bibkey: irvine2026a394431
authors: Jean-Marc Rebert; Sean A. Irvine
year: 2026
title: "OEIS A394431, The smallest base b < n where the sum of the digits for the number n in the base b is the largest"
doi: null
url: https://oeis.org/A394431
claim: "A394431 %F: Conjecture: a(n) = ceiling((n+1)/2) for n>8. - _Sean A. Irvine_, Mar 25 2026"
strata_touched:
  - D5/S1/Digit/MaxDigitSumBase
license: citation-only
triage: anchor

OEIS A394431

The entry records, for each n, the smallest base b with 1 < b < n at which the base-b digit sum of n is largest, with a(1) = a(2) = 1. Its terms begin 1, 1, 2, 3, 3, 4, 4, 3, 5, 6, 6, 7, 7, 8, 8, 9, ….

Verified locator

• URL: https://oeis.org/A394431
• Locator: NAME, “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.”; FORMULA, “Conjecture: a(n) = ceiling((n+1)/2) for n>8. - Sean A. Irvine, Mar 25 2026”.
• Sequence author Jean-Marc Rebert, Mar 20 2026; revision 20, Jul 10 2026. The conjecture line carries no answer, and the entry has no comment and no paper link.

Reading of the statement

Write s_b(n) for the base-b digit sum of n. The entry’s value is the least base among those in 1 < b < n attaining the largest s_b(n). The conjecture says this base is ceiling((n + 1) / 2) = floor(n / 2) + 1 for every n > 8. At n = 8 it fails, since s_3(8) = s_5(8) = 4 and the least maximizing base is 3.

Scope of the recorded answer

The conjecture holds. Above half of n the expansion of n in base b is the two digits 1 and n - b, so s_b(n) = n - b + 1, which is largest at the least such base floor(n / 2) + 1 and equals n - floor(n / 2) there. For a base b with 2 * b ≤ n, write n = q * b + r with q ≥ 2 and r < b, so s_b(n) = r + s_b(q). Every digit sum is at most its argument, and when q ≥ b a further expansion gives s_b(q) + (b - 1) ≤ q. In both cases 2 * s_b(n) < n once n > 8, so s_b(n) < n - floor(n / 2), strictly below the value at the first base above half. The single case of equality in these bounds is b = 3, q = r = 2, that is n = 8.

Bounded prior-resolution evidence

The entry was read in full and records no proof and no reference to one. Searching the catalogue by the first sixteen terms returns only this entry, and by the sequence of largest digit sums only the generic entry for floor(n / 2), whose comments do not state the cross-base maximum. No paper or note giving the result was found by web query or on MathDB. Citation indices were not exhaustively reachable, so this is a bounded negative finding and no worldwide priority claim is made.