bibkey: erdos1985consecutive authors: Paul Erdős year: 1985 title: “Problems and results on consecutive integers and prime factors of binomial coefficients” doi: 10.1216/RMJ-1985-15-2-353 claim: “Put (21) ∏_{i=1}^{k} (x + i) = u_k(x) v_k(x), (u_k(x), v_k(x)) = 1 where v_k(x) is squarefree and all prime factors of u_k(x) occur with an exponent greater than 1. The representation in (21) is clearly unique. Clearly for k > k_0(x), u_k(x) > v_k(x). Perhaps one can estimate the smallest k_0(x) so that u_k(x) > v_k(x) for all k > k_0(x) quite well. I have not done this. For small values of k usually v_k(x) > u_k(x). I thought that for every x there is a k for which v_k(x) > u_k(x). x = 7 seemed a likely counterexample but if k = 7, u_7(7) = 2^7 3^3 < v_7(7) = 5·7·11·13. On the other hand a simple computation shows that n = 23 is a counterexample, i.e., for every k, v_k(23) < u_k(23). The reason for this is the existence of 24, 25, 27 and 32. I would not be surprised if 23 is the only counterexample. Perhaps in fact there is a k_0 so that for every k > k_0 and all n > n_0(k) (22) v_k(n) > u_k(n). (22) is perhaps too optimistic.” strata_touched:

• D5/S3/Factorization/ErdosConsecutiveProductSquarefreeFactorRefutation license: citation-only triage: anchor

Erdős’s consecutive-product factorization question

The paper separates each consecutive product into the primes occurring once and the prime powers whose exponents exceed one, then discusses how the two factors compare.

Verified locator

• DOI: 10.1216/RMJ-1985-15-2-353
• URL: https://users.renyi.hu/~p_erdos/1985-27.pdf
• Locator: page 361 of the Rocky Mountain Journal of Mathematics article.
• COMMENT (verbatim): Put (21) ∏_{i=1}^{k} (x + i) = u_k(x) v_k(x), (u_k(x), v_k(x)) = 1 where v_k(x) is squarefree and all prime factors of u_k(x) occur with an exponent greater than 1. The representation in (21) is clearly unique. Clearly for k > k_0(x), u_k(x) > v_k(x). Perhaps one can estimate the smallest k_0(x) so that u_k(x) > v_k(x) for all k > k_0(x) quite well. I have not done this. For small values of k usually v_k(x) > u_k(x). I thought that for every x there is a k for which v_k(x) > u_k(x). x = 7 seemed a likely counterexample but if k = 7, u_7(7) = 2^7 3^3 < v_7(7) = 5·7·11·13. On the other hand a simple computation shows that n = 23 is a counterexample, i.e., for every k, v_k(23) < u_k(23). The reason for this is the existence of 24, 25, 27 and 32. I would not be surprised if 23 is the only counterexample. Perhaps in fact there is a k_0 so that for every k > k_0 and all n > n_0(k) (22) v_k(n) > u_k(n). (22) is perhaps too optimistic.

Only the sentence “I would not be surprised if 23 is the only counterexample” is refuted: the starting value 47 has u_k(47) > v_k(47) for every positive k; conjecture (22) and the statement about the starting value 23 are untouched.