The bonded-molecule density of one-dimensional random sequential adsorption is decreasing in k, and an exact reduction for the gap densities

The exact reduction of Pinsky's question: (l+1)(A_(k,l) - A_(k,l+1)) = 2 A_(k,l+k) for every k >= 1 and every l, hence (l+1) g_(k;l+1) - l g_(k;l) = g_(k;l) - 2 g_(k;l+k); so 'l g_(k;l) is increasing on 0,...,k-1' is exactly 'g_(k;l) > 2 g_(k;l+k) for l = 0,...,k-2'

The first two steps of Pinsky's conjecture for EVERY k >= 2: 0 * g_(k;0) < 1 * g_(k;1), and 1 * g_(k;1) < 2 * g_(k;2)

An exact-rational certificate scheme for one cell (an integer polynomial p with p(0)=0, p(1)>0 whose total derivative along the kernel is dominated coefficientwise by Q*psiₗ), and its 153 cells: the step l g_(k;l) < (l+1) g_(k;l+1) at every (k,l) with 4 <= k <= 20 and 2 <= l <= k-2

Pinsky's monotonicity conjecture in full at every k with 2 <= k <= 20: for all l < k-1, l * g_(k;l) < (l+1) * g_(k;l+1), i.e. l -> l g_(k;l) is strictly increasing on 0,1,...,k-1

Too-large control: the conjecture does NOT survive dropping Pinsky's range l in 0,...,k-1 – the step fails, (l+1) g_(k;l+1) < l g_(k;l), at (k,l) = (2,3), (3,4), (4,5), (5,6) and (6,7)

The source's own closed form re-proved from the definition: g_(k;k-1) = exp(-2 sumⱼ₌₁ᵏ⁻¹ 1/j)

Every gap density is positive, and l -> g_(k;l) is strictly decreasing – with no restriction on l

Two controls: int₀¹ (1-s) s² (3s-2) ds = -1/60 < 0, so no argument using only 'the weight is positive and increasing' can prove the step at l >= 2; and Wₖ is strictly increasing on [0, infinity) for k >= 2, so the hypothesis of the l = 1 argument is not vacuous

CORRECTION to arXiv:2608.04730v1 Table 1: the entry g_(10;5) is printed as 0.0059; the correct value is 0.005588..., i.e. 0.0056

CORRECTION to arXiv:2608.04730v1 Table 2: the entry m₅ is printed as 0.793; the correct value is 0.7922759..., i.e. 0.792

PROSE: for each fixed k the whole conjecture reduces to its single hardest case l = k-2, because (A_(k,l))_(l>=0) is a Hausdorff moment sequence, hence log-convex in l, hence rₗ = 2 A_(k,l+k) / A_(k,l) is non-decreasing in l

MEASUREMENT: the whole family costs about 1.5 CPU-hours and 7.81 GB peak RSS; the mₖ conjecture (mₖ decreasing in k, conjectured in Pinsky's 2014 book) is designed but PARKED at a measured price

The general two-sided certificate: for ANY integer coefficient list c, p(1) <= Q * int₀¹ c(s) Wₖ(s) ds whenever Q*c - p' - qqₖ*p is coefficientwise non-negative and p(0)=0 – so Q > 0 certifies an exact rational LOWER bound on the integral and Q < 0 an UPPER bound

The reduction of Pinsky's mₖ conjecture to one certificate against the UNCHANGED kernel: Wₖ₊₁(s) = Wₖ(s) exp(2(sᵏ-1)/k), hence mₖ - mₖ₊₁ = int₀¹ (k - (k+1) exp(2(sᵏ-1)/k)) Wₖ(s) ds; and the integer polynomial chiₖ(s) = 4k⁵ - (k+1)(2k² - 2k(1-sᵏ) + (1-sᵏ)²)² with chiₖ <= 4k⁴ (k - (k+1) exp(2(sᵏ-1)/k)) on [0,1], from exp x <= (1 + x/2 + x²/8)² for x <= 0

Pinsky's SECOND open conjecture – mₖ is decreasing in k – proved for every k with 2 <= k <= 20: mₖ₊₁ < mₖ

The chained form: k -> mₖ is strictly decreasing on the whole window 2,3,...,21, i.e. mⱼ < mᵢ for all 2 <= i < j <= 21

Certified exact-rational two-sided brackets on two of the source's transcendental integrals: 0.00279408 <= A_(10,5) = int₀¹ (1-s)s⁵ W₁0 <= 0.00279453, and 0.15845506 <= int₀¹ W₅ <= 0.15845530

CORRECTION to arXiv:2608.04730v1 Table 1, now KERNEL-BOUND: 0.00558816 <= g_(10;5) <= 0.00558906, so the printed 0.0059 is more than half a unit in the fourth decimal from the value and the correct entry is 0.0056

CORRECTION to arXiv:2608.04730v1 Table 2, now KERNEL-BOUND: 0.7922753 <= m₅ <= 0.7922765, so the printed 0.793 is wrong and the value rounds to 0.792; with the compute-first tie m₂ = 1 - e⁻² re-proved from (mk) itself

Controls for the mₖ route: the reverse inequality fails at the first index (NOT m₂ <= m₃); the certificate target changes sign (chi₂(1) = -64 < 0), so no pointwise argument can work; and the UNSQUARED quadratic bound exp x <= 1+x+x²/2 – a valid pointwise lower bound, proved so – yields a k=2 target whose integral against W₂ is at most -0.026323 < 0, so that route provably cannot prove m₃ < m₂

The CORRECTED first-failing-index table, cell by cell: 72 new certificates giving, for every k with 2 <= k <= 20, that the step l g_(k;l) < (l+1) g_(k;l+1) holds at every l < L(k) and FAILS at l = L(k), where L(k) = k+1 for 2 <= k <= 7, k+2 for 8 <= k <= 13, and k+3 for 14 <= k <= 20

The corrected table as three block theorems quantified over k (L(k) = k+1, k+2, k+3 on 2<=k<=7, 8<=k<=13, 14<=k<=20), and: Pinsky's range is NOT sharp – the step still holds at l = k-1 AND at l = k for every k with 2 <= k <= 20

MEASUREMENT: this pull costs about 0.25 CPU-h and 8.25 GB peak RSS (1.66 GB above the import Mathlib baseline); the extensions k >= 21 are priced and parked

Identity (I) for Pinsky's RSA kernel, kernel-bound: int₀¹ sᵏ Wₖ = (1/2) int₀¹ Wₖ for every k >= 1 – equivalently E[sᵏ] = 1/2 EXACTLY under the probability measure Wₖ ds / int Wₖ

The exact moment relation for Pinsky's RSA kernel: (m+1) Mₘ - m Mₘ₊₁ = 2 Mₘ₊ₖ for every k >= 1 and every m >= 0, with Mⱼ = int₀¹ sʲ Wₖ; and identity (II), 4 int₀¹ s²ᵏ Wₖ = int₀¹ Wₖ + k int₀¹ (1-s)² Wₖ, from its m = 1 and m = k instances

The uniform reduction of Pinsky's mₖ conjecture to ONE numeric inequality: 2k² (mₖ - mₖ₊₁) >= (k-1) Jₖ - k(k+1) Nₖ with no slack lost after exp x <= 1+x+x²/2, plus Nₖ <= Wₖ(0), Jₖ >= (1-e^(-2(k-1)))/(2(k-1)) and k² Wₖ(0) decreasing; hence ANY K >= 2 with 2K(K+1) W_K(0) < 1 - e^(-2(K-1)) proves mₖ₊₁ < mₖ for every k >= K – and this holds at K = 12 (312 W₁2(0) < 1-e⁻²², from e⁶ > 403) and at K = 5 (60 W₅(0) < 1-e⁻⁸, from e⁴ > 54 and e^(1/6) >= 7/6)

Pinsky's SECOND open conjecture, CLOSED OUTRIGHT: mₖ₊₁ < mₖ for every k >= 2, and indeed for every k >= 1. No window. Only the three cells k = 2, 3, 4 still rest on a certificate; every k >= 5 is an argument uniform in k

The chained form on all of k >= 1: mⱼ < mᵢ whenever 1 <= i < j – k -> mₖ is strictly decreasing on the whole of 1,2,3,..., which is the form the conjecture is stated in

Controls for the uniform route: the reduction is NEGATIVE at k = 2 in exact closed form, J₂ - 6 N₂ = 7 e⁻² - 1 = -0.0526530173..., with J₂ = (1-e⁻²)/2 and N₂ = 1/4 - (5/4) e⁻² both proved here; NOT (m₅ <= m₆) and NOT (m₁2 <= m₁3); mⱼ!= mᵢ for i < j; Mₘ > 0 and Mₖ < M₀, so identity (I) is not the trivial Mₖ = M₀

An explicit RATE, which the conjecture does not assert: mₖ - mₖ₊₁ >= 1/(100 k²) for every k >= 5, and mₖ - mₖ₊₁ >= 1/(20 k²) for every k >= 12

MEASUREMENT: this pull costs about 0.4 CPU-h and 7.15 GB peak RSS (0.56 GB above the family's import Mathlib baseline); nothing was parked for cost, and every remaining open item is mathematics, not compute