Totient comparisons preserving finitely many prime laws fill an interval

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_proof: Greedy rounding against complete smooth-orbit weights preserves finitely many prime multiplication laws; sparse composite rough parts protect every prime-power value, and a nearby target keeps all coefficients positive.

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For every fixed $y\ge2$, $M\ge1$ and $0<\varepsilon<1$, the values $\sum_{n\ge1}f(n)2^{-n}$ fill a nonempty open interval about $S=\sum_{n\ge1}\varphi(n)2^{-n}$ even when $f$ must satisfy all of:

• $f$ is positive and integer-valued, with $f(1)=1$ and $f(p^a)=\varphi(p^a)$ at every prime power;
• $f(n)\equiv\varphi(n)\pmod M$ and $(1-\varepsilon)\varphi(n)\le f(n)\le(1+\varepsilon)\varphi(n)$ for all $n$;
• for every prime $p\le y$, the exact laws $f(pn)=(p-1)f(n)$ when $p\nmid n$ and $f(pn)=pf(n)$ when $p\mid n$;
• $\#\{n\le X:f(n)\ne\varphi(n)\}=O(X/\log X)$.

In particular such a function can have a dyadic rational sum. The interval and the implied constant may depend on $y,M,\varepsilon$; there is no uniform assertion as $y$ grows. Full multiplicativity is not asserted.

The proof makes changes at selected integers with at least two prime factors above $y$, then propagates each change through multiplication by integers supported on primes up to $y$. Rounding uses the complete contribution of each such orbit, including all its multiples.

The result is Plasma AI's, recorded at tier 0 without independent review or a novelty claim. Its proof is elementary and consumes no native claim. It neither proves nor disproves [[problems/irrationality/E0249/_index|the irrationality asked for in Problem 249]]. The [[research/totient_binary_new_attack/partial_multiplicativity|research application]] proves an exponential cost bound for odd kernel vectors whose cancellation uses only these fixed-prime laws and prime-power values, on supports of arbitrary length. The distinction between fixed and growing sets of primes is essential.