Erdős Problem #345

declared status 'open'. Formalized: no. 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 $n\geq m$ are in$$P(A) = \left\{\sum_{n\in B}n : B\subseteq A\textrm{ finite }\right\}$$(the existence of $T(A)$ is guaranteed by completeness). Is it true that there are infinitely many $k$ such that $T(n^k)>T(n^{k+1})$? Current best: It is known that$$T(n)=1, T(n^2)=128, T(n^3)=12758,$$$$T(n^4)=5134240,\textrm{ and }T(n^5)=67898771.$$Erd\H{o}s and Graham remark that a good candidate for the $n$ in the question are $k=2^t$ for large $t$, perhaps even $t=3$, because of the highly restricted values of $n^{2^t}$ modulo $2^{t+1}$. Prize: no. OEIS: A001661. Tags: complete sequences, number theory.

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