What this attempt establishes
The paid-shortfall construction has an explicit, fully charged family: truncate a finite balanced seed lift at square-root support. Its repair bill is at most order sqrt(N) log N, with constants stated below. The factorial remainder is at most order log(N)^2 for a fixed seed.
This does not make the whole prime-counting error square-root sized. The leading term of the old seed survives. For the two seeds checked here it is C N with C > 1, so the excess over N is still linear.
A separate exact example at N=14, using the same paid-bound definition, shows why adding arithmetic information is actionable: a perfect-power cap removes an artificial-mass obstruction that the raw logarithmic cap cannot remove, even after optimizing all coefficients with support at most three. This is a finite example, not a uniform estimate.
Status: explicit mathematical derivations, independently checked algebra, and small deterministic checks recorded in RUNS.md. No Lean proof or external review has been completed. Novelty has not been assessed.
1. Definitions and the complete paid cost
Let N >= 2 be an integer. Put
\[ Q_N=\{\lfloor N/d\rfloor:2\le d\le N\},\qquad W_c(q)=\sum_j c_j\lfloor q/j\rfloor, \qquad B_N(c)=\sum_j c_j\log(\lfloor N/j\rfloor!). \]
Here the coefficients have finite support on positive integer indices. Let Lambda be the von Mangoldt function and define the true and raw available masses of a quotient cell by
\[ m_q=\sum_{\lfloor N/d\rfloor=q}\Lambda(d),\qquad w_q=\sum_{\lfloor N/d\rfloor=q}\log d. \]
For any cap U_q >= m_q, the paid cost is
\[ C_N^U(c)=B_N(c)+\sum_{q\in Q_N}U_q(1-W_c(q))_+. \]
The divisor identity log n = sum_{d|n} Lambda(d) gives B_N(c)=sum_q m_q W_c(q). Consequently the exact excess is
\[ C_N^U(c)-\psi(N)= \sum_q m_q(W_c(q)-1)_+ +\sum_q(U_q-m_q)(1-W_c(q))_+\ \ge 0. \tag{1} \]
The first term charges surplus coverage. The second charges slack in the available information precisely where coverage is missing. Neither is dropped in this study. Write C_N=C_N^w for the raw cap.
2. Early-truncation lemma
Fix an integer M >= 2 and finitely supported real coefficients a_j, with support in 1 <= j <= J, satisfying
\[ \sum_j\frac{a_j}{j}=0,\qquad g(t)=\sum_j a_j\lfloor t/j\rfloor,\qquad W_\infty(t)=\sum_{k\ge0}g(t/M^k)\ge1\quad(t\ge1). \]
Assume also g(t) <= H for t >= 0, with H >= 0. The lift is finite at every fixed t, since g(t)=0 for 0 <= t < 1. Define
\[ A=\sum_j|a_j|,\quad \kappa=-\sum_j\frac{a_j\log j}{j},\quad C=\frac{\kappa}{1-1/M},\quad L(x)=\log(\lfloor x\rfloor!). \]
Take any integer K >= 0, set R=M^(K+1), and use the explicit coefficients
\[ c_n=\sum_{\substack{jM^k=n\\0\le k\le K}}a_j, \qquad W_K(t)=\sum_{k=0}^K g(t/M^k). \]
Their support is at most J M^K. With
\[ E_N=A\sum_{k=0}^K\bigl(1+\log^+(N/M^k)\bigr), \]
the full bound is
\[ \psi(N)\le C_N(c) \le C(1-1/R)N+E_N+ H\sum_{\ell\ge0}L\!\left(\frac{N}{RM^\ell}\right). \tag{2} \]
Moreover,
\[ |B_N(c)-C(1-1/R)N|\le E_N, \tag{3} \]
and, writing x=N/R, the repair term is at most
\[ \frac{H}{1-1/M}\,x\log^+x. \tag{4} \]
These statements also cover N < R, when the repair bill is zero.
Proof, including the bill
First, the omitted lift controls the deficit:
\[ (1-W_K(q))+ \le\left(\sum{k>K}g(q/M^k)\right)+ \le H\sum{\ell\ge0}{\bf1}_{q\ge RM^\ell}. \]
No assumption that g is nonnegative is needed for this step. Since R M^ell is an integer, floor(N/d) >= R M^ell if and only if d <= floor(N/(R M^ell)). Thus
\[ \sum_q w_q(1-W_K(q))+ \le H\sum{\ell\ge0}\sum_{2\le d\le N/(RM^\ell)}\log d =H\sum_{\ell\ge0}L(N/(RM^\ell)). \tag{5} \]
All sums are finite. The elementary estimate L(v) <= v log^+ v bounds the last sum by the geometric series in (4).
For completeness, integral comparison for the increasing function log t gives, for every real v>0,
\[ |L(v)-(v\log v-v)|\le1+\log^+v. \]
One way to see the bound is to put n=floor v: the error at n lies between 1 and 1+log n; extending n log n-n from n to v<n+1 subtracts a number in [0,log v]. The case 0<v<1 follows directly. Apply this estimate to v=N/(jM^k) and use j>=1. Balance cancels both the N log N and the N terms inside each scale, leaving kappa N/M^k. Summing the scales proves (3). Adding (5) and using (1) proves (2).
Square-root support, and the surviving obstruction
For N >= J^2, take the largest K >= 0 with J M^K <= sqrt N. Then R > sqrt N/J, while
\[ E_N=A\left((K+1)(1+\log N) -\frac{K(K+1)}2\log M\right). \]
The repair bill is at most
\[ \frac{HJ}{1-1/M}\sqrt N\log(J\sqrt N). \tag{6} \]
For a fixed seed, (3), nonnegativity of the repair bill, and (6) give
\[ C_N(c)-N=(C-1)N+O_{a,M}(\sqrt N\log N). \tag{7} \]
This controls truncation, not the leading excess. Changing the seed with N requires controlling A, H, J, M, and C together. The fixed-seed big-O must not be reused with growing seed complexity hidden in its constant.
3. An explicit base-6 construction
Use
\[ g(t)=\lfloor t\rfloor-\lfloor t/2\rfloor -\lfloor t/3\rfloor-\lfloor t/6\rfloor. \]
It is balanced and periodic of period six. Its values on integer residues 0,1,2,3,4,5 are respectively 0,1,1,1,1,2. Therefore it is nonnegative, bounded by two, and at least one on 1 <= t < 6. Dividing any t>=1 by the appropriate power of six proves full-lift coverage.
For r=K+1 and R=6^r, the truncated lift telescopes to coefficients
\[ c_1=1,\quad c_R=-1,\quad c_{2\cdot6^k}=c_{3\cdot6^k}=-1\quad(0\le k<r). \]
At an integer q, each summand reads one base-6 digit. All weights are nonnegative and every nonzero digit has weight at least one. Hence
\[ (1-W_K(q))+={\bf1}{R\mid q}. \tag{8} \]
The exact repair bill is sum_{q in Q_N, R|q} w_q, bounded by L(N/R). The combined coefficient mass is 2r+2, giving the sharper explicit bounds
\[ C_6(1-1/R)N-(2r+2)(1+\log N) \le C_N(c) \le C_6(1-1/R)N+(2r+2)(1+\log N)+L(N/R), \]
where
\[ C_6=(4\log2+3\log3)/5>1. \]
This is an especially transparent example, not a leading-constant record. The existing pilot's seed a={1:1,2:-1,3:-1,5:-1,30:1}, with M=6, has A=5, J=30, H=1, and the smaller constant
\[ C_{30}=(14\log2+9\log3+5\log5)/25>1. \]
The generic early-truncation lemma applies to it too. It does not apply unchanged to an infinite nested adaptive lift masquerading as a finite seed.
4. A finite obstruction removed by arithmetic capacity
For d>=2, let r(d) be the largest integer r>=1 for which d=b^r with integer b>=2, and put
\[ u(d)=\log d/r(d),\qquad U_q=\sum_{\lfloor N/d\rfloor=q}u(d). \]
If d=p^k is a prime power, r(d)=k, so u(d)=Lambda(d). Otherwise Lambda(d)=0 <= u(d) <= log d. Thus m_q <= U_q <= w_q. Computing r(d) requires only exact integer-power tests, not a primality oracle.
At N=14, restrict coefficient support to j<=3. The quotient cells in ascending order are (1,2,3,4,7). Choose
\[ (c_1,c_2,c_3)=(1,-1,-3/2),\qquad (W_c(q))_q=(1,1,1/2,1/2,1). \]
Only the cells containing d=4 and d=3 need repair. Direct substitution in (1) gives
\[ C_{14}^{w}(c)=\psi(14)+\tfrac12\log2, \qquad C_{14}^{U}(c)=\psi(14). \tag{9} \]
The raw-cap excess is unavoidable for every choice of coefficients with this support, not just this candidate. To prove it, set
\[ \epsilon=\tfrac12\log2,\quad v=(-1,1,2,0,-1),\quad x=m+\epsilon v. \]
The three moments sum_q v_q floor(q/j) are zero for j=1,2,3, and sum_q v_q=1. The vector x is feasible for the raw caps: withdrawals in cells one and seven are smaller than the actual masses; cell two has slack log6 > epsilon; cell three uses its entire slack log2; cell four is unchanged. In particular 0<=x_q<=w_q.
For any real W and 0<=x<=w, x(W-1)+w(1-W)_+ >= 0. Summing and using the zero moments proves
\[ C_{14}^{w}(c')\ge\sum_qx_q=\psi(14)+\tfrac12\log2 \quad\text{for every }c'\text{ supported on }j\le3. \]
Together with (9), this proves both finite optima exactly. No assertion is made about the raw optimum with unrestricted support.
More generally, replacing the raw cap by the perfect-power cap saves exactly
\[ C_N^w(c)-C_N^U(c)=\sum_{d=2}^N \log d\left(1-\frac1{r(d)}\right) (1-W_c(\lfloor N/d\rfloor))_+. \]
This information excludes some artificial mass. It does not eliminate the slack from ordinary composite numbers, and no growth estimate follows from this example.
5. Provenance, checks, and the doors
The paid functional and capped dual were preserved from the September 7 research conversation. The balanced-seed factorial main term and remainder are inherited from the existing pilot. The delta here is the explicitly charged early truncation, its support-cost bound, and the exact raw-cap versus perfect-power-cap comparison. This study claims neither first discovery nor a new best prime-counting bound.
construction.py and tests/test_paid_shortfall.py check rational floor identities, exact prime-log coefficient identities, small cap inequalities, both seed families, and interval/logarithm refinements. The run manifest records the finite domain, output hashes, software, and non-claims. Finite checks can falsify an implementation or formula here; the all-N assertion rests on the displayed derivation, not on the size of that finite domain.
The result closes one practical gap: truncating a finite lift need not leave an unpriced deficit. The mathematical door still open is a uniform construction whose entire paid cost is N+O(sqrt(N) log(N)^2), including the leading term and any growing family constants. Improving a cap can help only where the candidate has a deficit, as (1) makes explicit. Improving coverage alone while retaining a fixed leading excess cannot deliver that target. Neither this upper-bound construction nor the finite example claims to establish RH, a two-sided error estimate, or a new exponent.