Consider the harmonic sequence $$H_n = \sum_{k=1}^{n} \frac{1}{k} $$ Set $$H_n = \frac{a_n}{n!}$$
Prove that $\gcd(a_n;n!)=1$
Gcd and harmonic sequence
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$\begingroup$
arithmetic
1 Answers
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Your conjecture is wrong from $n=4$ and onward.
Observe that
$$H_{4}=\frac{25}{12}$$
that is $a_{4}=50 \implies \gcd(a_4,4!)=\gcd(50,24)=2$