mreid>none of this speaks to jim's original question, which asked if, as N goes to infinity, floor(2^n / n) is even for asymptotically half the positive integers less than N . Well, at least we have an interesting remark about A090659, which was the only all-odd floor(2^n/n) sequence in Neil's collection as of a few months ago. This suggests that merely finding slower growing ones, let alone finite density ones, is a seriously hard problem. --rwg NEOSCHOLASTIC CHOCOLATINESS PS, What is closed form? We like to convert sums to products because products have fewer guises. And log of a closed form is a closed form. But log of a product is morally a sum! inf inf k k k d + 1 ==== k k + 1 ==== ==== (floor(-) - ceiling(-) + 1) z \ q z \ k \ d d
---------------- = > q > ---------------------------------- / k / / d + 1 ==== (k + 1) (1 - q ) ==== ==== k = 1 k = 0 d = 1
inf /===\ - z | | %e = log( | | ---------------), |q| < 1. | | - n n = 1 n q (1 - q z) It's kind of fun to see prod_n lim_n(f(n))/f(n). With n^2 instead of q^-n we get sqrt(z) / [ z/2 + 2 I t log(sin(%pi t)) dt ] inf / /===\ - z 0 | | %e %e | | ---------- = ---------------------------------------, | | 2 z n = 1 z n sin (%pi sqrt(z)) (1 - --) 2 n e.g., with z=1/4, 7 zeta(3) --------- + 1/8 inf 2 /===\ - 1/4 8 %pi | | %e %e | | ------------ = -----------------. | | 2 1/4 n = 1 1 n 2 (1 - ----) 2 4 n (With n for n^2 the product diverges.)