Curtiss 1922 kellogg s diophantine problem

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theorem_i: Proves that the least positive value of 1 minus a sum of n - 1 unit fractions is 1 over u_n, with u_1 = 1 and u_(k+1) = u_k(u_k + 1), attained only at the denominators u_k + 1.

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Curtiss, D. R., On Kellogg's Diophantine Problem. Amer. Math. Monthly 29 (1922), no. 10, 380-387, DOI 10.1080/00029890.1922.11986179.

Curtiss proves Kellogg's assertion that in any solution of 1 = 1/x_1 + ... + 1/x_n in positive integers the largest unknown is at most u_n, where u_1 = 1 and u_{k+1} = u_k(u_k+1), giving the sequence 1, 2, 6, 42, 1806, ... (each one less than Sylvester's sequence 2, 3, 7, 43, 1807, ...). Theorem I states more generally that, with f_{n-1}(x) defined by 1/f_{n-1}(x) = 1 - 1/x_1 - ... - 1/x_{n-1}, the maximum finite value of f_{n-1}(x) over positive integers x_1,...,x_{n-1} is u_n, so the least positive value of 1 - 1/x_1 - ... - 1/x_{n-1} is 1/u_n, attained only at x_k = u_k + 1 for k = 1,...,n-1; unlike Kellogg, Curtiss does not restrict to x's making f_{n-1}(x) an integer, and the maximum turns out to be one. The proof works by chains of inequalities on 'reduced sets', each step removing one variable: the best choice of the last variable is x_{n-1} = E(f_{n-2}(x)) + 1, and induction on the number of variables gives the bound. Two applications are given first (pp. 380--381): fitting regular-polygon tiles about a common vertex (a piece of a Riemann surface whose sheets form one cycle about a branch point), and perfect numbers, where Kellogg's bound gives (p. 381) that a perfect number with n divisors less than itself, unity included, is at most u_n; both reduce to solutions of 1 = sum 1/x_i. Erdos problem 206 asks whether, for almost every real x > 0, the best underapproximations of x by n distinct unit fractions are eventually built greedily. Theorem I gives the case x = 1 at every length: the x's in Theorem I need not be distinct, but the unique optimal set x_k = u_k + 1 has distinct members, so the best sum of n - 1 distinct unit fractions below 1 is the greedy sum 1/2 + 1/3 + 1/7 + ... + 1/(u_{n-1} + 1) = 1 - 1/u_n; this is a deduction recorded here, not a statement of the paper.

Source: https://archive.org/details/jstor-2299023.

The copy read for this card is the JSTOR Early Journal Content scan (nine physical pages: a cover sheet followed by printed pp. 380--387); its text layer garbles the formulas, and the statements here were read on the page images. Read status: claims checked for Theorem I and the definitions (1)--(4), compiled on Theorem I; the proof (pp. 382--386) was read for structure; no proof is rewritten in full and none has been independently reviewed. The author's note on p. 387 records Takenouchi's result, in a paper that had then just appeared (Proc. Phys.-Math. Soc. Japan (3) 3, 78--92), on the largest unknown for sums equal to b/a with a = (m+1)b - 1, and states, without proof, an upper bound B_n for the maximum finite value of the analogous f_{n-1}(x), n > 1, when b <= a. The printed pages 380--387 carry no copyright line; the scan's first page is JSTOR's Early Journal Content cover sheet, which states "Early Journal Content on JSTOR, Free to Anyone in the World" and "People may post this content online or redistribute in any way for non-commercial purposes." and names no license, so the term is recorded as reserved, the usage notice the file prints.

Bears on. #206: Theorem I implies the case $x=1$, where the best sums of $n$ distinct unit fractions below $1$ are the greedy ones for every $n$ (the deduction above, which the paper does not state); it says nothing about other $x$.

Results to transcribe.

• Theorem I: With 1/f_{n-1}(x) = 1 - 1/x_1 - ... - 1/x_{n-1}, the maximum finite value of f_{n-1}(x) over positive integers is u_n with u_1 = 1, u_{k+1} = u_k(u_k+1), attained only at x_k = u_k + 1; equivalently the least positive remainder is 1/u_n and the largest denominator in 1 = sum_{i<=n} 1/x_i is at most u_n.
• Necessary conditions (Sec. 3, pp. 382--384): for a maximum, the largest variable must satisfy x_{n-1} = E(f_{n-2}(x)) + 1 (4), where E(a) is the greatest integer not exceeding a; Theorem II (p. 384) strengthens this to the x's forming a 'reduced set', a compact set with a single largest member. Recorded inside the sketch on the Theorem I page.

No file of this source is held: the only terms on record, the cover sheet's notice, permit only non-commercial redistribution and name no open license, and the card cites the edition it names above.