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Erdős 699: effective bounds at indices 15 and 20


This is a retained earlier research note. The current reading guide gives the latest status and distinguishes completed proofs from remaining cases.

P699 research · Erdős problem 699 · Nonvanishing from surjective interpolation

Complete exclusions. Indices 15,2015,20 are fully excluded by the cutoff-15 proof; the bounds and constructions below are retained as earlier stages and inputs.

There is an effectively computable last bad row at each of i=15,20i=15,20. These are bounds for the entire index, including every possible assignment and collision of the small-prime maximum positions. No binomial rows or columns are enumerated.

Let XiX_i be the maximum of the effective symmetric BFT thresholds, with gap bound 100100, for the prime-pair weights used below. Then every bad triple satisfies

ieffective last-row bound15n<max⁡{X15+15,  101100000000},20n<max⁡{X20+20,  104000000}.\begin{array}{c|c} i&\text{effective last-row bound}\\\hline 15&n<\max\{X_{15}+15,\;10^{1100000000}\},\\ 20&n<\max\{X_{20}+20,\;10^{4000000}\}. \end{array}

The powers of ten are deliberately crude. The additional symmetric thresholds comprising XiX_i are effectively computable but not all numerically instantiated here. The two asymmetric thresholds are numerical and are already below the displayed algebraic bounds.

Cofactor inequalities

For each prime p<ip<i, choose a position rp∈{0,…,i−1}r_p\in\{0,\ldots,i-1\} maximizing vp(n−rp)v_p(n-r_p), and write

n−rp=Ppup,Pp=pvp(n−rp).n-r_p=P_pu_p,\qquad P_p=p^{v_p(n-r_p)}.

Use the symmetric weights in BFT Corollary 2.3 together with the twelve additional pair estimates. Include only pairs whose primes are less than ii. Add these two asymmetric inequalities, written for normalized cofactor exponents epe_p:

e5≥.3613ore7≥.2277,e7≥.0987ore13≥.2080.e_5\ge .3613\quad\text{or}\quad e_7\ge .2277, \qquad e_7\ge .0987\quad\text{or}\quad e_{13}\ge .2080.

Here ep=log⁡up/log⁡(n−i+1)e_p=\log u_p/\log(n-i+1). The exact asymmetric threshold evaluator gives:

PairTargetsSeed and c/dc/dL1,m0L_1,m_0Threshold for the smaller input
5,75,7(.3613,.2277)(.3613,.2277)2⋅25−49=12\cdot25-49=1, 8/58/51.5407,531.5407,5310240578210^{2405782}
7,137,13(.0987,.2080)(.0987,.2080)343−2⋅169=5343-2\cdot169=5, 7/57/51.4135,741.4135,741048436710^{484367}

Both use gap bound 100100. Their limiting exponents exceed, respectively, (.3613046269,.2277777234)(.3613046269,.2277777234) and (.0987539406,.2080310930)(.0987539406,.2080310930). The coefficient-gcd bounds L1,m0L_1,m_0 are the retained BFT Proposition 5.1 entries; no new asymptotic gcd assumption is made.

For each symmetric or asymmetric disjunction, choose an endpoint witnessing it. Keeping the largest chosen weight at each prime, and discarding componentwise dominated vectors, gives a finite list of rational vectors vv such that some vv satisfies vp≤epv_p\le e_p. The list has 39 vectors at i=15i=15 and 112 at i=20i=20.

When n≥2(i−1)n\ge2(i-1), each singleton maximum satisfies

Pp≤2vpn1−vp.P_p\le2^{v_p}n^{1-v_p}.

At a position occupied by at least two primes, their selected prime powers have product at most n−r≤nn-r\le n. Such a position therefore has exponent capacity 11. A singleton has capacity 1−vp1-v_p.

For a fixed singleton subset, the other primes occupy at most ⌊#other primes/2⌋\lfloor\#\text{other primes}/2\rfloor positions. Assign capacity 11 to that many positions and sort all capacities decreasingly. These vectors cover all collisions. For nonincreasing row multiplicities mtm_t, rearrangement puts the largest capacities at the earliest positions, giving the uniform upper exponent

∑amaca.\sum_a m_a c_a.

All entries vpv_p used here are less than 1/21/2, as checked exactly. The capacity construction and its proof are also retained in the weighted-ramp calculation.

Integer interpolation data

For i=15i=15, use

h=(69,63,57,51,43,34,24,11,0,0,0,0,0,0,0),m=(139,139,138,137,127,127,127,119,106,95,87,79,73,68,63),D=257.\begin{aligned} h={}&(69,63,57,51,43,34,24,11,0,0,0,0,0,0,0),\\ m={}&(139,139,138,137,127,127,127,119,106,95,87,79,73,68,63),\\ D={}&257. \end{aligned}

For i=20i=20, use

h=(34,32,30,27,25,22,19,16,13,9,5,1,0,0,0,0,0,0,0,0),m=(67,65,65,65,62,60,58,57,57,54,50,46,42,39,37,34,32,30,28,26),D=96.\begin{aligned} h={}&(34,32,30,27,25,22,19,16,13,9,5,1,0,0,0,0,0,0,0,0),\\ m={}&(67,65,65,65,62,60,58,57,57,54,50,46,42,39,37,34,32,30,28,26),\\ D={}&96. \end{aligned}

Both mm vectors are nonincreasing. At each grid point x,y≥0x,y\ge0, x+y<ix+y<i, prescribe residual multiplicity

Rxy=max⁡{0,mx+y−hx−hy}.R_{xy}=\max\{0,m_{x+y}-h_x-h_y\}.

The exact symmetric jet calculations give:

iiSymmetric equationsSymmetric monomials through degree DDModulusRank
15167681677025116768
2023992401655212399

Both moduli are prime. The matrix entries are the integer Hasse Taylor coefficients of the symmetric monomials. At off-diagonal support points, one point from each exchanged pair suffices; at diagonal points the Taylor coefficients with exchanged derivative indices coincide. These are exact identities, not numerical reductions.

Full row rank modulo the displayed prime implies surjectivity of the rational symmetric interpolation map. The multiplication-by-X+YX+Y lemma therefore supplies a finite family of integer polynomials of degree at most D+1D+1 with the prescribed residual multiplicities, having no common zero when X+Y>i−1X+Y>i-1. This uses the symmetric local jet algebra. In particular, independence of all nonsymmetric jet conditions is neither claimed nor needed.

Multiply these polynomials by

H(X,Y)=∏a=0i−1((X−a)(Y−a))ha.H(X,Y)=\prod_{a=0}^{i-1}((X-a)(Y-a))^{h_a}.

The resulting family has row multiplicities at least mtm_t, degree at most 2∑ha+D+12\sum h_a+D+1, and no common zero in the physical range i<j≤ci<j\le c. Exact capacity enumeration gives:

iiCapacity configurations checkedTotal degreeMinimum exponent gain
15184096243/500043/5000
2018479563713/5000713/5000

Here the gain is

∑tmt−∑amaca−(2∑aha+D+1).\sum_t m_t-\sum_a m_a c_a-\left(2\sum_a h_a+D+1\right).

Thus some nonzero polynomial evaluation gives a positive power bound uniformly for every bad pair. There is no remaining curve or finite base-locus exception.

Explicit algebraic constants

The following generous bounds make the effectivity assertion concrete without writing out a rational interpolation inverse. Let rr be the symmetric rank, N=(D+22)N=\binom{D+2}{2}, and put

C=r!(2iD)r,U=1+(i(i+1)2+i+4)N.C=r!(2i^D)^r,\qquad U=1+\left(\frac{i(i+1)}2+i+4\right)N.

A full-rank minor and its cofactors give integer interpolation sections qjq_j, with a common denominator-clearing factor Δ≠0\Delta\ne0, such that ∥qj∥1≤NC\|q_j\|_1\le NC and ∣Δ∣≤C|\Delta|\le C. Every symmetric jet matrix entry has magnitude at most 2iD2i^D.

In the standard local Taylor basis, multiplication by X+YX+Y has column coefficient norm at most t+3≤i+2t+3\le i+2 on row tt. At a diagonal point, the extra 11 allows the coincident derivative terms. The nonvanishing generators are

(X+Y)qj−∑kTkjqk,Δ−∑jujqj,(X+Y)q_j-\sum_kT_{kj}q_k, \qquad \Delta-\sum_j u_jq_j,

where TT represents that multiplication and uu is the jet of the constant 11. At most i(i+1)/2i(i+1)/2 entries of uu are nonzero, and they equal 11. Consequently every generator has coefficient norm at most UCUC.

Let ss be the number of primes less than ii, and let gg be the gain in the preceding table. The off-maximum small-prime contribution is at most ((i−1)!)m0((i-1)!)^{m_0}. For n≥2∑tmtn\ge2\sum t m_t, the product of the factors (1−t/n)mt(1-t/n)^{m_t} is at least 1/21/2. Combining the weighted Kummer divisibility with a nonzero generator gives

ng≤21+sm0((i−1)!)m0(∏a=0i−1(a+1)2ha)UC.n^g\le 2^{1+s m_0}((i-1)!)^{m_0} \left(\prod_{a=0}^{i-1}(a+1)^{2h_a}\right)UC.

Both indices are composite, so no rough denominator factor from i!i! needs to be added. These estimates are applied only after the source thresholds have been reached.

Use i<102i<10^2, r<105r<10^5, and U<107U<10^7. The right-hand side is strictly less than 10E10^E, where

E=r(6+2D)+1+sm0+2(i−1)m0+4∑ha+7.E=r(6+2D)+1+s m_0+2(i-1)m_0+4\sum h_a+7.

The two exact values are E=8725502E=8725502 and E=479024E=479024. Finally,

1100000000⋅435000>8725502,4000000⋅7135000>479024.1100000000\cdot\frac{43}{5000}>8725502, \qquad 4000000\cdot\frac{713}{5000}>479024.

The powers of ten also exceed the elementary row thresholds and both numerical asymmetric thresholds. This proves the bounds stated at the start.

Reproduction

The checker and recorded certificate retain the integer data, all source parameters used for the asymmetric thresholds, finite-vector counts, worst capacities, exact gains, and ranks. The floating-point optimization that suggested the integer data is not used to certify any assertion.

sh
.venv/bin/python wiki/research/erdos_699/computations/erdos_699_interpolation_extensions.py

Dependencies are pinned in the interpolation requirements. Install them in the chosen Python environment before running the checker.

The exact rank check for i=15i=15 takes about three minutes on the machine used here. The --arithmetic-only option checks the finite cofactor and capacity calculations without repeating the rank calculation.