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Do for every natural number $n$ exist finitely many nonzero integers $a_p$ such that $$\log(n) = \sum_{p \text{ prime}} a_p \log(p-1)?$$ equivalently: $$n = \prod_{p \text{ prime}} (p-1)^{a_p}$$

Edit This representation would be unique, if we allow only linearly independent primes in the summation: Linear independent prime numbers?

Do for every natural number $n$ exist (possibly negative) integers $a_p$, finitely many of them nonzero, such that $$\log(n) = \sum_{p \text{ prime}} a_p \log(p-1)\,?$$ Equivalently: $$n = \prod_{p \text{ prime}} (p-1)^{a_p}.$$

This representation would be unique, if we allow only linearly independent primes in the summation: Linear independent prime numbers?

Do for every natural number $n$ exist finitely many integers $0 \neq a_p \in \mathbb{Z}$ such that $$\log(n) = \sum_{p \text{ prime}} a_p \log(p-1)?$$ equivalently: $$n = \prod_{p \text{ prime}} (p-1)^{a_p}$$

Do for every natural number $n$ exist finitely many nonzero integers $a_p$ such that $$\log(n) = \sum_{p \text{ prime}} a_p \log(p-1)?$$ equivalently: $$n = \prod_{p \text{ prime}} (p-1)^{a_p}$$

Do for every natural number $n$ exist finitely many integers $0 \neq a_p \in \mathbb{Z}$ such that $$\log(n) = \sum_{p \text{ prime}} a_p \log(p-1)?$$

Do for every natural number $n$ exist finitely many integers $0 \neq a_p \in \mathbb{Z}$ such that $$\log(n) = \sum_{p \text{ prime}} a_p \log(p-1)?$$ equivalently: $$n = \prod_{p \text{ prime}} (p-1)^{a_p}$$