Is this number triangular?

Question

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

Answer 1

Uiua, 14 bytes ^SBCS

Outputs 0 for false and 1 for true. Uses the "8n+1 is Perfect Square" method.

&p=⌊.√+1×8⋕&sc

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Explanation

&p=⌊.√+1×8⋕&sc
           &sc # Read line from stdin
          ⋕    # Parse string as number
      +1×8     # Multiply by 8, then add 1
     √         # Square root
    .          # Duplicate
   ⌊           # Floor
  =            # Compare for equality
&p             # Print with newline

Answer 2

Uiua, 6 bytes

∊:\+⇡.

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Is the input a member of taking the cumulative sums of the range of 0 to the input?

Answer 3

QBIC, 21 19 bytes

[:|p=p+a~p=b|_x1}?0

Explanation

[ |     FOR a = 1 to
 :        b (read from the cmd line)
p=p+a   increment p by a (+1, +2, ...) (p is 0 at QBIC start)
~p=b|   IF p == b ; we've hit a triangular number!
_x1     THEN QUIT, printing 1
}       Close the IF and the FOR
?0      We didn't quit early, so print 0

Answer 4

Clojure, 54 bytes

#(and(some #{%}(reductions +(rest(range(max % 3)))))1)

I wish various truthy values were allowed so I didn't need to coerce them all to 1 with a penalty of 6 bytes.

Without caring about edge cases of 0 or 1, and allowing any truthy value this would have been just 35 bytes:

#(some #{%}(reductions +(range %)))

Answer 5

C (gcc), 47 bytes

a,i;f(n){for(a=i=0;i++<n+3;)a|=i*i==8*n+1;a=a;}

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Undefined behaviour?

Answer 6

Neim, 6 bytes

𝐈Γ𝐈𝐬)𝕚

Explanation:

𝐈       Inclusive range; [1 .. input]
 Γ      For each...
  𝐈       Inclusive range; [1 .. element]
   𝐬      Sum
    )   End for each
     𝕚  Check that the input is in the list

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Answer 7

VB.NET, 65

Sub T(n,j)
j=Math.Sqrt(8*n+1)
Console.Write(CInt(j)=j)
End Sub

Like other answers, determines if 8n + 1 is an integer or not.

Explanation:

Sub T(n,j)                     ' start the method, use n as the triangle number to check
                               ' declare j here as well for use later, saving 4 bytes over a "dim j" call

j=Math.Sqrt(8*n+1)             ' get the square root

Console.Write(                 ' write out the answer, either "True" or "False"
                               ' it saves two bytes to do "Sub/End Sub" with "Console.Write()"
                               ' over "Function/End Function" with "return "

              CInt(j)          ' round to the nearest integer (cast to int, but VB rounds instead of truncating)

                     =j)       ' compare against the sqrt

Answer 8

Octave, 23 bytes

@(n)sum(1:sqrt(2*n))==n

This is an anonyous function that returns true (displayed as 1) for triangular numbers and false (0) otherwise.

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Answer 9

Whitespace, 95 bytes

Visible representation:

SSNSNSTNTTNSSSNSSSTNTSSSSNSSNSSSSTNTSSSTSSNSSSTSNTSTSSSNTTTTSSTSNSNTSNNTTSNSSSTNTNSTNSSNSSNTNST

Reads the number from stdin. If it is triangular, it outputs 0 to stdout. Otherwise, it outputs 10.

Disassembly:

    push 0
     dup
      inum
loop:
     push 1
      add
     dup
      dup
       push 1
        add
       mul
      push 2
       div
      push 0
       get
       sub
      dup
       jz match
      jn loop
     push 1
      pnum
match:
     push 0
      pnum

The used technique is a simple brute force search through all possible triangular numbers, but for the given limits that is plenty. Using my own whitespace JIT compiler it takes 0.6 milliseconds to prove that 10⁶ is not a triangular number and about 20 seconds to prove that 10¹⁸ is not a triangular number.

Answer 10

Java 8, 58 48 23 bytes

n->Math.sqrt(8*n+1)%1>0

Port of @xnor's amazing Python answer.

-25 bytes by using a Java 8 lambda instead of Java 7 method (and Math.sqrt(...) instead of Math.pow(...,.5)).

Try it here.

Answer 11

JavaScript (ES8), 21 bytes

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

Answer 12

Japt, 11 9 7 bytes

Port of my JS answer. Outputs 1 for triangular numbers or 0 otherwise.

*8Ä ¬v1

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Answer 13

J, 10 9 Bytes

Triangular number of bytes as well :)

-1 Byte thanks to @FrownyFrog

0=1|2!inv

Explanation:

    2!       | n choose 2
      inv    | Inverse
0=1|         | Test if it's an integer

Could have been 7 bytes if the truthy/falsy values didn't have to be constant.

Works since n choose 2 is n!/2!(n-2)! = n*(n-1)/2

I don't know of any shorter ways to test for integers, previously I had been using (=<.)

Answer 14

cQuents, 7 6 bytes

?b$
;$

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Explanation

?b$     First program. Checks if the input is triangular.
?       Mode: query. Returns true if the input is in the sequence, and false otherwise.
 b$     Each item in the sequence is the secondary program, passed the current (1-based) index.

;$      Secondary program. Generates triangular numbers.
;       Mode: series. Calculates the sum of the sequence up to the input.
 $      Each item in the sequence is the current (1-based) index.

Answer 15

05AB1E, 4 bytes

ÅTI¢

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Explanation:

ÅT     Get list of triangle numbers less than or equal to the implicit input
  I¢   Count occurrences of input in list

Answer 16

Desmoslang, 24 Bytes

0^{\mod((8*I+1)^{.5},1OT

Answer 17

Perl 5 -p, 21 bytes

$_=sqrt(1+8*$_)!~/\./

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