Erdoes #261 - working report - Erdős Problem a Day

Erdős problem 261 — wave 8p report

Date of live-page access and computation: 2026-07-28 (UTC).

Claim labels used throughout:

(a) elementary-rigorous: proved below from definitions and exact integer algebra.
(b) rigorous-modulo-named-theorem: accurately attributed to the named primary source, without independently reproving the whole theorem.
(c) plausible/structural-unverified: an indicated direction, not a theorem.
(d) computational-only: established by the exact finite computation supplied here, not uniformly in $n$.

0. Mandatory live-page gate

(a, direct live-page transcription) I fetched the live problem page, its LaTeX view, and its discussion thread through the Bright Data browser path. The page was last edited 2025-12-01 and showed:

status OPEN;
0 claimed proofs;
“Currently working on this problem: None”;
“Interested in collaborating: None”;
three comments, all read and recorded below.

There was therefore no collision/skip condition.

(a, direct live-page transcription) The source line is #261: [Er74b][ErGr80][Er88c,p.104], and the displayed tag is number theory.

Verbatim current statement (copied from the live LaTeX view):

Are there infinitely many $n$ such that there exists some $t\geq 2$ and distinct integers $a_1,\ldots,a_t\geq 1$ such that\[\frac{n}{2^n}=\sum_{1\leq k\leq t}\frac{a_k}{2^{a_k}}?\]Is this true for all $n$? Is there a rational $x$ such that\[x = \sum_{k=1}^\infty \frac{a_k}{2^{a_k}}\]has at least $2^{\aleph_0}$ solutions?

(b, live-page cited results) The page lists the following established context.

The problem is related to problem 260.
Erdős reported that Cusick had a simple proof of infinitely many admissible

$n$, but did not record it.

Borwein and Loring gave the explicit identity

\[ n=2^{m+1}-m-2,\qquad \frac n{2^n}=\sum_{n<k\le n+m}\frac{k}{2^k} \tag{BL} \] for every positive integer $m$. For the stated $t\ge2$ condition take $m\ge2$.

Tengely, Ulas, and Zygadło verified the required property for

$n\le10{,}000$.

Erdős's 1988 formulation asks the weaker-looking question of a rational

$x$ having two representations.

(a, independent check of (BL)) Clearing $2^{n+m}$, the right side is

\sum_{i=1}^{m}(n+i)2^{m-i} =n(2^m-1)+(2^{m+1}-m-2)=n2^m,

which is the cleared left side. Also $n_{m+1}-n_m=2^{m+1}-1>0$, so this supplies infinitely many distinct $n$. The checker independently verifies this identity for $2\le m\le200$.

All three live comments

(a, exact arithmetic; comment itself is not treated as peer review) Vjeko Kovac's 2025-08-14 comment says that the $a_i$ must be mutually distinct, corrects an earlier attribution of (BL) to himself in favor of Borwein--Loring while retaining credit for Cusick, and says that the all-$n$ question remains open with $n\le10{,}000$ checked. The site says it was updated in response.

(a, exact arithmetic; comment itself is not treated as peer review) Zeraoulia Rafik's 2026-04-27 comment observes

\frac4{2^4}=\frac5{2^5}+\frac6{2^6}=\frac14,\qquad \sum_{m\ge1}\frac m{2^m}=2.

Thus, under the literal infinite-subset interpretation,

\frac74 =\sum_{a\in\mathbb N\setminus\{5,6\}}\frac a{2^a} =\sum_{b\in\mathbb N\setminus\{4\}}\frac b{2^b}.

This correctly gives two representations, but not continuum many.

(c, reported opinion only) Vjeko Kovac's 2026-05-05 comment says it is generally believed that “two” in the weakened formulation was a typo or misreading for $2^{\aleph_0}$. This is historical interpretation, not a mathematical result.

The live thread warns that comments are user-provided and unverified; I have accordingly used only the displayed elementary identities, not the comments as authority for open status.

1. Primary-source and later-literature audit

(b) Erdős's primary 1988 paper, “On the irrationality of certain series: problems and results”, pp. 102--109, states the finite equation on p. 104, asks “infinitely many $n$, or perhaps all $n$?”, and asks for a rational $x$ with two representations. Its footnote says Cusick communicated a proof of the infinitely-many assertion in June 1987 and that the all-$n$ question remained open.

(b) Borwein and Loring's primary paper, “Some questions of Erdős and Graham on numbers of the form $\sum g_n/2^{g_n}$”, Math. Comp. 54 (1990), 377--394, proves (BL). It formulates its stronger Conjecture 1 that every integer-initialized orbit

e_{j+1}=2(e_j\bmod j) \tag{BLC}

eventually reaches zero. It also proves that a dense set of irrational numbers has continuum many star-binary representations. Its rational consequences are conditional or give arbitrarily large finite multiplicity, not the requested rational continuum fiber.

(b) Tengely--Ulas--Zygadło, “On a Diophantine equation of Erdős and Graham”, J. Number Theory 217 (2020), 445--459 (arXiv:2008.01501), proves bounds for fixed term count, enumerates solutions through $k=8$, constructs high-multiplicity families, gives rational numbers with at least nine representations, and uses the exact greedy recurrence to verify $n\le10^4$. It explicitly leaves the all-$n$ question open.

(d, search audit rather than an exhaustiveness theorem) Exact-title, exact-equation, author, DOI-citation, arXiv, Crossref, OpenAlex, and Semantic Scholar searches on 2026-07-28 found no later primary paper extending the $10^4$ range or resolving either live question. The three citation indices each reported one citation of TUZ20: Hančl--Luca--Tachiya, “Irrationality and linear independence results for certain infinite series related to $r$-full numbers” (2026). Its publisher abstract concerns irrationality of series supported on $r$-full numbers, not the finite representation equation here. Citation indices are incomplete, so this is an honest search result, not proof that no uncatalogued literature exists.

2. Exact reduction to one integer orbit

Fix $n\ge2$. Scan candidate exponents $a=n+1,n+2,\ldots$ in increasing order, greedily taking $a$ whenever its term fits in the residual.

(a) Before examining $a$, let

R_a=\frac n{2^n} -\sum_{\substack{b<a\\b\ {\rm already\ selected}}}\frac b{2^b}, \qquad r_a=2^{a-1}R_a.

Then $r_{n+1}=n$. The condition that $a/2^a$ fits is exactly $a\le2r_a$, and clearing the next power of two gives

r_{a+1}= \begin{cases} 2r_a-a,&2r_a\ge a,\\ 2r_a,&2r_a<a. \end{cases} \tag{1}

Since $0\le r_{n+1}<n+1$, induction in (1) gives

0\le r_a<a,\qquad r_{a+1}=(2r_a)\bmod a, \tag{2}

where the remainder lies in $\{0,\ldots,a-1\}$.

(a) If the orbit first reaches $r_{A+1}=0$, the selected exponents form a finite, strictly increasing set and

n\,2^{A-n}=\sum_{a\ {\rm selected}}a\,2^{A-a}. \tag{3}

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, $0<r_a<a$ and (2) is zero exactly when $2r_a=a$.

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

e_{a+1}=2(e_a\bmod a),

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

\boxed{1\le n\le100{,}000}.

For $n=1$, where the greedy one-term equality is inadmissible because $t\ge2$, the separately checked witness is

\frac12=\frac3{2^3}+\frac6{2^6}+\frac8{2^8}.

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

\[ (n,k(n),A(n))=(5588,226913,460536). \]

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 $0\le r_a<a$ and the observed $A(n)\le659488$ make overflow 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 $0\le r_a<a$ and “termination occurs at $r_a=a/2$ for even $a$” do not 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.

This is the AI working report, labelled by outcome — not an independently verified claim unless marked PROVED. ← ledger

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 \(n=88,225\)