Is it in the sequence? (sum of the first n cubes)

According to Nicomachus' theorem, the sum of the first n cubes is equal to the square of the nth triangular number. See this question for a visualisation.

Your task today is to take an integer and determine if it is in the sequence of the sum of the first n cubes.

I take the opportunity to say Happy New Year for 2025! because 2025 is the 9th number in this sequence:

Input

An integer greater than zero. Your answer should be able to handle inputs at least up to 2^31-1 unless your language's maximum type size does not support this.

Output

Either of two distinct values: one if the number is in the sequence, another if it isn't.

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Explanation:

The nth square triangle number is \$x = (\frac{n(n+1)}{2})^2\$, so \$2\sqrt{x}=n^2+n\$, giving \$n=\frac{-1+\sqrt{1+8\sqrt{x}}}{2}\$. Since \$n\$ has to be an integer, that means \$\sqrt{1+8\sqrt{x}}\$ has to be an odd integer, i.e., \${\sqrt{1+8\sqrt{x}}}\mod{2}=1\$.

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Pad

-1 byte: Tbw - switched from checking \$n\in \sum k^3\$ to \$\sqrt{n}\in \sum k\$

Creates a list of \$\sum_{k=0}^n k \ (0\le n\le 999)\$, then checks membership of \$\sqrt{n}\$. Since the largest squared triangular number under \$2^{31}-1\$ is the \$303^{\text{st}}\$, this range must be 3 digits so we go up to the \$999^{\text{th}}\$.

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Edit

-3 thanks to att to shorten Sqrt

Try it online! Link includes some test cases. Explanation: Converts the input to unary, then searches for a triangular number whose square is the input.

Prompts for integer.

Try it online! Thanks to Dyalog Classic

Try it online! Link is to verbose version of code. Outputs a Charcoal boolean, i.e. - for a sum of cubes, nothing if not. Explanation: Port of my golf to @V_R's Python answer.

If potential floating-point accuracy is acceptable, then for 10 bytes:

Try it online! Link is to verbose version of code. Explanation: Port of @Eonema's R answer.

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Following the standard calling convention for Unix-like systems (from the System V AMD64 ABI), this takes a 32-bit integer in EDI and returns a 32-bit integer in EAX, which is 0 if the number is in the sequence and -1 if it is not.

In assembly:

-1 thanks to Neil

Takes quite a long time for inputs like 2^31-1...

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The xth triangular number is (x*x+x)/2, and we're looking for the square of that: ((x*x+x)/2)**2

We can save two characters by squaring the numerator and denominator separately: (x*x+x)**2/4

We need range(n+1) instead of range(n) to return True when n is 1

Expects the integer in cell A1. Returns true or false.



See A000537. Thanks to Arnauld.

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This is Arnauld's code, I'm just the messenger. Posting because of lack of a JavaScript answer.