Selfridge attack: complete classification of the first two complementary minimum layers

6 September 2026. Primary mode: ATTACK.

1. Scope and theorem

This packet continues the normalized primitive proper three-addition attack on Selfridge’s integer-complexity question

$$\lVert2^n\rVert\stackrel{?}=2n.$$

It closes the two layers left by the preceding exact lift lock. The setting is

$$\frac{1+3^b}{2^f}W+3^g=2^K, \tag{1}$$

where $f=1$ has $b\ge2$ even, the remaining split at $f=2$ has $b\ge3$ odd, and

$$W=1+2^x3^y\quad\hbox{or}\quad W=2^x+3^y.$$

The previous valuation theorems give

$$m:=\min(b,y,g)=1+v_3(K+f-h), \tag{2}$$

with $h=0$ in the first type and $h=x$ in the second.

Theorem. Subject to the inherited normalization and quadratic-reciprocity gate, every solution of (1) with $m\in\{1,2\}$ is one of seventeen explicit proper presentations. Every presentation costs strictly more than $2N$, where $N=K+f$; the minimum excess is six ones. Consequently these complete minimum layers contain no Selfridge counterexample.

This is a theorem about a restricted expression grammar, not about arbitrary integer-complexity formulas.

2. Input: the forty exact lift graphs

The preceding certificate classified 32 canonical minimum strata. Nine were empty. Every state in the other 23 lay on one of exactly forty coordinatewise lift graphs with common period

$$P=20{,}652{,}025{,}680, \qquad P=2^4 3^5\cdot5\cdot11\cdot13\cdot17\cdot19\cdot23. \tag{3}$$

Twenty-nine graphs were centered at seventeen proper presentations. Eleven were centered at nine improper rational identities involving $b=0,1$, $g=0$, $y=0$, or a negative ternary exponent. The lift-lock certificate and its independent reconstruction are included in the reproducible archive.

3. The next ternary digit removes every improper boundary graph

The order of 2 modulo $3^6=729$ is 486, and $486\mid P$. Thus the residues of $2^x$ and $2^{K+f}$ modulo 729 are unchanged along every lift graph. A positive ternary exponent congruent to a centered exponent modulo $P$ is either the centered positive exponent itself or is greater than five, in which case its power of three vanishes modulo 729. A nonpositive centered exponent can only lift to the latter case.

Therefore every graph has at most eight low/high patterns for $(3^b,3^y,3^g)$. Substitution in the equation multiplied by $2^f$ gives the following complete nonzero difference sets modulo 729 for the eleven improper graphs. Type II is abbreviated II, and the root is $(b,x,y,g,K)$.

$$\begin{array}{c|c|c|c|l} f&\text{type}&m&\text{root}&\text{possible LHS-RHS residues}\quad(\bmod729)\\ \hline 1&I&1&(0,1,1,0,3)&720\\ 1&I&1&(-2,3,0,1,3)&720\\ 1&I&2&(2,4,-1,-1,5)&675\\ 1&II&1&(0,2,1,0,3)&720\\ 2&I&1&(1,2,1,5,8)&447,690\\ 2&I&2&(1,1,1,2,4)&702,708\\ 2&II&1&(1,1,1,1,3),\ \min=y&702,714\\ 2&II&1&(1,2,1,2,4),\ \min=y&672,708\\ 2&II&1&(1,1,1,1,3),\ \min=g&711,714\\ 2&II&1&(1,2,2,1,4),\ \min=g&681,690\\ 2&II&1&(1,1,1,1,3),\ \min=y=g&714 \end{array} \tag{4}$$

None contains zero. Hence all eleven boundary-rooted lift graphs are empty. This is the higher 3-adic obstruction that the rational centered identities concealed: negative or zero ternary powers can satisfy the rational identity, but every positive lift loses exactly the low ternary contribution needed for the equality.

4. Twenty-six control graphs freeze completely modulo 729

Apply the same low/high enumeration to the 29 control-rooted graphs. On 26, the only surviving pattern keeps $b,y,g$ equal to their centered values. Once those three coordinates are fixed, $x$ cannot make a nonzero lift. Indeed, after multiplying (1) by $2^f$, reduction modulo $2^x$ requires

$$2^x\mid \begin{cases} (1+3^b)+2^f3^g,&W=1+2^x3^y,\\ (1+3^b)3^y+2^f3^g,&W=2^x+3^y. \end{cases} \tag{5}$$

The exact valuations of these fixed integers are smaller than the next possible lift $x+P$. If $x=1$ or 2, the lift-lock quotient already treats it as an exact preperiodic coordinate. Thus $x$ remains centered, and (1) then fixes $K$. These 26 graphs contain only their displayed controls.

5. The three residual graphs reduce to two equations

Modulo 729 leaves precisely three noncentered ternary patterns:

$$\begin{array}{c|c|c} \text{root }(b,x,y,g,K)&\text{possible lifted coordinate}&\text{reduction}\\ \hline (6,3,1,4,12)&b&2^{K+1}=11\cdot3^b+173\\ (6,1,2,4,12)&b&2^{K+1}=11\cdot3^b+173\\ (2,5,3,6,10)&g&2^K=3^g+295. \end{array} \tag{6}$$

For the first row, a lift of $x$ is impossible. Since $b\equiv6\pmod P$, also $b\equiv6\pmod{16}$, and

$$v_2(3^{b+1}+165)=4.$$

But reduction of the multiplied equation modulo $2^x$ would require $2^x\mid3^{b+1}+165$; a nonzero lift has $x\ge3+P$. Thus $x=3$, giving the first equation in (6). The second row has the exact coordinate $x=1$ and gives the same equation. In the third row, $g\equiv6\pmod{16}$, so

$$v_2(3^g+135)=5.$$

A lifted $x$ would require $2^x\mid3^g+135$, hence $x=5$, giving the second equation in (6).

It remains to show that each exponential equation has only its centered solution in the inherited progressions.

6. Explicit linear-forms cutoff

We use the standard rational specialization of Matveev’s explicit lower bound. If

$$\Lambda=\alpha_1^{c_1}\cdots\alpha_t^{c_t}-1\ne0,$$

the algebraic numbers are positive rationals, $B\ge\max|c_i|$, and $A_i\ge\max(\log\alpha_i,0.16)$, then

$$\log|\Lambda|> -1.4\,30^{t+3}t^{4.5}(1+\log(tB))A_1\cdots A_t. \tag{7}$$

For

$$2^{K+1}=11\cdot3^b+173, \qquad b=6+Pr,\quad K=12+Pt, \tag{8}$$

take $(\alpha_1,\alpha_2,\alpha_3)=(2,3,11)$. Then

$$\Lambda=\frac{2^{K+1}}{11\cdot3^b}-1=\frac{173}{11\cdot3^b}.$$

For $b\ge6$, the maximum coefficient is less than $3b$. The constant in (7) is

$$C_1=1.4\,30^6 3^{4.5}\log2\log3\log11 <2.615\cdot10^{11}. \tag{9}$$

If $b\ge10^{15}$, comparison of the exact upper value of $\log\Lambda$ with (7) would require

$$b\log3-\log(173/11)<C_1(1+\log(9b)). \tag{10}$$

At $10^{15}$, the left side exceeds the right by a factor greater than 111, and their difference is increasing thereafter. Hence $b<10^{15}$.

For

$$2^K=3^g+295, \qquad g=6+Pr,\quad K=10+Pt, \tag{11}$$

use $(2,3)$. The corresponding constant is

$$C_2=1.4\,30^5 2^{4.5}\log2\log3<5.862\cdot10^8. \tag{12}$$

At $g=10^{15}$, the analogue of (10) has safety factor greater than 50,000 and is increasing. Thus $g<10^{15}$.

The sole external proof dependency introduced in this run is E. M. Matveev, An explicit lower bound for a homogeneous rational linear form in the logarithms of algebraic numbers. II, Izvestiya: Mathematics 64 (2000), 1217–1269, https://doi.org/10.1070/IM2000v064n06ABEH000314.

7. Exact finite descent

The cutoff leaves

$$1\le r\le48{,}421. \tag{13}$$

The rational inequalities

$$\frac{15849}{10000}<\log_2 3<\frac{317}{200} \tag{14}$$

are checked exactly by

$$2^{15849}<3^{10000},\qquad 3^{200}<2^{317}.$$

Equation (8) implies

$$b\frac{15849}{10000}+3<K+1< b\frac{317}{200}+8, \tag{15}$$

because $8<11<16$ and $11\cdot3^b+173<184\cdot3^b<256\cdot3^b$. Equation (11) similarly gives

$$g\frac{15849}{10000}<K<g\frac{317}{200}+9. \tag{16}$$

Enumerating the integer lift indices $t$ allowed by the strict rational bounds (15) and (16) gives 117,363 pairs for each equation. Exact modular exponentiation leaves

$$\begin{array}{c|rrrrrr} \text{equation (8), after modulus}&\text{bounds}&29&43&59&83&101\\ \hline \text{pairs}&117363&7224&2412&47&1&0 \end{array}$$

and

$$\begin{array}{c|rrrrr} \text{equation (11), after modulus}&\text{bounds}&29&43&59&83\\ \hline \text{pairs}&117363&4815&2412&81&0. \end{array} \tag{17}$$

Thus neither equation has a nonzero lift. Their centered solutions are $(b,K)=(6,12)$ in (8) and $(g,K)=(6,10)$ in (11).

8. Complete list and cost

The seventeen distinct proper solutions are

$$\begin{array}{c|c|rrrrr|r} f&\text{type}&b&x&y&g&K&\text{cost}-2(K+f)\\ \hline 1&I&2&3&1&1&7&6\\ 1&I&2&5&1&3&9&12\\ 1&II&2&1&2&2&6&9\\ 1&II&2&3&1&2&6&10\\ 1&II&2&3&3&4&8&18\\ 1&II&2&4&2&1&7&10\\ 1&II&2&4&4&3&9&18\\ 1&II&2&5&1&4&8&16\\ 1&II&2&5&3&6&10&24\\ 1&II&6&1&2&4&12&15\\ 1&II&6&3&1&4&12&16\\ 2&I&3&3&1&4&8&16\\ 2&I&3&4&2&2&10&11\\ 2&II&3&3&2&2&7&14\\ 2&II&3&4&2&4&8&20\\ 2&II&3&6&4&2&10&20\\ 2&II&5&6&1&2&12&13. \end{array} \tag{18}$$

The smallest displayed excess is six. Therefore the complete $m=1,2$ complementary layers obey the desired Selfridge cost bound.

9. Evidence, status, and next question

The Python certificate reconstructs the forty input graphs, exhausts every low/high pattern modulo 729, checks the two 2-adic reductions, evaluates the Matveev constants, enumerates the strict rational windows, and reproduces all counts in (17). A separately written C++20 program starts from the frozen graph list and independently reconstructs the 11/26/3 split and both finite modular descents.

This is an exact computer-assisted restricted theorem with one classical external dependency. It has not received external review, and novelty has not been assessed. It completes two layers inside the normalized three-addition grammar; it does not prove Selfridge’s conjecture.

The next useful move is not a third minimum layer one at a time. Equations (2) and (7) suggest a uniform argument: combine Matveev’s logarithmic cutoff with the exponentially growing congruence period $2\cdot3^{m-1}$ to seek a single inequality excluding every sufficiently large $m$, then classify only the remaining finite band. If that uniform comparison is too weak, move to the complementary $f=0$ six-row quotient.

NEXT MODE: ATTACK — test whether the valuation period and Matveev cutoff cross uniformly in $m$, aiming to replace layer-by-layer work by a finite minimum bound; otherwise open the complementary $f=0$ quotient.