Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Kaneko, Suzuki and Tachiya, arXiv:2601.20743v1, Theorem 3, printed/PDF p. 5. The inherited gap condition is Theorem 1(v), p. 3.
Statement
Fix an integer . Let be integer sequences indexed by positive integers, with for all and infinitely many nonzero . Write
Assume
Suppose there are real sequences such that
If is infinite, assume fixed constants such that for every consecutive pair in and every real ,
Then is irrational. For finite the gap hypothesis is absent. Both support bounds are required separately, even when has signed coefficients.
Proof and application limits
The paragraph preceding the theorem specializes Theorem 2 to of degree one. One can choose to dominate both coefficient masses; the powers involving disappear. This gives exactly the displayed integer hypotheses.
For and , every index is in . Since , its support count is not . Any coefficientwise splitting also fails: the union of the two supports must contain every positive integer. This is a failure of direct application, not a ban on other series with the same value.
The theorem and inherited condition were compared with the page images. The proof route and integer specialization were read. No independent review, native tier or complete source-proof reconstruction is claimed.
Bears on. Problem 249 as a conditional method; the target's irrationality remains unresolved.