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The [[research/leads/totient_binary_series/integer_interpolation/phase_transference|phase comparison]] equates a small odd-kernel cost with a large phase error on a specified family. Its reverse implications from irrationality on other supports do not, by themselves, give a reverse implication on $[J,2J]$. Here the identity $\varphi(3k)=2\varphi(2k)$ for odd $k$ coprime to $3$ supplies the missing arithmetic on that short support.

The result is an equivalence along a subsequence of powers of two, and an exact finite sufficient test for phase separation. It proves no unbounded sequence of successful tests and does not resolve Problem 249. These are author-recorded, unreviewed arguments. The eventual structural assertions use the fixed-modulus prime number theorem at the disclosed standing below.

Notation and the structural hypotheses

Let $J=2^m\ge4$, use all indices $J\le n\le2J$, and put

$$H=3J+2,\quad a_t=\varphi(t)/2\quad(t\ge3),\quad A_n=\Phi_n/2,\quad V_n=U_J(n)/2,\quad w_n=n+J+2.$$

Here $\Phi_n$, $U_J(n)$ and the complete positive tail $T_n$ are defined on the [[research/leads/totient_binary_series/integer_interpolation/two_depth_carries|two-depth page]]. In particular

$$2^JT_n=U_J(n)+T_{n+J},\qquad 0<T_{n+J}\le w_n,\qquad 0<V_n<2^{J-1}(n+2). \tag{1}$$

Write $\|x\|=\operatorname{dist}(x,\mathbb Z)$, and let $\epsilon_J=\min_\theta\max_{J\le n\le2J} |\theta V_n-A_n/2|/w_n$. This minimum is attained, as on the phase comparison page. Define the finite set

$$\mathcal K_J=\{k:\ k\text{ odd},\ 3\nmid k,\ J<2k<3k\le2J\}, \quad D_k=a_{3k+J}-2a_{2k+J}.$$

The two structural hypotheses used below are

$$\gcd_{J<t\le2J}a_t=1,\qquad G:=\gcd_{k\in\mathcal K_J}D_k>0\ \hbox{and odd}. \tag{2}$$

The gcd is taken positive, ignoring zero entries; (2) requires a nonzero entry. Set $D_*=\max_{k\in\mathcal K_J}|D_k|$, so $1\le G\le D_*\le H$. Also $a_*:=\max_{J<t\le2J}a_t\ge a_{2J}=J/2$, hence $H/a_*\le8$.

Lemma 1. Conditions (2) hold for every sufficiently large power of two $J$. They also give the odd feasibility and row-rank hypotheses of Theorem 4 on the phase comparison page.

Proof. The integer $a_{2J}=J/2$ is a power of two. A prime $p\in(J,2J)$ with $p\equiv3\pmod4$ supplies the odd value $a_p=(p-1)/2$, proving the first gcd assertion. For the second, choose a prime

$$5J/2<P<3J,\qquad P\equiv3\pmod4,\qquad P\equiv J+3\pmod9.$$

These congruences specify a reduced class modulo $36$, since $J$ is coprime to $3$. Then $k=(P-J)/3$ is odd, is $1$ modulo $3$, and belongs to $\mathcal K_J$. Its $D_k=a_P-2a_{2k+J}$ is odd.

Both prime choices follow separately from the fixed-modulus theorem $\pi(x;q,r)\sim x/(\varphi(q)\log x)$ for fixed coprime $q,r$. The counts in these intervals are asymptotic to $J/(2\log J)$ and $J/(24\log J)$, respectively. Only six reduced classes modulo $36$ arise, so their thresholds can be made common. No growing modulus or simultaneous prime pair is used.

More generally, (2) implies that some $D_k$ is odd and some $a_t$ is odd. For such $k$, since it is odd and $3\nmid k$, $a_{3k}=2a_{2k}$ is even, while $a_{3k+J}$ is odd. Together with an odd $a_t$ from the first gcd assertion this gives an odd minor of the two rows $(a_t)$ and $(a_{t+J})$. Thus they have rank two. For $n\ge4$, $A_n\equiv a_n\pmod2$ and $V_n\equiv a_{n+J}\pmod2$. The same two columns have an odd minor and include an odd $V_n$. Dividing their scalar cross-product by their odd gcd produces an odd kernel vector. $\square$

The external prime-theorem statement was inspected in NIST DLMF §27.11, closing paragraph, version 1.2.8 (2026-09-15), and the introductory fixed-$q$ formula in Tao's 2009 exposition. Its analytic proof is neither reconstructed nor independently reviewed here. This use changes the fixed moduli from the earlier $20$ to $4$ and $36$; it makes no uniformity assertion for unbounded moduli. At an individual $J$, (2) can instead be checked directly without this asymptotic input.

Localizing a small phase

Proposition 2. Assume (2). Suppose $\delta>0$, $18\delta H^3<1$, and some real $\theta$ satisfies

$$\|\theta V_n-A_n/2\|\le\delta w_n \qquad(J\le n\le2J). \tag{3}$$

Then there is a nonzero odd integer $q$ with

$$|q|\le26\delta2^JH,\qquad \left\|\frac{qV_n}{2^{J+1}}-\frac{A_n}{2}\right\| \le31\delta H w_n\quad(J\le n\le2J). \tag{4}$$

There is also an integer $p$ such that

$$\left\|q2^{J-2}S\right\|\le38\delta H^2 2^{-J},\qquad \left|S-\frac{p}{q2^{J-2}}\right| \le\frac{152\delta H^2}{|q|2^{2J}}. \tag{5}$$

Proof. Choose integers $m_n$ so that the errors in (3) have absolute value at most $\delta w_n$. The recurrences give, for $J<t\le2J$,

$$\theta(a_{t+J}-2^Ja_t)-a_t/2-\ell_t=e_t, \quad \ell_t=m_t-2m_{t-1}\in\mathbb Z, \quad |e_t|\le3\delta H. \tag{6}$$

Subtract twice the relation at $2k$ from the one at $3k$. The terms involving $a_t$ cancel, so integers $h_k$ satisfy $|\theta D_k-h_k|\le9\delta H$. If $s$ realizes $|D_s|=D_*$, every integer $D_kh_s-D_sh_k$ has absolute value at most $18\delta HD_*<1$, and is zero. Thus $h_k/D_k$ has one common value on the nonzero entries. Because all $h_k$ are integers and their corresponding $D_k$ have gcd $G$, this value is $r/G$ for some integer $r$. Consequently

$$|\theta-r/G|\le E,\qquad E=9\delta H/D_*. \tag{7}$$

This integer-minor argument includes the small-error threshold; a gcd assertion alone would not give a quantitative approximation.

Put $\lambda=2^{J+1}G\theta$. Multiplying (6) by $2G$ and using (7) yields integers

$$q_t=2r a_{t+J}-Ga_t-2G\ell_t, \qquad |\lambda a_t-q_t|\le\eta, \qquad \eta=GH(E+6\delta). \tag{8}$$

For any $t,u$, the integer $a_uq_t-a_tq_u$ has absolute value at most $H\eta$. Moreover

$$H\eta=9\delta(G/D_*)H^3+6\delta GH^2 \le15\delta H^3<1.$$

These integers vanish. The first gcd assertion in (2) now gives an integer $Q$ with $q_t=Qa_t$ for every $t$. Hence $|\lambda-Q|\le\eta/a_*$. Since $q_t\equiv Ga_t\pmod2$ and some $a_t$ is odd, $Q$ is odd. Set $q=Q-2^{J+1}r$, also odd. Then

$$|q|\le2^{J+1}GE+\eta/a_* \le18\delta2^JH+120\delta H \le26\delta2^JH,$$

where $J\ge4$, $G\le D_*\le H$ and $H/a_*\le8$ were used.

The dyadic phase $q/2^{J+1}$ differs from $G\theta-r$ by at most $\eta/(a_*2^{J+1})$. Because $G$ is odd, multiplying (3) by $G$ preserves the half-integer target modulo integers. Equation (1) gives

$$\left\|\frac{qV_n}{2^{J+1}}-\frac{A_n}{2}\right\| \le\left(G\delta+\frac{\eta}{4a_*}\right)w_n \le(13G\delta+2GE)w_n\le31\delta H w_n.$$

This proves (4), including its parity requirement. Oddness of $G$ cannot be dropped from this step.

For any odd integer $q$, the exact complete-tail identity (1) implies

$$\frac{qV_n}{2^{J+1}}-\frac{A_n}{2} \equiv q2^{n-2}S-\frac{qT_{n+J}}{2^{J+2}}\pmod1. \tag{9}$$

The omitted algebraic term is $(q+1)\Phi_n/4\in\mathbb Z$. Thus (1) and (4) give $\|q2^{n-2}S\|\le38\delta H^2$ throughout the window. This bound is less than $1/3$, since $18\delta H^3<1$ and $H\ge14$. Choose the nearest-integer residuals $f_n$ for these quantities. Each $f_{n+1}-2f_n$ is an integer of absolute value less than $1$, hence zero. Therefore $|f_J|=2^{-J}|f_{2J}|\le38\delta H^2 2^{-J}$. Dividing by $|q|2^{J-2}$ proves (5). $\square$

The subsequence equivalence on the short support

Theorem 3. Along the powers of two $J$, the following statements are equivalent:

1. $S$ is irrational.
2. $\limsup_{J\to\infty,\ J=2^m}\epsilon_J2^J=\infty$.
3. On some unbounded subsequence of those $J$, there are odd kernel vectors supported on $[J,2J]$ with cost $o(2^J)$.

Proof. By Lemma 1 and Corollary 5 of the phase comparison page, 2 implies 3. The complete-tail criterion on the two-depth page, with lower cutoff $B=J\to\infty$, shows that 3 implies 1.

For the remaining implication, suppose 2 fails. Choose a fixed $M\ge1$ such that $\epsilon_J\le M2^{-J}$ for every sufficiently large power of two. Apply Proposition 2 with $\delta=M2^{-J}$, which eventually meets its threshold, and write

$$r_J=\frac{p_J}{q_J2^{J-2}},\qquad 0<|q_J|\le26MH_J,\qquad |S-r_J|\le\frac{152MH_J^2}{|q_J|2^{3J}}, \quad H_J=3J+2.$$

If $r_J\ne r_{2J}$, their common dyadic factors give $|r_J-r_{2J}|\ge1/(|q_Jq_{2J}|2^{2J-2})$. Multiplying the sum of their approximation errors by that denominator would imply

$$1\le988M^2\left( \frac{H_J^2H_{2J}}{2^J} +\frac{H_JH_{2J}^2}{2^{4J}}\right), \tag{10}$$

which is false for every sufficiently large $J$. Thus consecutive powers eventually give the same rational number. Their errors tend to zero, forcing $S$ to equal that rational number. This proves the contrapositive of 1 implying 2. $\square$

The dyadic factors in the denominator comparison are essential: using the product of the two denominators loses a factor $2^J$. This proves a subsequence assertion, not $\epsilon_J2^J\to\infty$ along all powers of two and not an all-depth polynomial cost bound. The earlier wider-support shadow construction still has its original scope; it was not restricted to the short support to obtain this theorem.

A finite exact sufficient test

Fix an integer $M\ge1$ and a power of two $J\ge4$ satisfying (2) and $18MH^3<2^J$. If, for each positive odd integer $q\le26MH$, there is an $n\in[J,2J]$ such that

$$\operatorname{dist}\left(qU_J(n)-2^J\Phi_n, 2^{J+2}\mathbb Z\right) >124MH(n+J+2), \tag{11}$$

then $\epsilon_J>M2^{-J}$. Indeed otherwise Proposition 2 supplies an odd $q$ in this range satisfying the opposite inequality at every index, by multiplying (4) by $2^{J+2}$. Negative $q$ gives the same phase distances as positive $q$ because $A_n$ is integral, so positive $q$ suffice.

This is a finite test on exact integers using totients only through $3J$, with $O(MJ)$ candidate numerators and $O(J)$ coordinates for each. All the infinite-tail reasoning is in Proposition 2, not in a truncation of (11). A failure of this sufficient test says nothing decisive about the optimal phase error. Successful tests for a finite set of parameters prove no asymptotic claim. An unbounded sequence of successful pairs with $M\to\infty$ would establish Theorem 3(2); no such sequence is proved here.

For a small structural check, at $J=32$ one has $\mathcal K_J=\{17,19\}$, $D_{17}=41-2\cdot10=21$ and $D_{19}=44-2\cdot12=20$, hence $G=1$. Also $a_{64}=16$ and $a_{43}=21$ give the other gcd in (2). These exposed computations check the index and parity conventions; they are not a phase certificate or evidence for the required growth.

The new finite test offers a concrete way to examine the arithmetic left by the phase comparison. Proving its success in a growing family, or obtaining separation by a different arithmetic argument, remains open. Neither this equivalence nor its connection with runs of binary digits rules out Fourier, lattice, signed-multiplier or other methods.