The Binary Totient Series

Observation 6.106 (The sign of the actual endpoint representative). Assume $a\ge 8$, $H=H_{2^a}$ and $J+K+a+6<2\cdot 2^a$, as in Proposition 6.84. If $d=R_{2H+J}-R_{H+J}$ is an integer, define

The tail recurrence gives $c_K=R_{2H+J+K}-R_{H+J+K}>0$. Its negative is an integer representative satisfying

Thus the representative corresponding to the actual tail is negative. The stated size bound allows it to remain in the endpoint interval, so its sign alone does not contradict integrality. Other integers in that interval with the same residue need not represent the actual tail. Proposition 6.85 gives the further residue separation that would exclude this case. actualLcm_trueEndpointSurvivor_neg

Observation 6.107 (The complete divisibility pattern is impossible). Let $a\ge 8$, $H=H_{2^a}$ and $J,K,m\in \mathbb{N}$. Assume

The simultaneous divisibilities

cannot hold. The terminal difference would be positive by Proposition 6.80, smaller than $2^m$, and divisible by $2^m$, a contradiction. The obstruction applies to this complete pattern under the stated size bounds. Omitting its last divisibility condition leads to the necessary restrictions in Proposition 6.86; the residue inequality in Proposition 6.64 is another sufficient condition. Neither statement here supplies such witnesses at arbitrarily large scales. not_actualLcmTerminalDyadicStaircase_of_room

Observation 6.108 (An exact valuation in a dyadically scaled family). For every integer $n\ge 0$, take $H=2^{n+2}$ and $j=3\cdot 2^{n+1}$. Then

Indeed, the three arguments are $9\cdot 2^{n+1}$, $7\cdot 2^{n+1}$ and $5\cdot 2^{n+1}$. Their totients are respectively $6\cdot 2^n$, $6\cdot 2^n$ and $4\cdot 2^n$, which give the identity. Thus the difference has $2$-adic valuation exactly $n+1$. This example limits a valuation bound for arbitrary even heights and scaled odd offsets. Here $j>H$, whereas the offset in Proposition 6.88 is at most $2^{a/2}$ and the height is $H(2^a)$. The example therefore does not exclude additional divisibility on that LCM family. fixedRankSecondDifference_dyadic_fixture_exact

Observation 6.109 (A rational sequence with matching parity).

Start with the coefficient sequence $0,1,1,2,4,4,4,\dots$, indexed from $0$. For each $m=2^{k+3}$, $k\ge 0$, replace the pair at positions $m,m+1$ from $(4,4)$ to $(6,0)$, leaving all other coefficients unchanged. These pairs are disjoint. Each replacement preserves the binary sum because

The resulting sequence has the same sum $3/2$ and the same parity at every index as $\phi$. It is not eventually periodic, since the gaps between its occurrences of $6$ tend to infinity. This construction uses the particular initial parity pattern and the binary weights; it does not establish an analogous result for arbitrary prescribed parities or other summand weights. parityCoboundaryWeight

Observation 6.110 (What the parity example disproves). The sequence $c$ in Observation 6.109 satisfies

and is not eventually periodic. More precisely, for every $N,G,K\in \mathbb{N}$, it has $K$ of the displayed $(6,0)$ pairs starting beyond $N$, with successive starting positions separated by more than $G$. Choose sufficiently late powers of two in the construction to obtain these pairs. Hence the growth bound, the exact parity pattern, nonperiodicity and these arbitrarily separated occurrences are compatible with a rational sum. They cannot by themselves prove irrationality. The construction does not preserve the complete totient sequence or its multiplicativity, and does not specify which additional arithmetic hypothesis would suffice. exists_totientParity_arbitrarilyManySeparatedCarry_rational_countermodel

Observation 6.111 (A Mersenne factor does not force tail-difference divisibility). For the completely multiplicative sequence $c(n)=n$, the scaled tail is

If $q>K>0$ and $q\mid 2^K-1$, then every such tail difference is integral but is not divisible by $q$. For example, take $K=3$ and $q=7$; the multiplicative order of $2$ modulo $7$ is $3$. Thus even this prime divisor of the homogeneous recurrence multiplier does not force divisibility of the inhomogeneous tail difference. An application to the totient sequence needs a further relation between the prime and the coefficient residues or the tail’s size. This example supplies no general obstruction for Problem #257 merely because that problem also involves Mersenne denominators. Compare the specific totient construction in Proposition 6.91. idCoeff_mersenneModulus_does_not_annihilate_tailShift

Observation 6.112 (A lower bound for the stated reconstruction cost). Let $I$ be finite, $w_i\in \mathbb{Q}$ and $x_i\in \mathbb{N}$, and assume $\sum _{i\in I}{w}_i\phi (x_i)=1$. Then

Indeed, $1\le \sum _i|w_i|\phi (x_i)\le \sum _i|w_i|x_i$, and the displayed cost is $3\sum _i|w_i|x_i+4\sum _i|w_i|$. For $x\ge 1$, the reconstruction $\phi (x)=2R_{x-1}-R_x$ and the separate estimates $R_M\le M+2$ give exactly this cost. Terms at $x=0$ contribute nothing to the normalisation, since $\phi (0)=0$. This particular termwise triangle bound cannot be smaller than one. The conclusion is about that bound, not the true combined tail error; cancellation or a different estimate is not excluded. The same elementary calculation applies to any nonnegative sequence $c(n)\le n$ under the corresponding normalisation $\sum _i{w}_{i}c(x_i)=1$. three_le_totientAdjugateTailCost

Observation 6.113 (A second-difference test need not use less depth). For $h,N,L\in \mathbb{N}$, put

The second-difference certificate is the finite inequality

It implies $R_{N+2h}-2R_{N+h}+R_N\notin \mathbb{Z}$. The two first-difference truncation errors have bounds $N+2h+L+2$ and $N+h+L+2$ after multiplication by $2^L$; their sum is at most the stated $B_2$. This explains the radius used by the linked test, without claiming it is optimal. At $(h,N)=(1,8)$, the first-difference test holds at depth $8$, whereas the second-difference test fails at every depth at most $8$ and first holds at $9$. The earlier 40-cell probe for $t\le 20$ reported the first test at least as shallow in 30 cells; that aggregate is a reported computation, not part of the single-cell theorem linked below. These comparisons do not give a uniform ordering of the depths. totient_tail_rank_two_kill_sound_but_not_shallower_cell

Observation 6.114 (The prime-power restriction in the unit-gap refinement). The ordinary interval test excludes a rational denominator when its possible numerator interval contains no integer. Reduced fractions permit a refinement: every numerator is coprime to its denominator, so an interval containing only nonunits can also be excluded. For a candidate $S=a/(2^{c}u)$ in lowest terms, with $u>0$ odd and $N\ge c$, the integer $u{R}_N=2^{N-c}a-u{\mathrm{\Phi }}_N$ is coprime to $u$. For such an $N$ and any $K\ge 0$, define

After subtracting an integer multiple of $u$, the finite prefix of $u{R}_N$ is $J+W/2^K$ and its remaining tail lies in $(0,u(N+K+2)/2^K]$. Its possible integer values are therefore $J+1,\dots ,J+L$. The refined condition is $gcd(J+t,u)>1$ for all $1\le t\le L$, contradicting the required coprimality of $u{R}_N$. For a prime $p$ and $e\ge 1$, nonunits modulo $p^e$ are exactly the multiples of $p$. Since two consecutive integers cannot both be multiples of $p$, the condition for $u=p^e$ is equivalent to

Thus a successful refined certificate at a prime-power denominator has at most one candidate integer. This is not a bound for an arbitrary window: $(u,N,K)=(3,3,3)$ gives $V=2$, $J=0$, $W=6$ and $L=3$, and the test fails because $1,2,3$ are not all nonunits modulo $3$. The prime-power restriction also matters. With the actual totient window $(u,N,K)=(15,13,8)$, one gets

Both candidates $9,10$ are nonunits modulo $15$. This interval passes the refined test but not the empty-interval test. The prime-power bound therefore neither applies to all composite denominators nor bounds the possible improvement in the final denominator cutoff. The prime-power criterion is proved in Lean (reducedDenominatorCandidateCount_prime_pow_le_one).

Observation 6.115 (A synthetic sequence preserving the stated LCM differences). Here is the recurrence behind the comparison. Fix $t\ge 3$, put $H=\mathrm{lcm}(1,\dots ,t)$ and $A=\phi (H)$, and define

The index $i$ here corresponds to the argument $H+i+1$ of a totient difference. For $2\le q<t$ this gives

The equality follows because all prime divisors of $q$ and $q+1$ divide $H$: hence $\phi (qH)=qA$ and $\phi ((q+1)H)=(q+1)A$. Every $a_i$ is divisible by $A$, and $|c_i|\le A$, $|a_i|\le 2A$. For finitely supported integer weights $({\lambda }_j{)}_{j\ge 0}$, put $d_i=\sum _j{\lambda }_j{c}_{i+j}$ and $b_i=\sum _j{\lambda }_j{a}_{i+j}$. Linearity and telescoping give

The triangle inequality gives the precise uniform bounds

These are uniform in $i$, not in the weights. The factor $\sum _j|{\lambda }_j|$ cannot be omitted: taking only ${\lambda }_0=2$ doubles the nonzero state and coefficient maxima. Divisibility by $A$ and by $\phi (d)$ for $d\mid H$ is preserved, since $\phi (d)\mid \phi (H)$.

Finite shift polynomials therefore preserve the recurrence, but not the original numerical bounds independently of their coefficients. This comparison does not identify $a_i$ with actual totient differences away from the prescribed indices, or bound the terms discarded by a proposed projection. The construction and the weight-dependent bounds are in lcm_factorIdeal_finiteRank_shiftAlgebra_not_sufficient. The decomposition of the actual totient into divisor contributions is proved in Lean (totient_eq_sum_mobiusTotientChannel).

Observation 6.116 (Compatible carries for fixed-precision local symbols). For example, prescribing an odd coefficient modulo $8$ permits any coefficient of the form $a+8z$ with $a$ odd and $z\in \mathbb{Z}$. Whatever the incoming integer carry $e$, one can choose $z$ so that $2e+a+8z$ lies between $-4$ and $3$. More generally, fix $u\ge 1$ and a finite list of integers $v_j\ge 0$ and odd integers $a_j$. Starting from any integer $e_0$, there are integers $z_j$ and coefficients

At each step, reduce $2e_j+2^{v_j}{a}_j$ to its centred representative modulo $2^{v_j+u}$, choosing the interval

Solving for $z_j$ gives the stated next carry. Iterate this choice. The bounds apply to the resulting states, not to the arbitrary initial state $e_0$. The unrestricted higher quotients $z_j$ are essential: the construction prescribes a valuation and a finite unit residue, not the full coefficients or their signs. It therefore does not realise a prescribed totient-difference sequence. A contradiction using only these local symbols cannot exclude all compatible integer carries; further restrictions on the coefficients need a separate argument. fixedPrecisionTropicalNoGo

Observation 6.117 (Correction-free factorisation does not ensure nonvanishing). Consider the two finite differences

They arise from the square-congruence choice $52\equiv 52(\mathrm{mod}{13}^2)$ and $52\equiv 2(\mathrm{mod}{5}^2)$. The four arguments factor as $13\cdot 5$, $5\cdot 13$, $13\cdot 6$ and $5\cdot 14$; in each product the prime and cofactor are coprime. Thus all four totients have the correction-free form $\phi (pm)=(p-1)\phi (m)$, yet both differences vanish. Nonvanishing is also possible: the choices $27\equiv 0(\mathrm{mod}{3}^2)$ and $27\equiv 2(\mathrm{mod}{5}^2)$ give

again with coprime factors at every term. The square-CRT construction removes the non-coprime multiplicativity correction on the prescribed finite range. That property alone does not imply nonvanishing. These are explicit witnesses of the two possibilities, not a claim that the first is a smallest counterexample. squareCRT_clean_block_can_vanish

Observation 6.118 (No fixed integer clears every primitive denominator). Let $A=\phi \ast \mu$ be the Dirichlet convolution used in Proposition 6.102. At prime powers,

These follow by subtracting consecutive prime-power values of $\phi$. For every odd prime $p$, the reduced denominator of $A(p)/p$ is $p$. For every prime $p$, the reduced denominator of $A(p^2)/p^2$ is $p^2$, including the denominator $4$ at $p=2$. Consequently any positive integer $D$ for which $DA(n)/n\in \mathbb{Z}$ for all $1\le n\le N$ must be divisible by every odd prime $p\le N$, every prime square $p^2\le N$, and by $4$ once $N\ge 4$. No fixed $D$ can do this for all $n$, since there are infinitely many odd primes. In periodic-point counting, the quantities $A(n)/n$ would count orbits of length $n$; their integrality and nonnegativity characterise realisable periodic-point sequences [28]. Here only the denominator obstruction is used. It excludes a fixed integer clearing all these particular coordinates, not every possible Euler-product representation or an argument with a multiplier growing with $N$. It is not an irrationality proof for $S$. no_fixed_integral_primitive_euler_index

Observation 6.119 (A finite exclusion is not a membership test). A successful residue certificate proves the exclusion stated in its soundness theorem. Failure of that sufficient condition at finitely many depths does not prove integrality of the tail difference. In particular, a finite unsuccessful search cannot be substituted for the eventual-period assertion needed to prove rationality. The same logical distinction occurs for the linked greedy test in a separate Mersenne achievement-set problem: its finite inequality implies nonmembership. The linked object is a definition, with a soundness theorem in the same file, not a theorem asserting a universal impossibility of finite rationality proofs for Problem #249. Membership or rationality established by some additional exact identity is not excluded by this observation. Likewise, the successful finite ranges in Propositions 6.94 and 6.104 remain finite ranges until an unbounded supply is proved. CertifiedGreedyMersenneDeath