discussion: It studies the same dissociated or distinct-subset-sum condition as #1, but replaces the extremal maximum element by a quadratic energy and asks for a finite-defect inverse theorem. There is an exact elementary lower bound behind the normalization. If Z is the sum of a uniformly random subset of A, then Var(Z)=(1/4)sum a_i^2. Its support consists of 2^n distinct integers. Among any M distinct integers, consecutive integers minimize variance, with minimum (M^2-1)/12. Taking M=2^n gives sum a_i^2 >= (4^n-1)/3. Equality forces the subset sums to be {0,1,...,2^n-1}, and then induction forces A={1,2,4,...,2^{n-1}}. The new conjecture asks whether a bounded excess above this minimum can only alter boundedly many low binary digits. For example, {2,3,4,8,16,...} has a fixed energy excess while agreeing with the binary sequence from a bounded index onward.

domain: Additive combinatorics

inspiration: Erdős Problem #1: if an n-element set A contained in {1,...,N} has all subset sums distinct, must N be bounded below by a positive constant times 2^n?

license: MIT

problem:
Bounded-defect rigidity for distinct subset sums. Let A={a_1<...<a_n} be positive integers such that all 2^n subset sums are distinct, and define

D_2(A) = sum_{i=1}^n a_i^2 - (4^n-1)/3.

For every fixed integer D>=0, does there exist r=r(D) such that, whenever D_2(A)<=D, one has

a_i=2^{i-1} for every i>r?

More precisely, for each D are there only finitely many possible defective initial segments (a_1,...,a_r), after which the sequence is forced to continue 2^r,2^{r+1},...,2^{n-1}?


discussion:
It retains the same sunflower configuration but imposes an adversarially colored, equitable-petal condition. The question is not about the exponential growth rate alone, but whether the extra requirement costs asymptotically nothing even at the sharp threshold. There are immediate bounds. Monochromatically coloring the ground set shows f_bal(n,k,r)>=f(n,k). On the other hand, an n-set has one of C(n+r-1,r-1) possible color-count profiles. If a family has more than C(n+r-1,r-1)(f(n,k)-1) members, one profile class has at least f(n,k) sets and hence contains an ordinary sunflower. Since those sets have equal total profiles and share the same core, their petals automatically have equal profiles. Thus

f_bal(n,k,r) <= C(n+r-1,r-1)(f(n,k)-1)+1.

The known elementary comparison loses a polynomial factor; the new problem asks whether that factor can be removed completely.

domain: Extremal set theory

inspiration: Erdős Problem #20, the sunflower conjecture: for each fixed k, is the threshold f(n,k) for a k-sunflower in an n-uniform family at most c_k^n?

problem:
Profile-balanced sunflower threshold. Fix integers k>=3 and r>=2. Color the ground set of an n-uniform family with r colors. Call a k-sunflower S_1,...,S_k with core C profile-balanced if the r-vectors

( |(S_i minus C) intersect color j| )_{j=1}^r

are the same for all i. Let f_bal(n,k,r) be the least M such that every r-colored ground set and every n-uniform family of M sets contains a profile-balanced k-sunflower. If f(n,k) is the ordinary sunflower threshold, is

f_bal(n,k,r) = (1+o(1)) f(n,k)

for every fixed k and r?

title: Profile-balanced sunflower threshold


discussion:
Similarity: a gap between consecutive squarefree numbers is exactly an interval in which every integer has a square prime divisor. Erdős #208 studies the length of such an interval at a given location; the new problem studies the minimum number of distinct prime-square obstructions needed to manufacture an interval of a given length.

New feature: κ(h) measures certificate complexity rather than the location or maximum length of a gap. It is an optimization over both the interval and the collection of prime squares.

Basic first-order argument: fix a cutoff y. Impose M≡0 mod p² for every prime p≤y. The positions i≤h divisible by one of these p² are then covered. For each remaining position i, choose a fresh prime q_i>h and impose M≡−i mod q_i². The Chinese remainder theorem gives one M satisfying all conditions. This costs

π(y)+R_y(h),

where R_y(h) counts integers i≤h divisible by no p² with p≤y. Taking y→∞ slowly gives

κ(h)≤(6/π²+o(1))h.

Conversely, for any fixed y, the residue classes supplied by primes p≤y cover at most

(1−∏_{p≤y}(1−1/p²))h+O_y(1)

positions. Primes larger than y contribute at most h/p²+1 positions each. Letting y→∞ yields

κ(h)≥(6/π²−o(1))h.

Hence the first-order term is already forced: κ(h)=(6/π²+o(1))h. The proposed second term comes from the heuristic optimization

π(y)+h∏_{p≤y}(1−1/p²),

using ∑_{p>y}p⁻²∼1/(y log y). The optimum occurs near y≈√((6/π²)h) and predicts the constant 4√6/π≈3.11879. Controlling finite-interval sieve errors sharply enough to confirm or refute that constant is the new problem.

domain: Analytic and combinatorial number theory

inspiration: Erdős Problem #208, which asks for sharp upper bounds on gaps between consecutive squarefree numbers, including the conjectural scale (π²/6)·log x/log log x.

problem:
Second-order certificate complexity of a squarefree-free interval. Let κ(h) be the smallest cardinality of a set Q of primes for which there exists an integer M such that every one of

M+1,M+2,…,M+h

is divisible by q² for at least one q∈Q. Thus Q is a square-divisor certificate that the whole interval contains no squarefree integer.

Determine the second-order asymptotic of κ(h). Is

κ(h)
=
(6/π²)h
+
(4√6/π+o(1))·√h/log h?

At minimum, is

κ(h)−(6/π²)h = Θ(√h/log h)?


discussion: It uses exactly the pinned-distance statistic from #604, but replaces one exceptional good pin by a positive proportion of good pins. Thus it is a distributional or robust version rather than a change of exponent. The m by m integer grid, with n=m^2, gives the expected upper scale for every pin: every squared distance is a sum of two squares of size O(m^2), and the classical count of integers representable as two squares is O(m^2/sqrt(log m))=O(n/sqrt(log n)). For a possible lower-bound route, suppose more than half of the pins have at most D distance classes. At each such pin, Cauchy-Schwarz forces on the order of n^2/D equal-distance pairs, hence many isosceles triangles. Summing over the bad pins converts the problem into a global perpendicular-bisector incidence estimate. Problem #604 only needs that estimate to produce one good pin; the new conjecture requires enough control to rule out a large population of bad pins and also suggests a stability theorem for near-extremizers.

domain: Discrete geometry

inspiration: Erdős Problem #604: must every n-point set in the plane contain at least one point from which there are n^{1-o(1)} distinct distances, perhaps as many as a constant multiple of n/sqrt(log n)?

problem:
Median pinned-distance conjecture. Let A be a set of n distinct points in the Euclidean plane, and for x in A write

d_A(x) = |{||x-y|| : y in A, y != x}|.

Define the upper median pinned-distance count by

q(A) = max{D : at least ceil(n/2) points x in A satisfy d_A(x) >= D},

and define q(n) = min_{|A|=n} q(A). Is

q(n) = Theta(n/sqrt(log n))?

A stronger version asks whether q(n) has an asymptotic constant, and whether the extremal configurations are, after deleting o(n) points and applying a Euclidean similarity (translation, rotation, reflection, and uniform scaling), essentially two-dimensional lattice patches.

title: Median pinned-distance conjecture


discussion:
Similarity: both problems compare the size of a Littlewood polynomial on the unit circle with its Parseval or root-mean-square scale √(n+1). Erdős #1150 asks for one point with a fixed excess above that scale. The new problem asks how much of the circle must remain near the RMS scale, even when the polynomial is allowed to concentrate its energy.

New feature: this is a distributional concentration problem rather than a supremum problem. A single very high spike may settle a maximum question while occupying negligible measure; μ_η(n) distinguishes narrow spikes from genuinely spread-out magnitude.

Basic universal lower bound: Parseval gives

(1/2π)∫|P(e^{iθ})|²dθ=n+1,

while |P(e^{iθ})|≤n+1. Put a=1−η and μ=μ_η(P). Bounding |P|² by a²(n+1) off the superlevel set and by (n+1)² on it yields

1 ≤ a²(1−μ)+(n+1)μ,

so

μ_η(n) ≥ (1−a²)/(n+1−a²) = Θ_η(1/n).

Basic upper construction: for the all-plus polynomial P(z)=1+z+⋯+z^n, the Dirichlet-kernel formula gives |P(e^{iθ})|≤1/|sin(θ/2)| away from θ=0. Hence its RMS superlevel set has measure O_η(n^{-1/2}), and μ_η(n)≤O_η(n^{-1/2}).

The new problem is to close the exponent gap between n^{-1} and n^{-1/2}. An n^{-1/2} answer would say the Dirichlet-kernel concentration pattern is essentially extremal under the ±1 coefficient constraint; an n^{-1} answer would require much sharper spike constructions. Targeted searches found extensive work on L^q norms, flatness, and subarc behavior of Littlewood polynomials, but no exact minimization of this RMS-superlevel measure.

domain: Harmonic analysis and polynomial inequalities

inspiration: Erdős Problem #1150, which asks whether there is an absolute c>0 such that every sufficiently high-degree polynomial with coefficients in {−1,1} has max_{|z|=1}|P(z)|>(1+c)√n. Parseval gives only the baseline √(n+1).

problem:
RMS-superlevel concentration for Littlewood polynomials. Fix 0<η<1. For a Littlewood polynomial

P(z)=∑_{j=0}^n ε_j z^j,  ε_j∈{−1,1},

define

μ_η(P)= (1/2π) · meas{θ∈[0,2π] : |P(e^{iθ})| ≥ (1−η)√(n+1)}

and

μ_η(n)=min_P μ_η(P),

where the minimum is over all degree-n Littlewood polynomials.

Determine the order of μ_η(n). Is μ_η(n)=n^{-1/2+o(1)} for every fixed η, or can Littlewood polynomials concentrate their L² mass so efficiently that μ_η(n)=n^{-1+o(1)}? Does the exponent depend on η?