incentives in code golfing: that 1-byte savings is more important in a code-golf question than all the efficiency in the world :) May 23, 2017 at 2:44

# Pari/GP, 18 bytes

n->issquare(8*n+1)


Try it online!

There is a built-in to test if a number is a polygonal number, but it is one byte longer.

# Pari/GP, 19 bytes

n->ispolygonal(n,3)


Try it online!

# Python - 52 bytes

Note: I know that the other two Python answers are much shorter, but this is the old-school way, more of a by-hand algorithm

n=input();i=s=0
while s<n:s=(i*i+i)/2;i+=1
print s>n


# Excel, 31 22 bytes

9 bytes saved thanks to Octopus

Outputs TRUE for triangular numbers. Else FALSE. Checks if 8*n+1 is a perfect square.

=MOD(SQRT(8*B1+1),1)=0

• =MOD(SQRT(8*A1+1),1)=0 saves a few bytes May 23, 2017 at 23:39

# Alchemist, 80 bytes

_->In_n+s
f+b->f+a
f+0b->a
0f+n+a->b
0f+0a+n->n+f
0n+a+s->Out_n
0n+0a+s->Out_"1"


Try it online!

Test cases

Subtracts increasing numbers from the input until it reaches 0, and checks if the counter is zero.