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Let \(d(n)\) represent the total number of divisors of \(n\) including negative integers. Find all the natural numbers such that \[\large d(d(\ldots d(n)))=n\] where the composition is arbitrarily finite Kainui's derivation for lower bound of \(n\) : https://www.overleaf.com/read/rvgfgxkwbmnc
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I found one solution by trial and error : d(8) = 8 d(d(8)) = 8 ...
there coule be a multiple repetition case
4 is another one
there could be cases like alternating like maybe something like 6 to 8 to 6 to 8
4 doesn't work
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oh my bad 4 is also a divisor xD
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