Pihko 2010 remarks greedy odd egyptian fraction algorithm ii

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corollary_3_6: For an odd prime p and 1 < a < p, a residue class of t modulo p makes the greedy odd algorithm on a/(2k+1), with k = −1 + t·a(a+1)⋯(p−1), run through the numerators a, a+1, …, p−1, 1.

open_problem_p202: The 2010 paper's introduction restates as open whether the greedy odd Egyptian fraction algorithm always stops after finitely many steps.

theorem_2_3: When k + 1 is a positive multiple of a(a+1)⋯(a+n), the greedy odd algorithm on a/(2k+1) has numerators starting a, a+1, …, a+n and the next unreduced numerator a+n+1.

theorem_3_5: For an odd prime p and k = −1 + (p−1)!, the greedy odd algorithm on 2/(2k+1) has the numerator sequence 2, 3, …, p−1, 1.

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Jukka Pihko, Remarks on the "greedy odd" Egyptian fraction algorithm II, Fibonacci Quart. 48 (2010), no. 3, 202--208; DOI 10.1080/00150517.2010.12428097. MSC2010 11D68. The paper is "a direct continuation" of the 2001 paper pihko_2001_remarks_greedy_odd_egyptian_fraction_algorithm (its [5]), from which it repeats the setup.

The copy read for this card is a seven-page typeset file ("Pihko.dvi"; physical PDF p. $n$ is printed p. $201+n$) whose text layer renders each prime mark as a 0 (so $a_1'$ reads "a01"); the statements below were checked on the page images. Provenance: obtained from the repository's survey download set of September 2026 (file dated 2026-09-05; the download URL was not recorded); 98,365 bytes. No copyright line is printed on the pages (pp. 202 and 208 read); the journal's issue page that lists the article shows the footer "Copyright © 2010 The Fibonacci Association. All rights reserved." (https://www.fq.math.ca/48-3.html), and the publisher's DOI page, which had returned HTTP 403, was not retried, every other right reserved.

Read status: claims checked. The abstract, the introduction's statement of the open problem, Lemma 2.1, Theorem 2.3, Theorem 3.5 and Corollary 3.6 were read clause by clause on the page images; the proofs (pp. 203--206) were read for structure and not checked; Section 4 (pp. 207--208) was read at statement level only.

Contents

• Setup (p. 202): $a<b$ positive integers with $\gcd(a,b)=1$ and $b$ odd; the greedy odd algorithm takes the greatest Egyptian fraction $1/x_1$ with $x_1$ odd and $1/x_1\le a/b$, forms $a/b-1/x_1=a_1/b_1$ in lowest terms and continues while the remainder is nonzero, giving $a/b=1/x_1+1/x_2+\cdots$ (display (1.2)). "A well-known open problem is whether the greedy odd algorithm always stops after finitely many steps, i.e., whether the sum in (1.2) is always finite", cited to Guy's problem book (2nd ed., 1994), Guy's Monthly article of 1998 and Klee--Wagon's problem book (1991), the paper's [2], [3] and [4]. Result page: open_problem_p202.
• Section 2 (pp. 202--204): with $b=2k+1$ and $x_1=2n_1+1$, consecutive numerators have opposite parity (display (2.3)) and $k_1'=2kn_1+k+n_1$ (display (2.4)). Lemma 2.1 (p. 203): for $a>1$, $k\in\mathbb N$ and $k\equiv-1\pmod a$ the first step gives $n_1=(k+1)/a$ and the next unreduced numerator $a_1'=a+1$. Theorem 2.3 (p. 203): for $a>1$, $n\ge0$ and $k=-1+t\,a(a+1)\cdots(a+n)$, with $t\in\mathbb N$ as in Lemma 2.2, the numerator sequence of $a/(2k+1)$ starts $a,a+1,\dots,a+n$, the next unreduced numerator is $a+n+1$, and for $n\ge1$ the first $n$ steps need no reduction. Result page: theorem_2_3. Example 2.5 (Table 1, p. 204) lists the numerator sequences for $a=2$, $n=4$ and $1\le t\le7$, for instance $2,3,4,5,6,1$ at $t=1$.
• Section 3 (pp. 204--206): sequences of numerators $a,a+1,\dots,2m,1$. Lemma 3.1 characterizes them by a congruence modulo $2m+1$; Corollary 3.3 shows the outcome depends only on $t\bmod(2m+1)$; Theorem 3.5 (p. 205): for $a=2$ and $2m+1=p$ an odd prime, $t=1$ gives the sequence $2,3,\dots,p-1,1$ (proof by Wilson's theorem; result page theorem_3_5). Corollary 3.6 (p. 206): for every odd prime $p$ and $1<a<p$ there is $t\in\{1,\dots,p\}$ such that $k=-1+t\,a(a+1)\cdots(p-1)$ gives the numerator sequence $a,a+1,\dots,p-1,1$ for $a/(2k+1)$, and the same sequence results for all $K=-1+T\,a(a+1)\cdots(p-1)$ with $T\equiv t\pmod p$; hence, as the abstract states, infinitely many odd $b$ with $\gcd(a,b)=1$ and $a<b$ with these numerators, for which the algorithm stops after exactly $p-a+1$ steps (the step count is deduced on the result page corollary_3_6). Tables 2--4 (pp. 205--206) give computed examples.
• Section 4 (pp. 207--208): all solutions $t$ modulo $p$ when $n=p-1-a\in\{0,1,2\}$: $t\equiv m$ for $a=p-1$ (Theorem 4.1), the inverses of $2$ and $4$ for $a=p-2$ with $p\ge5$ (Theorem 4.2), and two or four solutions for $a=p-3$ with $p\ge5$ according to $p\bmod8$ (Theorem 4.3).

Compiled scope

The statements above were checked on the page images of a typeset file; the proofs were read for structure only. Nothing here is independently reviewed.

Bears on. #282: the introduction (printed p. 202; open_problem_p202) restates the odd-denominator termination question as a "well-known open problem" in 2010, in the 2001 paper's convention (the greatest odd unit fraction not exceeding the remainder); Corollary 3.6 (p. 206; corollary_3_6) constructs, for each odd prime $p$ and $1<a<p$, infinitely many odd $b$ whose numerator sequence is $a,a+1,\dots,p-1,1$, so the algorithm stops after $p-a+1$ steps for them (a count deduced on the result page), and Theorem 3.5 (p. 205; theorem_3_5) is its case $a=2$, $t=1$; neither decides anything about termination in general.

No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.