The Least Base Maximizing the Digit Sum of n Is ceiling((n+1)/2) for n > 8

OEIS A394431, by Jean-Marc Rebert, Mar 20 2026:

The smallest base b < n where the sum of the digits for the number n in the base b is the largest, with 1 < b < n and a(1) = a(2) = 1.

The formula line carrying the question, by Sean A. Irvine, Mar 25 2026:

Conjecture: a(n) = ceiling((n+1)/2) for n>8.

The frozen theorem D5/S1/Digit/MaxDigitSumBase.result settles the conjecture for every n > 8, stating both that no base in 1 < b < n beats ceiling((n + 1) / 2) and that every smaller base does strictly worse.

Issue 9617 records the screen carried out before the work. The A-number appears nowhere under Problems/, D5/, Library/ or Blueprint/ in this repository, nor in the three screening records, nor under FormalConjectures/OEIS/ in google-deepmind/formal-conjectures. The entry records no proof and no reference. A catalogue search by the terms returns only this entry, and by the largest digit sums only the generic entry for floor(n / 2). Web and MathDB queries returned no paper or note giving the result. This is a bounded negative finding.

Write s_b(n) for the base-b digit sum and b0 = floor(n / 2) + 1, which equals ceiling((n + 1) / 2).

One. If b is a base with n < 2 * b and b < n, then n / b = 1 and n mod b = n - b, so s_b(n) = n - b + 1. In particular s_b0(n) = n - floor(n / 2), and every larger base gives less.

Two. Every digit sum is at most its argument, s_b(q) ≤ q. If q ≥ b then expanding once more, s_b(q) = (q mod b) + s_b(q / b) ≤ (q mod b) + q / b, while q = (q mod b) + b * (q / b) with q / b ≥ 1; hence s_b(q) + (b - 1) ≤ q.

Three. Let 2 ≤ b with 2 * b ≤ n and n > 8. Write n = q * b + r, so q ≥ 2, r < b and s_b(n) = r + s_b(q). If q ≥ b, step two gives 2 * s_b(n) ≤ 2 * r + 2 * q - 2 * (b - 1), which is below r + b * q = n because r ≤ b - 1 and b ≥ 2. If q < b, then b ≥ 4, because b = 2 or b = 3 with q < b would force n ≤ 8; then 2 * s_b(n) ≤ 2 * r + 2 * q < r + b * q, since (b - 2) * q ≥ 2 * b - 4 > b - 1 ≥ r. In both cases 2 * s_b(n) < n.

Four. Step three gives s_b(n) < n - floor(n / 2) = s_b0(n) for every base at most half of n, and these are exactly the bases below b0; step one covers the bases from b0 up to n - 1. So b0 attains the maximum and every smaller base falls strictly short.

Every step is elementary: the one-step digit expansion, the bound of a digit sum by its argument, a case split on the base, and integer arithmetic.

The statement would fail at n = 8, where s_3(8) = 4 = s_5(8) makes 3 the least maximizing base; this is the equality case b = 3, q = r = 2 of step three, and the conjecture excludes it by requiring n > 8. Reading ceiling((n + 1) / 2) as floor((n + 1) / 2) would give n / 2 for even n, a base with two digits 2, 0, whose digit sum 2 is not maximal for n > 8.

The least maximizing base was recomputed from the definition for every n from 1 to 1499 and agrees with the entry's b-file term by term; the b-file's 10000 terms satisfy the conjecture at every n > 8. The strict gap between the largest digit sum over bases at most n / 2 and ceiling(n / 2) was checked for every n from 9 to 19999, with no exception, and the exception at n = 8 was reproduced.

theorem; Tier 1 named external open question from 2026, preregistered in issue 9617 before the work. The computational use is none: no declaration is a bounded enumeration, a checker, a numeric reduction or a certified instance, and the delivered statement is universally quantified over every n > 8.

The literature screen is bounded. The entry was read in full and records no proof and no reference to one; the conjecture has stood since March 2026. Citation indices and printed sources were not exhaustively reachable, so no worldwide priority claim is made. The weight is stated plainly: the argument is elementary, and what is settled is that the sentence sat on the entry unjudged.