The Binary Totient Series

An impossible complete divisibility pattern

Let $a\ge 8$, $H=H_{2^a}$ and $J,K,m\in \mathbb{N}$, with

It is impossible that

Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.not_actualLcmTerminalDyadicStaircase_of_room Indeed, the last requirement would make ${\delta }_{2^a}(J+K)$ a multiple of $2^m$, whereas the short-window bounds give $0<{\delta }_{2^a}(J+K)<B<2^m$. The positivity input is

Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.lcmRayArithmeticLetter_pos_of_lt_two_mul, without a rationality hypothesis.

Omitting only the last divisibility requirement leaves the partial pattern in Proposition 6.86. It additionally requires ${\delta }_{2^a}(J+K)\le 2^m-B$. The resulting necessary equality is penultimate:

Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.puncturedDyadicStaircase_penultimate_eq_half The terminal difference is not pinned to $2^{m-1}$. Neither this necessary restriction nor the full-pattern impossibility supplies an unbounded family of successful partial patterns.

The remaining upper-endpoint residue

Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.ActualLcmTopEdgeResidueGap No unbounded family satisfying this condition is known.

The positivity theorem gives

Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.actualLcmTailDiff_shift_pos For $a\ge 8$, the residue test at depth $K$ uses the stronger bound $J+K+a+6<2\cdot 2^a$. Under these hypotheses, an integral initial tail difference would give an integer terminal carry $e=R_{2H+J+K}-R_{H+J+K}$ with $0<e<2H+J+K+2$. The tail identity is

Hence, when $2H+J+K+2<2^K$, it is the residue of $D(H,H+J,K)$, not of $e$, that equals $2^K-e$ (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.actualLcm_integral_forces_topEdgeResidue). More generally, the bounds $m\le K$ and $2H+J+K+2<2^m$ give

because $2^m$ divides $2^K$ and $0<e<2^m$. This contradicts ${\mathcal{G}}_a(J,K,m)$. Consequently, for $a\ge 8$ and $J+K+a+6<2\cdot 2^a$,

Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.actualLcmTailDiff_notMem_int_of_topEdgeResidueGap The one-sided inequality excludes the upper-endpoint interval forced by integrality. Positivity alone does not establish that inequality.

The cofinal target built from it is

Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.PowerTwoActualLcmTopEdgeResidueGapSupply This condition is unproved; its implication is proved:

Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.irrational_totientSeries_of_topEdgeResidueGapSupply

Five sufficient conditions for that residue inequality

The next five quantified conditions each imply irrationality. Conditions (i)–(iv) reach the upper-endpoint test; (v) gives nonintegrality directly. Their proven relations are summarised in Proposition 6.65.

1. For every $a_0$, there are $a\ge max(a_0,8)$ and $m\ge 0$ with $H=H_{2^a}$ such that

   $$\begin{array}{c}m+1+a+6<2\cdot 2^a,2H+m+3<2^m,\\ 2H+m+2\le (D(H,H,m)+{\delta }_{2^a}(m+1))mod{2}^m\le 2^m-(2H+m+3),\end{array}$$

   where ${\delta }_{2^a}(j)=\phi (2H+j)-\phi (H+j)$. The size condition includes the larger depth $m+1$, so the conclusion remains valid whichever of the two adjacent depths gives the certificate. This condition is unproved (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.PowerTwoAdjacentSuffixMidbandSupply); its sufficiency is proved (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.powerTwoActualLcmTopEdgeResidueGapSupply_of_adjacentSuffixMidband, Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.irrational_totientSeries_of_adjacentSuffixMidbandSupply).

2. For every $a_0$, there is $a\ge max(a_0,14)$ with the following property. Put $H=H_{2^a}$ and choose $q$ so that $2q+1$ is the least odd integer not smaller than $\lfloor {\log }_{2}H\rfloor +10$. Define integers $W_{a,j}$ recursively by

   $$\begin{array}{rl}{W}_{a,0}& =0,\\ W_{a,j+1}& =4W_{a,j}+\frac{{\delta }_{2^a}(2j+3)-2{\delta }_{2^a}(2j+2)+{\delta }_{2^a}(2j+4)}{2}.\end{array}$$

   The increments ${\delta }_{2^a}(s)$ are even, so the recurrence is integral. The required inequality is

   $$H+q+2\le W_{a,q}mod{4}^q\le 4^q-(H+q+2).$$

   At depth $m=2q+1$, the residue in (i) equals $2(W_{a,q}mod{4}^q)$. Dividing its two endpoint bounds by $2$ and using integrality gives exactly the threshold $H+q+2$. This condition is unproved (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.PowerTwoOddGuardTopEdgeHalfWordBandSupply). The width is $H+q+2$ and the depth is the prescribed depth; both must be retained when comparing this with a fixed-fraction band. Its sufficiency is proved (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.powerTwoAdjacentSuffixMidbandSupply_of_oddGuardTopEdgeHalfWordBand).

3. Use exactly the quantifiers and prescribed depth of (ii). Let $r=W_{a,q}mod{4}^q$ and define its centred representative by

   $$u_{a,q}=\{\begin{array}{ll}r,& 2r\le 4^q,\\ r-4^q,& 2r>4^q.\end{array}$$

   Thus $-4^q/2<u_{a,q}\le 4^q/2$; when a midpoint occurs, the positive representative is used. The equivalent inequality is

   $$H+q+2\le |u_{a,q}|.$$

   This condition is unproved (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.PowerTwoActualFinalTopEdgeMagnitudeSupply). At the prescribed depth the size bound $2(H+q+2)\le 4^q$ holds. For a residue in $[0,4^q)$, the band in (ii) is therefore exactly this absolute-value condition. The equivalence, including its quantifiers, is proved (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.topEdgeHalfWordBandSupply_iff_actualFinalCenteredMagnitudeSupply).

4. Allow the odd depth to vary, but retain both size bounds. For every $a_0$, require $a\ge max(a_0,8)$ and $q\in \mathbb{N}$ with

   $$\begin{array}{c}2q+2+a+6<2\cdot 2^a,2(H+q+2)\le 4^q,\\ H+q+2\le |u_{a,q}|,H=H_{2^a}.\end{array}$$

   This condition is unproved (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.PowerTwoFlexibleActualTopEdgeMagnitudeSupply). The first bound keeps the adjacent depth inside the short window; the second makes the required residue band nonempty. One implication passes through (i), and another gives the two-sided carry exclusion below (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.powerTwoAdjacentSuffixMidbandSupply_of_flexibleActualTopEdgeMagnitude, Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.irrational_totientSeries_of_flexibleActualTopEdgeMagnitudeSupply).

5. Retain the full quantifiers and the two size bounds in (iv), but replace its absolute-value inequality by

   $${\delta }_{2^a}(2q+2)\le 2u_{a,q}.$$

   This compares the signed centred residue with the terminal totient difference. It is not an unrestricted assertion at an arbitrary $q$. This condition is unproved (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.PowerTwoFlexibleActualTerminalDominanceSupply); its sufficiency is proved (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.irrational_totientSeries_of_flexibleActualTerminalDominanceSupply).

A further sufficient terminal-carry condition

With $H=H_{2^a}$, the unproved quantified condition is

Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.PowerTwoFlexibleActualTerminalCarryCorridorEscapeSupply Either alternative gives nonintegrality at that scale, so the quantified condition implies irrationality (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.irrational_totientSeries_of_flexibleActualTerminalCarryCorridorEscapeSupply).

The proof assumes an integer $z={\mathrm{\Omega }}_a$ and uses

Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.two_mul_actualOddHalfCenteredLift_eq_terminal_sub_trueCarry The carry is then a positive tail difference, strictly smaller than $2H+2q+3$. The negative endpoint representative is minus this carry (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.actualLcm_trueEndpointSurvivor_neg). The first alternative contradicts the upper bound; the second contradicts positivity, which is the terminal-dominance argument (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.actualLcmTailOrbit_notMem_int_of_actualTerminalDominance).

Both arguments are conditional contradictions, not evidence against either unproved estimate. No converse from nonintegrality to either finite inequality is asserted. In particular, the exact identity must not be described as making this sufficient condition necessary.

Certificates in short windows

For every $t,j\in \mathbb{N}$, the arithmetic difference on the LCM progression is exactly

No coprimality restriction on $j$ is needed (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.lcmRayArithmeticLetter_eq_deltaTotient). Consequently the corresponding window numerator is

and applying the same residue and error bounds gives precisely the test $\mathcal{C}(H_t,H_t,L)$ (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.lcmDiagonalArithmeticKill_iff_certifiedKill). Thus the quantified assertion

implies irrationality (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.PowerTwoActualLcmShortArithmeticKillSupply, Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.actualLcmOrbitNonintegralitySupply_of_shortArithmeticKill); the assertion itself is unproved. It restricts the diagonal indices to powers of two and imposes the additional bound $L<2\cdot 2^a$. Pointwise completeness allows an unspecified finite depth; it does not imply a witness within this short window.

The displayed certified examples have $a=4$, $L=23$ and $a=6$, $L=93$ (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.actualLcmTailOrbit_four_and_six_notMem_int); these are finite computations. The bounded statement of Theorem 9.22 uses the single witness $(a,L)=(6,93)$ to cover all thresholds $a_0\le 6$. It makes no assertion for larger thresholds, or that these are the only successful exponents.

A sufficient approximation bound

Put $H=H_{2^a}$ and define the finite rational and its error bound by

For every $a,q\in \mathbb{N}$,

Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.abs_actualLcmTailOrbit_sub_rawApprox_lt Let $2q_a+1$ be the least odd integer not smaller than $\lfloor {\log }_{2}H\rfloor +10$, as in condition (ii). The sufficient quantified separation condition is

This condition is unproved; its sufficiency is proved (Erdos249257.DiagonalFreshLossBridge.PowerTwoOddWindowAffine.irrational_totientSeries_of_actualLcmOrbitSeparationSupply). The triangle inequality then gives $|{\rho }_{a,q_a}-z|>1/32$ for every integer $z$. The hypothesis concerns the real tail ${\mathrm{\Omega }}_a$, not merely its computable approximant. The truncation estimate transfers the stated separation, but does not itself establish it. Also, the depth is prescribed by $a$; it is not a free parameter in this hypothesis.