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Source. J. Geneson, Deletion thresholds and exponential examples for complete sequences, arXiv:2609.25107v1 (20 September 2026): Section 6 "Two sequences with a common base", Corollary 12 with its proof, physical pp. 12--13. Read in the canonical conversion beside the held PDF; the artifact is identified on the library source card, Geneson (2026), and the corollary on its result page.

Standing. This is an author-recorded reconstruction. It is not an independent review, changes no status and assigns no tier. It rests on Theorem 9 and Proposition 11 and through them on the two inputs imported from Dubickas.

Definitions

Salem numbers, {x}\{x\}, φ\varphi and the completeness convention are as on the Theorem 9 page. The interleaving of two sequences (xn)n≥0(x_n)_{n\ge0}, (yn)n≥0(y_n)_{n\ge0} is x0,y0,x1,y1,…x_0,y_0,x_1,y_1,\ldots, with repeated values kept as separate occurrences; it is complete if every sufficiently large integer is a sum of occurrences with distinct positions. A sequence (yn)(y_n) is a tail of (xn)(x_n) if yn=xn+ky_n=x_{n+k} for all n≥0n\ge0 and some k≥0k\ge0.

Statement

Corollary 12. There are 1<γ<φ1<\gamma<\varphi and real α,β>0\alpha,\beta>0 such that every term of both (⌊αγn⌋)n≥0(\lfloor\alpha\gamma^n\rfloor)_{n\ge0} and (⌊βγn⌋)n≥0(\lfloor\beta\gamma^n\rfloor)_{n\ge0} is even, and

βα≠rγk(r∈Q, k∈Z).\frac\beta\alpha\ne r\gamma^k\qquad(r\in\mathbb Q,\ k\in\mathbb Z).

In particular α/β\alpha/\beta is irrational, neither sequence is a tail of the other, and their interleaving is not complete.

Proof

Let γ\gamma be the Salem number of Theorem 9, with minimal polynomial P(x)=x18−x12−x11−x10−x9−x8−x7−x6+1P(x)=x^{18}-x^{12}-x^{11}-x^{10}-x^9-x^8-x^7-x^6+1 and P(1)=−5P(1)=-5. Proposition 11 with q=5q=5 gives η>0\eta>0 with

320<{ηγj}<14(j≥1).\frac3{20}<\{\eta\gamma^j\}<\frac14\qquad(j\ge1).

Set α=2ηγ\alpha=2\eta\gamma and β=α(1+γ)=2ηγ+2ηγ2\beta=\alpha(1+\gamma)=2\eta\gamma+2\eta\gamma^2.

All floors even. For n≥0n\ge0, αγn=2ηγn+1\alpha\gamma^n=2\eta\gamma^{n+1} with n+1≥1n+1\ge1, and {ηγn+1}<1/4<1/2\{\eta\gamma^{n+1}\}<1/4<1/2, so ⌊αγn⌋=2⌊ηγn+1⌋\lfloor\alpha\gamma^n\rfloor=2\lfloor\eta\gamma^{n+1}\rfloor is even. Also βγn=2(ηγn+1+ηγn+2)\beta\gamma^n=2(\eta\gamma^{n+1}+\eta\gamma^{n+2}), and writing each summand as its integer part plus its fractional part, ηγn+1+ηγn+2=I+σ\eta\gamma^{n+1}+\eta\gamma^{n+2}=I+\sigma with I∈ZI\in\mathbb Z and

310<σ={ηγn+1}+{ηγn+2}<12,\frac3{10}<\sigma=\{\eta\gamma^{n+1}\}+\{\eta\gamma^{n+2}\}<\frac12 ,

so ⌊βγn⌋=2I+⌊2σ⌋=2I\lfloor\beta\gamma^n\rfloor=2I+\lfloor2\sigma\rfloor=2I is even.

The ratio condition. The polynomial PP is reciprocal: its coefficient list is symmetric, so x18P(1/x)=P(x)x^{18}P(1/x)=P(x), and γ−1\gamma^{-1} is a root of PP. Since PP is the minimal polynomial of γ\gamma, there is a field isomorphism Q(γ)→Q(γ−1)\mathbb Q(\gamma)\to\mathbb Q(\gamma^{-1}) sending γ\gamma to γ−1\gamma^{-1}; the two fields coincide, each generator being the inverse of the other, so it is an automorphism τ\tau of Q(γ)\mathbb Q(\gamma) with τ(γ)=γ−1\tau(\gamma)=\gamma^{-1}. Suppose β/α=1+γ=rγk\beta/\alpha=1+\gamma=r\gamma^k with r∈Qr\in\mathbb Q, k∈Zk\in\mathbb Z. Then r≠0r\ne0, and applying τ\tau gives 1+γ−1=rγ−k1+\gamma^{-1}=r\gamma^{-k}. Dividing the first identity by the second,

γ=1+γ1+γ−1=rγkrγ−k=γ2k,\gamma=\frac{1+\gamma}{1+\gamma^{-1}} =\frac{r\gamma^k}{r\gamma^{-k}}=\gamma^{2k},

and γ>1\gamma>1 forces 2k=12k=1, impossible for an integer kk. So β/α≠rγk\beta/\alpha\ne r\gamma^k for all r∈Qr\in\mathbb Q, k∈Zk\in\mathbb Z.

Consequences. With k=0k=0, β/α∉Q\beta/\alpha\notin\mathbb Q, so α/β\alpha/\beta is irrational. If (⌊βγn⌋)(\lfloor\beta\gamma^n\rfloor) were a tail of (⌊αγn⌋)(\lfloor\alpha\gamma^n\rfloor), then for some k≥0k\ge0, ∣βγn−αγn+k∣<1|\beta\gamma^n-\alpha\gamma^{n+k}|<1 for every n≥0n\ge0 (two reals with equal floors differ by less than 11), that is ∣β−αγk∣γn<1|\beta-\alpha\gamma^k|\gamma^n<1 for all nn, which as n→∞n\to\infty forces β=αγk\beta=\alpha\gamma^k, excluded above with r=1r=1. Symmetrically a tail relation the other way would force β/α=γ−k\beta/\alpha=\gamma^{-k}, excluded with r=1r=1 and exponent −k-k. Finally every occurrence in the interleaving is even, so every finite sum of occurrences is even, no odd integer is represented, and the interleaving is not complete; the same holds for the set union of the two value sets, since discarding repeated occurrences cannot create representations.

Scope. The corollary answers the second question of Problem 354 negatively under the reading "for every γ∈(1,2)\gamma\in(1,2)" (the source, p. 2, states it answers the "variable-base extension" of the two-sequence question; p. 13 adds that the corollary does not resolve the original question with base 22); it says nothing about base 22, and nothing about the reading "for some γ∈(1,2)\gamma\in(1,2)". The coefficients are not explicit. The Salem base lies in (6/5,13/10)(6/5,13/10); whether other bases in (1,2)(1,2), in particular bases at which the single sequence ⌊tγn⌋\lfloor t\gamma^n\rfloor is always complete, admit such pairs is not addressed.