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For example,$ f (2) = 2$, and $f(3) = 6$.

I don't know what kind of technique I should use.

By putting $n = 1, 2, 3, \cdots$ it's even hard to know if $f(2) = 2$ for ALL integer $n$.

an4s
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Ikuyasu
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    Hint: $f(k)$ equals the product of all prime natural numbers $p$ such that $p-1$ divide $k-1$. – Batominovski Feb 05 '20 at 01:45
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    https://oeis.org/A027760 – Ivan Neretin Feb 05 '20 at 07:55
  • Thank you. I've been taking a look at it. – Ikuyasu Feb 06 '20 at 01:16
  • Hi, WE Tutorial School. That f(k) works. I plug k = 2, 3, 4,.. and I get the correct sequence. But can you elaborate a little more on that idea? It's hard to see how (p-1) and (k-1) come up. I reviewed the Fermat's Little Theorem, but it only states that n^k - n is divisible by k when k is prime, but it's hard to deduce anything from it to your statement. – Ikuyasu Feb 07 '20 at 00:42
  • Does this answer your question: https://math.stackexchange.com/a/1387249/72152? – Batominovski Mar 20 '20 at 18:39

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