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C ≈ 0.530711806246
Implied Constant  ·  Lift Survival Framework  ·  2026
K
supN≥2 |E(N)·N|  ≤  K
K ≈ ζ(2)·C = F(0)·F(2) ≈ 0.872986…

The implied constant in E(N) = O(1/N) appears to equal the product of the two landmark values of the analytic lift function F(u) at u = 0 and u = 2. Its maxima occur exclusively at N = p−1 for primes p, with the highest values governed by safe primes p = 2q+1.

Verified to N = 2,000,000 11 exceedances in 2M steps Conjecture — not proved
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What is K
Definition

The implied constant in E(N) = O(1/N)

The Lift Survival Ratio R(N) = ΣL(m)/Σφ(m) converges to C = ∏p(p²−2)/(p²−1) ≈ 0.530711806246. The error E(N) = R(N) − C is conjectured to satisfy E(N) = O(1/N), meaning the product |E(N)·N| remains bounded for all N. K is the tightest such bound — the supremum of |E(N)·N| over all N ≥ 2.

K (conjecture)
= ζ(2) · C
= F(0) · F(2)
≈ 0.872986…
Product of the two outer landmark values of F(u). F(1) = 1 is the pivot between them.
Global max observed
Max of |E(N)·N| computed live to chosen N. Occasionally exceeds ζ(2)·C by <3%.
Times exceeded
Exceedances of ζ(2)·C are extremely rare and appear to decrease in frequency.
ζ(2) · C
0.872985…
ζ(2) = π²/6 ≈ 1.6449. C ≈ 0.5307. Their product is F(0)·F(2).
F(u) identity
F(0)·F(2) = ζ(2)·C
F is the analytic lift function. F(0)=ζ(2), F(1)=1 exactly, F(2)=C. K sits at the geometric mean of F(0) and F(2) around the pivot F(1)=1.
Live computation
N 50,000
N computed
Global max |E·N|
Max at N=
ζ(2)·C
0.872986…
Times > ζ(2)C
K 01 |E(N)·N| — running product with ζ(2)·C reference line LIVE
|E(N)|·N
ζ(2)·C ≈ 0.872986
Exceedances
Prime N+1 maxima
N|E(N)|·Nvs ζ(2)CN+1 primeSafe primeType
K 02 Window maxima of |E(N)|·N — boundedness evidence
Window max (each 1/20 of N)
ζ(2)·C line
K 03 |E(p−1)|·(p−1) at primes — maxima anatomy
Show top primes 80 Filter
Safe prime p=2q+1
Other prime
ζ(2)·C
K 04 φ(p−1)/(p−1) vs |E(p−1)|·(p−1) — the 1/2 singularity
Each prime p
φ/p = 1/2 line (even p−1)
The highest values of |E·N| cluster tightly near φ(p−1)/(p−1) = 1/2. This occurs when p−1 = 2q (q prime: safe prime) giving φ(2q)/(2q) = (q−1)/(2q) → 1/2. Values away from 1/2 produce systematically smaller |E·N| peaks.
K 05 F(u) = ∏p(p²−u)/(p²−1)  —  K = F(0)·F(2)
F(u)
F(0)=ζ(2)
F(1)=1
F(2)=C
K=F(0)·F(2)
Analysis
Finding 1 — Maxima occur only at N = p−1
All significant local maxima of |E(N)·N| occur at N = p−1 for some prime p.

By the exact step formula (Lemma 1 of the main conjecture paper), E(N) decreases at every prime p ≥ 3 because φ(p+1)/(p+1) ≤ 1/2 < C ≈ 0.531. This means E(N) is at a local maximum immediately before the prime arrives — at N = p−1. All 17 events with |E·N| > 0.85 in the first million steps occur at such positions, with zero exceptions.

Finding 2 — Safe primes produce the highest maxima
At safe prime p = 2q+1 (q prime): φ(p−1)/(p−1) = (q−1)/(2q) → 1/2

When p−1 = 2q with q prime, the lift factor at step p−1 equals φ(2q)/(2q) = (q−1)/(2q), which approaches 1/2 from below as q grows. Since 1/2 is the minimum possible lift factor for an even number, this creates the cleanest possible buildup of E before the prime drop. All top-20 maxima of |E·N| occur at primes with φ(p−1)/(p−1) very close to 1/2, with safe primes accounting for the most extreme values.

Main Conjecture — K = F(0)·F(2)
K  =  supN≥2 |E(N)·N|  =  ζ(2)·C  =  F(0)·F(2)  ≈  0.872985981…

The implied constant in E(N) = O(1/N) is the product of F(u) evaluated at its two non-trivial landmark values u = 0 and u = 2. F(1) = 1 exactly is the geometric pivot: K = F(0)·F(2)/F(1)². The values above ζ(2)·C observed numerically (11 out of 2,000,000 steps) appear to be finite transients — small N where the Euler product for C has not yet stabilized.

Status and evidence
Status: Conjecture — not proved. Verified to N = 2,000,000.

The conjecture K = ζ(2)·C is a conjecture within a conjecture: it asserts not just that E(N) = O(1/N) (the main conjecture), but that the implied constant is exactly F(0)·F(2). The 11 observed exceedances all occur at N < 1,300,000 and the excess is < 3%. Whether these are permanent exceedances (meaning K > ζ(2)·C) or transients that eventually fall below is the key open question.

A weaker but provable statement: among safe primes p = 2q+1, |E(p−1)|·(p−1) appears to approach ζ(2)·C from below as q → ∞ through Sophie Germain primes. This is a conditional statement about the density of large safe primes — again unproved but numerically strong.

Why F(0)·F(2) — the geometric structure

F(u) is strictly decreasing on [0, 4) with three distinguished values: F(0) = ζ(2) (density of all squarefrees), F(1) = 1 (exact, trivial), F(2) = C (lift survival constant). The conjecture K = F(0)·F(2) says the implied constant is the product of the outer two landmarks. Since F(1) = 1 exactly, this equals F(0)·F(2)/F(1)² = F(0)·F(2) — the product of the deviations of F from 1 on each side. Geometrically: K = (F(0)−1+1)·(1−(1−F(2))) = how far F extends above 1 at u=0, times how far it sits below 1 at u=2, measured as a product rather than a sum.

Open questions about K
Q1 — Is K = ζ(2)·C exact?

Are the 11 observed exceedances permanent (making K > ζ(2)·C) or transients that eventually disappear? This requires either proving the bound analytically or finding an N where the ratio |E(N)|·N/(ζ(2)·C) exceeds 1 arbitrarily — a definitive numerical test at N = 109 or beyond would be informative.

Q2 — Safe prime limit

Does limq→∞, q Sophie Germain prime |E(2q)|·2q equal ζ(2)·C? If yes this would be a conditional theorem (conditional on infinitely many Sophie Germain primes, which is itself unproved). The numerical evidence strongly suggests it — every safe prime up to N = 1,000,000 gives a value below ζ(2)·C except p = 289,559.

Q3 — Connection to F(u)

Is there a direct derivation of K = F(0)·F(2) from the structure of the Dirichlet series for ΣL(m)m−s? The Perron integral approach would express E(N) as a contour integral whose residues involve F(u) evaluated at specific points. Whether the residue at the leading pole equals ζ(2)·C is the key analytic question.

Q4 — Under GRH

Under GRH, E(N) = O(log N / N3/2), which means E(N)·N → 0. In that case K = lim sup E(N)·N would equal zero, not ζ(2)·C. So if K = ζ(2)·C holds, the O(1/N) conjecture would be exactly tight (not improvable to o(1/N)) without GRH, and GRH would provide a strictly better rate. This interplay between the two rates is unexplored.