GPT 6 Astra Ultra attempt on Math-5680

Erdős Problem #482: nonlinear recurrences that extract digits

Complete reconstruction of two known recurrence families, 24 September 2026.

Statement and digit convention

The Graham--Pollak recurrence $u_1=1$, $u_{n+1}=\lfloor\sqrt2(u_n+1/2)\rfloor$ yields the binary digits of $\sqrt2$ through $u_{2n+1}-2u_{2n-1}$. The question asks for analogous results for square roots and other algebraic numbers. We give a binary construction for every positive real number, recovering that recurrence, and then an arbitrary-base construction. For $w>0$ and integer base $g\ge2$, let $m=\lfloor\log_g w\rfloor$ and $t=w/g^m\in[1,g)$. Its digits, starting with the leading nonzero digit, are \[d_n=\lfloor tg^{n-1}\rfloor-g\lfloor tg^{n-2}\rfloor\quad(n\ge1).\tag{1}\] The radix point is restored using $m$. The floor convention chooses terminating expansions when available. The first digit of $\sqrt2$ here is its leading $1$, not the first digit after the point.

A binary recurrence recovering Graham--Pollak

Normalize $t=w/2^{\lfloor\log_2w\rfloor}$ and set \[a=\frac{2(t+1)}{t+2},\qquad b=\frac{t+2}{t+1}=\frac2a. \] Define \[u_1=1,\qquad u_{2k}=\lfloor a(u_{2k-1}+1/2)\rfloor, \qquad u_{2k+1}=\lfloor b(u_{2k}+1/2)\rfloor.\tag{2}\] We prove the exact invariants \[u_{2k}=2^{k-1}+\lfloor t2^{k-1}\rfloor,\qquad u_{2k+1}=2^k+\lfloor t2^{k-1}\rfloor.\tag{3}\]

The odd formula holds for $k=0$, since $\lfloor t/2\rfloor=0$. For the first induction step put $F=\lfloor t2^{k-2}\rfloor$, $r=\{t2^{k-2}\}$, $\delta=\lfloor2r\rfloor$, and $G=2F+\delta=\lfloor t2^{k-1}\rfloor$. Subtracting the desired integer from the unrounded expression gives \[a(2^{k-1}+F+1/2)-(2^{k-1}+G) =\frac{2r+t+1-(t+2)\delta}{t+2}. \] If $\delta=0$, the numerator lies in $[t+1,t+2)$; if $\delta=1$, it lies in $[0,1)$. The error is therefore in $[0,1)$ in both cases, proving the even formula. For the reverse step put $s=\{t2^{k-1}\}$. The corresponding error is \[b(2^{k-1}+G+1/2)-(2^k+G)=\frac{-s+(t+2)/2}{t+1}\in[0,1). \] Indeed its numerator is positive and at most $(t+2)/2<t+1$. This proves the odd formula and completes the induction. Subtracting twice the preceding odd formula in $3$ proves $u_{2n+1}-2u_{2n-1}=d_n$. For $t=\sqrt2$, both $a,b$ equal $\sqrt2$, recovering the original recurrence. For any positive algebraic $w$, both coefficients are algebraic as well.

Every integral base

For arbitrary $g\ge2$ use its normalized $t\in[1,g)$ and put \[a=\frac{g}{(g-1)(t+g)},\qquad b=\frac ga=(g-1)(t+g). \] Define the nonnegative integer sequence \[v_1=1,\qquad v_{2k}=\lfloor a(v_{2k-1}-1/g)\rfloor, \qquad v_{2k+1}=\lfloor b(v_{2k}+1/(g-1))\rfloor.\tag{4}\] Here $v_2=0$ is allowed. The exact formulas are \[v_{2k}=\frac{g^{k-1}-1}{g-1},\qquad v_{2k+1}=g^k+\lfloor tg^{k-1}\rfloor.\tag{5}\] The first even value is zero because $a(1-1/g)=1/(t+g)\in(0,1)$. Given the even formula, the unrounded next expression is exactly $(t+g)g^{k-1}$, proving the odd formula. Conversely, if the odd formula holds, write $s=\{g^{k-1}(t+g)\}$. The unrounded expression for $v_{2k+2}$ equals \[\frac{g^k-1}{g-1}+E,\qquad E=\frac1{g-1}-as-\frac ag. \] Since $0<a\le g/(g^2-1)$ and $0\le s<1$, \[E>\frac1{g-1}-a(1+1/g)\ge0, \qquad E\le\frac1{g-1}-\frac ag<1. \] This proves $5$. Equation $1$ now yields $v_{2n+1}-gv_{2n-1}=d_n$ in every base.

Outcome and attribution

SOLVED in the requested constructive sense. These families cover every positive square root and algebraic number, and in fact every positive real. They are known Rabinowitz--Gilbert/Stoll constructions, not new digit-generation results. The exact invariants and all floor inequalities used here are proved above. Sources: https://www.erdosproblems.com/482; T. Stoll, On families of nonlinear recurrences related to digits, Theorems 1.1 and 1.3, https://cs.uwaterloo.ca/journals/JIS/VOL8/Stoll/stoll56.pdf. The all-base instance uses the permitted parameter $\epsilon=-1/g$. No claim about numerical efficiency or formal verification is made.

COMPLETION ESTIMATE

COMPLETION: 100%