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We define $$S(n)=\sum_{a=2+(n\pmod 2)}^{n-2} \sharp(\{j,1\leq j<n \pmod{a},(a,j)=1\})\ .$$

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For $n>2$ we have the following experimental observations (checked up to $n=1000$) on $S(n)\pmod 4$:

For $n>2$ the number $S(n)$ seems to be even for $n\equiv -1,0\pmod 4$ and odd for $n\equiv 1,2\pmod 4$.

We have seemingly the following congruences modulo $4$:

$S(n)\equiv 0\pmod 4$ for all $n\equiv 0\pmod 8$.

If $n\equiv 1\pmod 8$ then $S(n)\equiv 1\pmod 4$ if and only if $n=p^{2e}$ is a prime-power (of necessarily even exponent) for some prime $p\equiv 3\pmod 4$ and $S(n)\equiv 3\pmod 4$ otherwise (i.e. if $n\equiv 1\pmod 8$ is not a prime power of a prime $\equiv 3\pmod 4$).

If $n\equiv 2\pmod 8$ then $S(n)\equiv 3\pmod 4$ if and only if $n=2p^{2e}$ is twice a prime-power (of necessarily even exponent) for some prime $p\equiv 3\pmod 4$ and $S(n)\equiv 1\pmod 4$ otherwise (i.e. if $n\equiv 2\pmod 8$ is not twice a prime power of a prime $\equiv 3\pmod 4$).

If $n\equiv 3\pmod 8$ then $S(n)\equiv 0\pmod 4$ if and only if $n=p^{2e+1}$ is a prime-power (necessarily of odd exponent $2e+1$) of a prime $p\equiv 3\pmod 8$ and $S(n)\equiv 2$ otherwise.

If $n\equiv 4\pmod 8$ and $n\geq 12$ then $S(n)\equiv 2\pmod 8$.

If $n\equiv 5\pmod 8$ then $S(n)\equiv 1\pmod 4$.

If $n\equiv 6\pmod 8$ then $S(n)\equiv 1\pmod 4$ if and only if $n=2p^{2e+1}$ for some prime $p\equiv 3\pmod 4$ and $S(n)\equiv 3\pmod 4$ otherwise.

If $n\equiv 7\pmod 8$ then $S(n)\equiv 2\pmod 4$ if and only if $n=p^{2e+1}$ is a prime-power (necessarily of odd exponent $2e+1$) of an odd prime $p\equiv 7\pmod 8$ and $S(n)\equiv 0\pmod 7$ otherwise.

Is there an explanation (or counterexample) for these congruences?

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    $\begingroup$ In $S(n+1)-S(n)$ almost all is cancelled, this looks like a more tractable sequence $\endgroup$ Commented 2 days ago
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    $\begingroup$ @FedorPetrov : Indeed, we get $S(n)-S(n-1)=\phi(n)+\phi(n-1)-n$ leading to $S(n)=2\sum_{j=1}^n\phi(j)-{n+1\choose 2}-\phi(n)$. This proves the congruences easily. $\endgroup$ Commented 2 days ago
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    $\begingroup$ @RolandBacher So you do have an answer to your question? $\endgroup$ Commented 2 days ago

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