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The short-support argument in the kernel rigidity page leaves a possible opening: on longer supports the fixed-prime relations can have nonzero first shadow $u$. The result here shows that this opening does not supply odd vectors of cost $o(2^J)$ using only those relations. The proof uses a comparison sequence preserving the relations exactly, rather than a dimension count. It is Plasma AI's author-recorded work, without independent review, and does not resolve Problem 249.

Rational comparisons with the exact prime-power values

The claim L220 proves the following stronger form of a fixed-identity comparison. For any fixed $y\ge2$, $M\ge1$ and $0<\varepsilon<1$, a nonempty interval about $S=\sum\varphi(n)2^{-n}$ consists of binary sums of positive integer functions $f$ such that

• $f$ agrees with $\varphi$ at every prime power and at $1$;
• $f\equiv\varphi\pmod M$ and $(1-\varepsilon)\varphi\le f\le(1+\varepsilon)\varphi$ everywhere;
• $f(pn)=(p-1)f(n)$ for $p\nmid n$ and $f(pn)=pf(n)$ for $p\mid n$, for every prime $p\le y$ and every $n$;
• only $O(X/\log X)$ integers up to $X$ have changed values.

In particular the comparison can have a dyadic rational binary sum. Its construction changes selected composite integers $e$ with no prime factor up to $y$, and propagates each change to the entire orbit $se$, where $s$ is supported on the primes up to $y$. The rounding weight is the complete positive series $\sum_s\varphi(s)2^{-se}$, not just $2^{-e}$. The proof controls that weight and all its later contributions.

The exception count also preserves every empirical distribution of windows of length $o(\log N)$ up to $o(1)$ in total variation: a union bound over the exceptional set proves this exactly as in the [[research/totient_binary_new_attack/perturbation_barrier|sparse comparison note]]. These are statements about empirical distributions with a vanishing error, not equality of every individual window or of finer error terms.

A cost bound without a support-length restriction

For the fixed $y$, write $n=s_y(n)k_y(n)$ with $s_y(n)$ supported on primes up to $y$ and $k_y(n)$ coprime to their product. Let $\mathcal F_y$ be the class of functions

$$f_h(n)=\varphi(s_y(n))h(k_y(n)),$$

where $h$ is any real function on the integers coprime to that product, with $h(1)=1$ and $h(p^a)=\varphi(p^a)$ for every prime $p>y$ and $a\ge1$. This class obeys exactly the prime laws just stated and fixes all prime-power values. It includes $\varphi$ and the comparisons supplied by L220. No bound on the other values of $h$ is imposed in the definition.

For a function $f$ and integer $J\ge1$, set $U_J^f(n)=\sum_{j=1}^Jf(n+j)2^{J-j}$. A finitely supported integer vector $z=(z_n)$ is robust here when

$$\sum_n z_n U_J^f(n)=0\qquad\hbox{for every }f\in\mathcal F_y. \tag{1}$$

This is the precise scope of cancellation deduced solely from these fixed-prime laws and prime-power values. Coefficients chosen using further values of $\varphi$ need not satisfy (1).

Proposition. For every fixed $y\ge2$ and $0<\varepsilon<1$, there is $B_0$ such that, for every $J\ge1$, every integer vector $z$ supported on arbitrarily many indices $n\ge B_0$ and satisfying (1), if

$$v_\varphi(z):=\sum_nz_n\sum_{m=1}^n\varphi(m)2^{n-m}\equiv2\pmod4,$$

then its complete-tail cost satisfies

$$C_J(z):=\sum_n|z_n|(n+J+2)\ge\frac{2^{J+1}}{1+\varepsilon}. \tag{2}$$

Proof. Apply L220 with modulus $M=4$. Its interval contains a dyadic rational $r=a/2^w$, with $a\in\mathbb Z$ and $w\ge0$, and hence a positive comparison $f\in\mathcal F_y$ with this binary sum and $f(n)\le(1+\varepsilon)n$. Fix it and take $B_0=\max(3,w+2)$. Put

$$\Phi_n^f=\sum_{m=1}^nf(m)2^{n-m},\quad T_n^f=2^nr-\Phi_n^f=\sum_{j\ge1}f(n+j)2^{-j},\quad u=\sum_nz_n2^n,\quad v_f=\sum_nz_n\Phi_n^f.$$

Absolute convergence and (1) give the complete-tail identity

$$ur-v_f=\sum_nz_nT_n^f =2^{-J}\sum_nz_nT_{n+J}^f. \tag{3}$$

Since $n\ge w+2$ on the support, $ur$ is an integer divisible by four. Since $f\equiv\varphi\pmod4$, one has $v_f\equiv v_\varphi\equiv2\pmod4$. Thus the left side of (3) has absolute value at least two. The right side has absolute value at most $(1+\varepsilon)2^{-J}C_J(z)$, because $0<T_m^f\le(1+\varepsilon)(m+2)$. This proves (2). $\square$

Consequently, if $y$ is fixed and the lower support cutoff tends to infinity, every family of robust odd vectors has $\liminf C_J(z)/2^J\ge2$. In particular it cannot meet the sufficient $o(2^J)$ criterion. This conclusion holds even when the support length grows without bound. The finite examples of robust vectors with $u\ne0$ on the kernel rigidity page are consistent with it: nonzero $u$ does not supply a cheap odd vector.

A quantitative bound when the prime set grows

There is also a weaker bound with an explicit constant, useful when $y$ varies with $J$. Choose any distinct primes $r_0,t_0>y$, and put

$$a=r_0t_0,\qquad P=\prod_{p\le y}p,\qquad Q=aP.$$

For every robust odd vector supported on positive indices, with no restriction on the lower cutoff or support length,

$$C_J(z)\ge 2^{J-Q-3}. \tag{4}$$

Here odd still means $v_\varphi(z)\equiv2\pmod4$. To prove (4), use the same orbit weights $A_e$ as in the proof of L220, but choose $e_j=a+(j-1)Q$, target zero, and modulus four. Set $\eta_0=-S$ and round $\delta_j=4\lfloor\eta_{j-1}/(4A_{e_j})\rceil$. Then $|\eta_j|\le2A_{e_j}$. Since $1<S<2$ and $a\le Q$, the first change satisfies $|\delta_1|\le2^{a+1}+2<2^{Q+3}$. Each later change satisfies

$$|\delta_j|\le2\left(\frac{A_{e_{j-1}}}{A_{e_j}}+1\right) \le2^{Q+2}+2<2^{Q+3}.$$

Propagate these changes through the smooth orbits as before. The result $f\in\mathcal F_y$ is integer-valued, congruent to $\varphi$ modulo four, has binary sum zero, and satisfies $|f(n)|\le2^{Q+4}n$. It need not be positive, which is permitted in the definition of $\mathcal F_y$. Absolute convergence and the identity (3), now with $r=0$, give $2\le|v_f|\le2^{Q+4-J}C_J(z)$, proving (4).

Thus if $Q=o(J)$ along a growing-prime construction, its robust odd vectors have cost at least $2^{J-o(J)}$. In particular a bound $C_J(z)=2^{o(J)}$ cannot hold in that regime. This is weaker than ruling out every $o(2^J)$ bound: $2^{J-\sqrt J}$ satisfies both asymptotic inequalities. The explicit $Q$ makes that limitation visible, without an estimate for primes in a varying range.

What remains open

The constant $B_0$ depends on the fixed prime set and on $\varepsilon$. There is no uniform bound for $y=y(J)\to\infty$. Such a growing-prime construction, or one using exact totients at rough composite arguments, is outside the fixed-$y$ proposition, and (4) applies only at its stated quantitative strength. The normalized prime-dilation method likewise uses primes growing with its scale and is not excluded by (2).

The comparisons are not fully multiplicative. Keeping every prime-power value while imposing full multiplicativity would determine $f=\varphi$. Thus L220 does not provide the fully multiplicative comparison proposed in the sparse comparison note, which remains a separate question. Nor do these comparisons rule out a proof using exceptional windows, quantitative distribution errors smaller than the altered proportion, or other arithmetic information they fail to preserve.

Finite checks and the cutoff distinction

The evidence entry point verifies five retained integer vectors against every affine constraint defining $\mathcal F_y$, as well as their kernel equation, odd shadow and cost. They were found by weighted LLL; none is claimed to minimize cost.

The small costs for larger $y$ at these finite $J$ do not contradict (2): its threshold $B_0$ depends on $y$ and $\varepsilon$ and is not asserted to be below these supports. Nor do the large returned costs prove a minimum; the asymptotic lower bound rests on the comparison proof.

The same entry point certifies a finite prefix of the signed target-zero construction used for (4), with $(a,Q)=(15,30)$ and $(35,210)$. Rational intervals bound $S$ and the complete orbit weights, and each interval determines one nearest-integer choice uniquely. The resulting sequences obey the prime laws, fixed prime-power values, congruences and signed size bound through $n=10000$. A control omitting propagation to the smooth multiples fails a prime law. This check does not instantiate L220's positive interval at its much larger sufficient starting point or prove an infinite equality by finite computation; those conclusions use the written arguments.