(a) If the orbit first reaches \(r_{A+1}=0\), the selected exponents form a

finite, strictly increasing set and

Thus they are an exact witness to the original equation. The first candidate

\(n+1\) is selected, after which the residual state is \(n-1>0\); hence every

terminating orbit for \(n\ge2\) selects at least two terms. No floating point

or tolerance enters (1)--(3).

(a) A nonzero orbit can terminate at exponent \(a\) only when \(a\) is even

and \(r_a=a/2\). Indeed, \(0

\(2r_a=a\).

(a) On putting \(e_a=2r_a\), (1) becomes

which is precisely the Borwein--Loring recurrence (BLC), here on the special

initial data \(e_{n+1}=2n\). Consequently, the exact missing sufficient lemma

for the greedy route is:

> For every \(n\ge2\), the orbit \(r_{n+1}=n\),

> \(r_{a+1}=(2r_a)\bmod a\) reaches zero.

Proving this diagonal case would settle the all-\(n\) question affirmatively.

The full Borwein--Loring conjecture is stronger because it permits arbitrary

integer initial data. Conversely, failure of one greedy orbit would not

disprove the original problem, since a non-greedy finite representation might

still exist.

3. New exact finite computation

(d) The recurrence (1) terminates for every

For \(n=1\), where the greedy one-term equality is inadmissible because

\(t\ge2\), the separately checked witness is

For every \(2\le n\le100{,}000\), the terminating greedy orbit supplies the

required distinct exponents. This extends the primary-source range

\(n\le10{,}000\) by a factor of ten.

(d) Aggregate certificate for \(2\le n\le100{,}000\):

| quantity | exact value |

|---|---:|

| total recurrence updates \(\sum(A(n)-n)\) | 17,929,666,741 |

| total selected terms \(\sum k(n)\) | 8,946,326,760 |

| FNV-1a digest of all ordered triples \((n,k(n),A(n))\) | 9f0a3e00dd8f0008 |

| largest term count | 285,549 at \(n=88,225\) |

| largest terminal exponent | 659,488 at \(n=88,225\) |

Here \(A(n)\) is the final selected exponent. The digest serializes each of

\(n,k(n),A(n)\) as eight little-endian bytes, in increasing order of \(n\).

(d) The simultaneous record holders are:

| \(n\) | number of terms \(k(n)\) | final exponent \(A(n)\) |

|---:|---:|---:|

| 2 | 3 | 8 |

| 5 | 5 | 14 |

| 7 | 7 | 24 |

| 8 | 13 | 32 |

| 10 | 189 | 392 |

| 56 | 6,092 | 12,230 |

| 3,113 | 13,370 | 29,752 |

| 3,817 | 76,072 | 155,942 |

| 5,325 | 227,040 | 460,536 |

| 88,225 | 285,549 | 659,488 |

Correction to TUZ20 Table 2

(a) The published last row

\((n,k(n),a_k(n))=(5588,460536,226913)\) cannot be correct as printed:

460,536 distinct exponents all greater than 5,588 cannot fit below 226,913,

since there are only \(226913-5588=221325\) available integers.

(d) Exact recurrence and cleared-denominator recomputation give instead

Thus the last two entries were swapped. There is a second consequence:

\(n=5588\) is not the term-count record claimed by that table, because the

earlier \(n=5325\) has 227,040 terms and the same final exponent 460,536.

(d) The standalone checker reconstructs every listed record witness and

the \(n=5588\) witness in Python, then independently checks (3) as an exact

big-integer equality. The two largest highlighted checks have cleared integers

of 455,224 bits (\(n=5325\)) and 571,280 bits (\(n=88225\)), respectively.

4. Reproduction

The entire mathematical kernel is:

def greedy_terms(n):
    r, a, terms = n, n + 1, []
    while r:
        r *= 2
        if r >= a:
            r -= a
            terms.append(a)
        assert 0 <= r < a
        a += 1
    return terms

def exact_check(n, terms):
    A = terms[-1]
    assert n << (A - n) == sum(a << (A - a) for a in terms)

The standalone checker is

erdos261_wave8p_verify.py.

The transparent serial discovery implementation is

erdos261_sweep.c.

Run:

python runs/erdos261_wave8p_verify.py

Observed output on this VM:

PASS: exact greedy witnesses for every 1 <= n <= 100000
total recurrence steps (n>=2): 17929666741
total selected terms (n>=2):   8946326760
FNV-1a certificate:             9f0a3e00dd8f0008
record holder:                  (88225, 285549, 659488)
TUZ20 corrected row:            (5588, 226913, 460536)
elapsed wall time:               17.528 s

(a) The C worker embedded in the checker uses unsigned 64-bit integers, but

the invariant \(0\le r_a

impossible in this run. Each job has a five-billion-update cap; no orbit came

remotely close to it. The checker then uses a separately written Python

implementation and arbitrary-precision integers for the displayed exact

identities.

5. What remains, precisely

(a) No finite computation, including \(10^5\), supplies the quantifier

“for every \(n\).” The exact outstanding lemma for this approach is uniform

termination of the diagonal orbit in Section 2. Current elementary facts

\(0\le r_a

force such a hit. The record jumps from \(A(3817)=155942\) to

\(A(5325)=460536\) and then \(A(88225)=659488\), illustrating why a bound

in terms of the checked range was not inferred.

(c, cost estimate only) A naive compiled sweep to \(10^6\), if the present

average cost merely scaled, would require roughly \(1.8\times10^{11}\) state

updates, about 0.2 core-hour at the measured \(2.1\)--\(2.6\times10^8\)

updates/second (well under \$0.05 at a typical \$0.05--\$0.10/core-hour

commodity rate). Heavy-tail record orbits could make it substantially larger.

Such a sweep would still not address the uniformity lemma, so I did not spend

that computation.

(b) For the continuum question, Borwein--Loring already obtain independent

finite replacements from (BL), hence continuum many representations of

certain explicitly defined numbers \(B_M\); they prove those \(B_M\) are

irrational. TUZ20 obtains infinitely many rational numbers with at least nine

representations, still finite multiplicity.

(a) The precise constructive gap is therefore not “find another finite

identity.” One needs infinitely many independent choices whose common total is

rational (equivalently, a perfect branching family inside one rational fiber).

The known disjoint (BL) trades provide the independence but have an irrational

lacunary total; the common-tail trick in the live comment provides a rational

total but only finitely many independent choices. Neither ingredient currently

supplies both properties at once.

PARTIAL: exact greedy constructions now verify every 1<=n<=100000, correct TUZ20's n=5588 table row, and reduce the uniform problem to a specific modular orbit; the all-n and rational-continuum questions remain open.