cp's OEIS Frontend

From Jianing Song, Jun 16 2022: (Start)

Conjecture: a(2^k-1) < a(2^k-2) for all k >= 3. Checked up to k = 263. Note that Catalan(2^k-1) is odd and Catalan(2^k-2)/Catalan(2^k-1) = 2^(k-1)/(2^(k+1)-3). Suppose that 2^(k+1)-3 = Product_{i=1..r} (p_i)^(e_i), let r_i be the (p_i)-adic valuation of binomial(2*(2^k-1),2^k-1), then a(2^k-2)/a(2^k-1) = k * Product_{i=1..r} (e_i-r_i+1)/(e_i+1). This seems unlikely to be less than 1. Actually, it seems that a(2^k-2)/a(2^k-1) tends to infinity as n goes to infinity.

Conjecture: a(2^k-1) != a(2^k) for all k. Checked up to k = 265. Note that Catalan(2^k)/Catalan(2^k-1) = 2 * (2^(k+1)-1)/(2^k+1). Suppose that (2^(k+1)-1)/(2^k+1) = Product_{i=1..r} (p_i)^(e_i), let r_i be the (p_i)-adic valuation of binomial(2*(2^k-1),2^k-1), then a(2^k)/a(2^k-1) = 2 * Product_{i=1..r} (e_i+r_i+1)/(e_i+1). This seems unlikely to be equal to 1. Among the numbers k <= 265, the number k for which a(2^k)/a(2^k-1) is closest to 1 is k = 70, where a(2^k)/a(2^k-1) = 104/105. (End)

a(n) <> 2 iff n = 2^k - 1 (A000225). In fact for k>1, a(2^k-1): 5, 3, 3, 7, 3, 3, 7, 3, 3, 3, 3, 3, 3, ..., . (A120275) - Robert G. Wilson v, Nov 14 2015