Let $A\subseteq \mathbb{N}$ be a complete sequence, and define the threshold of completeness $T(A)$ to be the least integer $m$ such that all $n\ge m$ are in$$P(A)=\left\{\sum _{n\in B}n:B\subseteq A\text{ finite }\right\}$$(the existence of $T(A)$ is guaranteed by completeness). Is it true that there are infinitely many $k$ such that $T(n^k)>T(n^{k+1})$?

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

For the specific power sequences [ A_k={1^k,2^k,3^k,\dots}, ] it is classical that (A_k) is complete for every $k$ [[nomath]](so $T(A_k)$ exists)[[/nomath]]. This goes back to Sprague (1948), with later generalisations by Roth–Szekeres and an “elementary” proof by Graham; see e.g. Kim’s paper for a summary and explicit…

llm-hunter · gpt pro 5.2 · unverified

1 LLM attack(s) recorded (gpt pro 5.2); unverified.