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Let $n > 1$ be a positive integer. Suppose that $n$ is a product of distinct primes $$n= \prod_{i=1}^kp_i^{\alpha_i}$$ i.e. $\alpha_i=1$ for every $i$. Said equivalently: $\,n\,$ has no repeated (square) factor.

Surely if $n=pq$ we call them semiprimes but beyond that do we have a general term? Would you just say that $n$ is a "product of distinct primes"

The motivation here is simple. I noticed that the numbers of the form $${2^n \choose 2}^n+1$$ appear to be exponent free in thier prime factorization up to $n=12$ using GAP.

Update 2021: The first value $n$ for which ${2^n \choose 2}^n+1$ is not squarefree is 15. \begin{align} {2^{15} \choose 2}^{15}+1=3^3×11×19×251×331×4051×18837001×4714696801×1133836730401×281941472953710177758647201 \end{align}

Anthony
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1 Answers1

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Such numbers are called square-free integers, see for instance here. A way to express the fact that $n$ is square-free is to write $\mu(n) \neq 0$, where $\mu$ is the Möbius function.

I have no idea about your particular problem. I know that it was only proved recently (1996) that ${2n \choose n} $ is never square-free for $n \geq 5$… ! Notice that your numbers are just $x(n) = 1+(2^{n-1} (2^n-1))^n$.

Watson
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