We see the fundamental theorem of arithmetic and the sieve of Eratosthenes in the decomposition into weight × level + jump of natural numbers. Applied to prime numbers this decomposition leads to a new classification of primes.

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I've always been wary of "division".

Can I do this? ....

Given \(r \not \equiv 1 \pmod p\) and \(p \not \mid r \) ,

\[ 1 + r + r^2 + \ldots + r^{p-3} + r^{p-2} \equiv \frac{1-r^{p-1}}{1-r} \equiv \frac{1-1}{1-r} \equiv 0 \pmod p \]

This uses Fermat's Little Theorem, \(r^{p-1} \equiv 1 \pmod p\), and noting that the denominator \(1 - r \not \equiv 0 \pmod p\).

The alternative way is to multiply the series on the left by \(1-r\), we know is not congruent to 0, to get \(1-r^{p-1}\) and use FLT to get 0.

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My first, favorite and most important sequence, the weights of prime numbers: A117078
We see prime numbers classified by level and by weight on the graph.

A117078: a(n) is the smallest k such that prime(n+1) = prime(n) + (prime(n) mod k), or 0 if no such k exists ➡️ https://oeis.org/A117078

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If \(a \pmod n\) has order ϕ(𝑛) then 𝑎 is called a primitive root of 𝑛 .

This is the definition in the textbook I'm following, and if differs from what is common on eg wikipedia.

This definition doesn't tell me that a primitive root, if it exists, is always a prime number.

That is, 4 could never be a primitive root but 3 could be.

Is it true? And is the explanation easy?

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