GPT 6 Astra Ultra attempt on Math-5445
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FORMAL RESTATEMENT
Erdős #125; reconstruction, 2026-09-23. Let $A$ and $B$ be the nonnegative integers whose base-$3$ and base-$4$ digits, respectively, belong to $\{0,1\}$. Does $A+B$ have positive lower asymptotic density? We prove that its lower density is zero. This does not assert that its upper density is zero.QUICK LITERATURE AND CONTEXT CHECK
The current problem page records a negative answer and a Lean proof. The linked formal development uses aligned powers of $3$ and $4$ and a contraction of counting densities. We reconstruct that argument in ordinary mathematical notation. This is an exposition of the known resolution, with no claim of novelty or of having run the Lean checker.WORK
Put $S=A+B$, $S(X)=|S\cap[0,X)|$, and $d(X)=S(X)/X$ for integral $X>0$. Splitting the digit strings after $k$ and $m$ positions gives \[ a=3^k a'+a_0,\quad b=4^m b'+b_0, \qquad a'\in A,\ b'\in B, \] with $0\le a_0\le(3^k-1)/2$ and $0\le b_0\le(4^m-1)/3$. Let $q=\min(3^k,4^m)$ and $r=\max(3^k,4^m)$. Consequently every $x=a+b<qM$ can be written \[ x=qy+c,\qquad y=a'+b'\in S,\quad 0\le y<M, \] where \[ 0\le c\le C:=(3^k-1)/2+(4^m-1)/3+(r-q)M. \tag{1} \] Indeed the excess term is $(r-q)a'$ or $(r-q)b'$, and either quotient is at most $y<M$. Counting possible pairs $(y,c)$, without requiring unique representations, yields \[ S(qM)\le S(M)(C+1),\qquad d(qM)\le d(M)\frac{C+1}{q}. \tag{2} \]For any fixed $M$, there are arbitrarily large $k,m$ with $r/q\to1$. Here is the precise elementary input. The number $\theta=\log4/\log3$ is irrational: a rational equality would imply $4^u=3^v$ for positive integers $u,v$. Dirichlet approximation supplies unbounded $m$ and integers $k$ with $|m\theta-k|\to0$; the denominators must be unbounded since no fixed nonzero $m$ gives equality. Exponentiating gives $4^m/3^k\to1$, with both powers tending to infinity. Along these pairs, $1$ gives \[ \frac{C+1}{q}\longrightarrow\frac12+\frac13=\frac56. \] If $\delta=\liminf_{X\to\infty}d(X)$, $2$, with $M$ fixed, therefore implies \[ \delta\le\frac56d(M)\quad\hbox{for every positive integer }M. \] Choose a sequence $M\to\infty$ on which $d(M)\to\delta$. Then $\delta\le5\delta/6$, so $\delta=0$.
There is also an explicit iterative formulation. Starting with $M_0=1$, at stage $j$ select aligned powers with $q_j\ge100$ and $(r_j-q_j)/q_j\le1/(100M_j)$. Formula $1$ gives \[ (C_j+1)/q_j\le (5/6)(1+1/(100M_j))+1/100+1/100<9/10. \] Set $M_{j+1}=q_jM_j$. Then $M_j\to\infty$ and $d(M_j)\le(9/10)^j$. Thus the proof produces a subsequence of explicitly searchable scales at which the density tends to zero.$\square$