Plectis

The drawing uses two scales because the roots are about one unit from 0 while the passage is only about a trillionth of a unit away. Hatching always means allowed points, and red always marks the two arms containing A and B.

1. Start with seven petals. The polynomial $z^7-1$ has seven roots evenly spaced on the unit circle. Its allowed region has seven petals whose boundaries touch at 0; 0 itself is forbidden. ani changes its coefficients slightly and moves every root to distance $\rho =1-{10}^{-96}$ from 0, strictly inside the circle. These changes are invisible in the left view but alter how the petals meet.
2. Find the only pair that can be joined. The paper verifies that five connected pieces contain one root each and the sixth contains A and B. Paths cannot jump between pieces, so only this pair needs a length estimate.
3. Locate every possible crossing. On the two-root piece, each ordinary polynomial value has two preimages. Cutting the disc of values along a suitable segment separates these into a side containing A and a side containing B. Any joining path must cross the preimage of the cut. A local quadratic estimate places that whole passage within $\rho \epsilon /3750$ of P, where $\epsilon ={10}^{-12}$. Enclosing the whole passage rules out an alternative crossing farther away.
4. Pay for reaching the passage. The paper’s root enclosures give $|A-P|+|B-P|>\rho (2+0.143\epsilon )$. Moving the crossing away from P can reduce each distance by at most the passage’s enclosing radius. Therefore every allowed joining path has length greater than $\rho (2+(0.143-2/3750)\epsilon )>2$. The excess is tiny but strictly positive.

The dashed route through 0 would cost $|A|+|B|=2\rho <2$, but part of it crosses blank ground. The close-up shows why moving from 0 to the compulsory passage matters. Its contours approximate the region at drawing precision; exact rational inequalities in the paper establish the bound. The result panel below replaces the tiny passage by a larger enclosing disc so that this same length obstruction is visible.

For any positive integer $N$, divide it by an exponent $n$: write $N=un+r$ with $0\le r<n$. Then

$$\frac{2^N-1}{2^n-1}=\sum _{j=1}^u{2}^{N-jn}+\frac{2^r-1}{2^n-1}.$$

The first sum is an integer and the final fraction lies between 0 and 1. It vanishes exactly when $n$ divides $N$. Adding these fractions gives

$$\mathrm{\Delta }(N)=\sum _{n\in A}\frac{2^{Nmodn}-1}{2^n-1}=(2^N-1)X-J(N),J(N)\in \mathbb{Z}.$$

Here $Nmodn$ denotes the remainder on division by $n$, and $J(N)$ is the sum of the integer parts removed. Only finitely many are nonzero, since they vanish for $n>N$. The total $\mathrm{\Delta }(N)$ is a sum of fractional parts; it can exceed 1.

If $X=p/q$ with integers $p$ and $q>0$, then $q\mathrm{\Delta }(N)$ is an integer. It is positive because an infinite $A$ contains an exponent larger than $N$, whose remainder is positive. Hence $\mathrm{\Delta }(N)\ge 1/q$. This explains why arbitrarily small positive totals contradict rationality.

For the pictured set, choosing $N=6^m$ cancels every row $2^i{3}^j$ with $i,j\le m$. The first four totals are $0.586\dots ,0.0595\dots ,0.00779\dots ,0.0000152\dots$. These values illustrate the mechanism; the reciprocal-summability theorem justifies its use for the infinite set. At the displayed cut $N=12$, the total is $0.1509\dots$.

This particular test has limits. For all positive exponents, the terms above $N$ alone give $\mathrm{\Delta }(N)>1-2^{-N}\ge 1/2$; for prime exponents the short paper proves $\mathrm{\Delta }(N)>1/3$. Both sums are known to be irrational by other arguments. The weighted theorem explained in the result panel makes small totals for further classes of exponents, but arbitrary infinite choices remain open.

Assume the total is $p/q$ with positive integers $p,q$. Write the remaining sum and its clearing denominator as

$$x_n=\sum _{k\ge n}1/a_k,D_n=q\prod _{j<n}{a}_j,C_n=D_n{x}_n.$$

The empty product at $n=1$ is 1. Subtracting the finite prefix from $p/q$ shows that $C_n$ is a positive integer: $D_n$ is divisible by every denominator in that prefix. Removing the current reciprocal gives the exact updates

$$C_{n+1}=a_n{C}_n-D_n,D_{n+1}=a_n{D}_n.$$

These are Koizumi's integer tail coordinates. Their change is

$$C_{n+1}-C_n=D_n((a_n-1)x_n-1).$$

Thus $C_n$ is unchanged precisely when the tail is $1/(a_n-1)$. If $C_n=C_{n+1}=C_{n+2}=C$, the first update gives $D_n=(a_n-1)C$; applying it once more gives $a_{n+1}-1=a_n(a_n-1)$. Eventual constancy is therefore equivalent to the eventual recurrence in the question.

Exact squaring shows how these coordinates can prove irrationality. For $2,4,16,256,\dots$, each tail satisfies

$$x_n=1/a_n+1/a_n^2+1/a_n^4+\cdots <1/a_n+1/a_n^2+1/a_n^3+\cdots =1/(a_n-1).$$

If the sum were rational, its positive integer $C_n$ would strictly decrease at every step, which is impossible.

The short paper proves the eventual rule when upward steps of these integers are bounded by a fixed constant. It also proves irrationality under the separate condition

$$a_n^2/a_{n+1}=1+3/n+o(n^{-3}).$$

Here the error after subtracting $1+3/n$, multiplied by $n^3$, must tend to zero. This forces eventual cubic tail numerators; the result diagram shows the final contradiction. A possible counterexample to the full problem must evade both conclusions. Its tail numerators would have unbounded upward steps, with a divergent sum of their relative increases.

For $t=a/b$, the divisor-count form is $F(t)=\sum _{n\ge 1}\tau (n)b^n/a^n$. Multiplying by $a^N$ makes its first N terms integers. Remove those terms and call what remains $E_N$:

$$a^{N}F(a/b)=\underset{\text{integer}}{\underbrace{\sum _{n=1}^N\tau (n)b^n{a}^{N-n}}}+E_N,E_N=b^N\sum _{m\ge 1}\tau (N+m)(b/a{)}^m.$$

The graph follows $E_N$ as $N$ increases from 0 to 20. It is a multiplied tail, not the original sum’s tail. If $F(t)=r/s$, then $s{E}_N$ is a positive integer. Making $E_N$ arbitrarily small would therefore contradict rationality.

At base 2, the blue points stay between 1 and 5 in the displayed range. For example, $E_1=2F(2)-1=2.21339\dots$. These points alone provide no contradiction. Erdős’s integer-base proof instead uses divisibility of consecutive divisor counts to force special patterns in the base expansion; it needs more than this simple tail.

At base $3/2$, the clearing multiplier is $3^N$. The factor $2^N$ survives in the tail formula. Since every divisor count is at least 1,

$$E_N\ge 2^N\sum _{m\ge 1}(2/3{)}^m=2^{N+1}.$$

This is the red dashed lower bound. At $N=10$ it is 2048, while $E_{10}=7165.1\dots$; by $N=20$ the red points exceed 8.5 million. Each equal vertical step in the graph is a factor of 10. Thus clearing this straightforward expansion makes the leftover larger, rather than giving the small positive expression an irrationality proof would need.

Zudilin’s 2004 construction uses carefully chosen polynomial combinations of the same total. Their small positive remainders compete with the cost of making the coefficients integers. Cancelling common polynomial factors first lowers that cost. The result panel follows this operation at 31/4, where the remainders do tend to zero.

For coprime $a>b\ge 1$, the paper proves irrationality when $\log b/\log a<{\theta }^{\ast }=0.4056830213840605\dots$. The simpler strict cutoff $81/200=0.405$ is rigorously below ${\theta }^{\ast }$; $4^{200}<{31}^{81}$ places 31/4 inside it. A common positive power preserves the logarithmic ratio. The forms and constants are Zudilin’s; the paper supplies the rational-base degree calculation. Zudilin previously noted rational-base extensions and regards this consequence as hard to classify as a new result. The base 3/2 and the universal question remain open.