Let $C_n=\frac1{n+1}\binom{2n}n$ be the well-known Catalan numbers. Here is a curiosity.
QUESTION. Are there infinitely many $C_n$ that are "radical", i.e. that are square-free?
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asked Jul 10, 2021 at 21:29
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5$\begingroup$ Seems unlikely based on the main theorem of Sárközy, A. (1985). On divisors of binomial coefficients, I. Journal of Number Theory, 20(1), 70-80: the square root of the largest square number dividing $\binom{2n}{n}$ is in the range $(c \pm \varepsilon)\sqrt n$ for all $n$ greater than a lower bound which depends on $\varepsilon$, where $c \approx 0.855$. $\endgroup$– Peter TaylorCommented Jul 10, 2021 at 22:47
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4$\begingroup$ In fact, Granville, A., & Ramaré, O. (1996). Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients. Mathematika, 43(1), 73-107 go further: when $n > 2^{1617}$, $\binom{2n}{n}$ is divisible by the square of some prime $p > \sqrt{n}$. $\endgroup$– Peter TaylorCommented Jul 10, 2021 at 22:57
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1$\begingroup$ @PeterTaylor You seem to have dropped an exp or log in the first result. The correct statement is that the square root of the square part is $e^{ (c\pm e) \sqrt{n}$. This answers the question: As soon as this is larger than $2n+1$, as it will be for all large $n$, $C_n$ cannot be squarefree. $\endgroup$– Will SawinCommented Jul 10, 2021 at 23:10
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1$\begingroup$ It may take a while for these estimates to kick in. benvitalenum3ers.wordpress.com/2012/03/10/catalan-num3ers factors the first $35$ Catalan numbers, and they are squarefree for $n=1,2,3,4,5,7,8,9,11,17,19,31,35$. $\endgroup$– Gerry MyersonCommented Jul 10, 2021 at 23:53
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4$\begingroup$ It is reported at primepuzzles.net/problems/prob_043.htm that $C_n$ is not squarefree for any $n$, $35<n\le2^{30}-1$, so it seems quite plausible that $C_{35}$ is the last squarefree Catalan number. $\endgroup$– Gerry MyersonCommented Jul 11, 2021 at 0:00
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1 Answer
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If $s(n)$ is the largest integer such that $s(n)^2 \mid \binom{2n}{n}$, then the main result of Sárközy, A. (1985). On divisors of binomial coefficients, I. Journal of Number Theory, 20(1), 70-80 is
If $\varepsilon > 0$, $n > n_0(\varepsilon)$ then we have $$e^{(c-\varepsilon) \sqrt{n}} < s(n)^2 < e^{(c+\varepsilon) \sqrt{n}}$$ where $$c = \sqrt{2} \sum_{k=1}^\infty \frac{1}{\sqrt{2k-1}} - \frac{1}{\sqrt{2k}}$$
Numerically, $c \approx 0.855$.
Define $s'(n) = \frac{s(n)}{\textrm{gcd}(s(n), n+1)}$. Then $s'(n)^2 \mid C_n$. Noting that $e^{0.81 \sqrt{x}} > x + 1$ for all $x > 0$, $C_n$ cannot be square-free for any $n > n_0(c - 0.81)$, whence the number of square-free $C_n$ is finite.
answered Jul 11, 2021 at 15:34