Re: [math-fun] parity of floor(2^n / n)
mreid>> (This may be asking way too much: I don't even know
how to prove that the floor of 2^n / n is odd for infinitely many values of n.)
Note that n = A090659 plus its conjectured continuation, 25, 91, 703, 1891, 12403, 38503,79003 ,88831 ,146611,188191, 218791, 269011, 286903, 385003, 497503, 597871, 736291, 765703, 954271, 1056331, 1314631, 1869211, 2741311, 3270403, 3913003, 4255903 ,4686391 ,5292631
gives all odds. --rwg
i didn't quite follow this. apart from the first term (25), the others are all of the form pq where p is a prime = 3 mod 4 , and q = 2 p - 1 . there is a comment in the online encyclopedia that later terms in the sequence are conjectured to be of this form (although it does not mention that p = 3 mod 4 ). for such numbers, we have 2^(pq) = 0 mod 2 , 2^(pq) = 2^q = 2 mod p , and 2^(pq) = 2^p = -2 mod q (using the fact that 2 is a quadratic non-residue mod q ). put these together to get 2^(pq) = (2p - 4) q - 2 mod 2pq . if p > 3 , then the remainder, (2p - 4) q - 2 , is in the range [pq, 2pq) , so indeed floor(2^(pq) / pq) is odd. (this also explains why 15 doesn't work.) conjecturely, there are infinitely many primes p = 3 mod 4 , for which 2p - 1 is also prime, but i don't believe it's been proven. nevertheless, these terms should grow slower (i.e. there are more of them) than those in the provably infinite "sequence" i gave, whose plateaus grow doubly exponentially. 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 . mike
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.)
rwg>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 Or with z = 1/9, inf /===\ - 1/9 | | %e | | ------------ = | | 2 n = 1 1 n (1 - ----) 2 9 n sqrt(3) Psi (1/3) 13 zeta(3) 1 2 sqrt(3) %pi ---------- - ----------------- + ------------- + 1/18 2 27 %pi 81 18 %pi %e ------------------------------------------------------- 1/18 3
participants (2)
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Michael Reid -
R. William Gosper