|
%I #37 Jun 29 2019 19:47:41
%S 1,2,4,5,10,10,20,22,37,40,80,72,144,158,278,283,566,548,1096,1120,
%T 2106,2162,4324,4210,8389,8584,16650,16772,33544,33194,66388,66968,
%U 131882,132690,265222,263607,527214,530138,1052078,1054254,2108508,2103282,4206564,4216760
%N a(n) = Sum_{k=1..n} 2^(n mod k).
%C A more precise asymptotic formula is given in the link.
%C From _David Morales Marciel_, Oct 19 2015: (Start)
%C If n is prime then a(n)=2*a(n-1).
%C It appears that for every (deficient, abundant)-pair of numbers (11+6x, 11+6x+1), a(11+6x) > a(11+6x+1).
%C (End)
%H Amiram Eldar, <a href="/A198383/b198383.txt">Table of n, a(n) for n = 1..1000</a>
%H B. Cloitre, <a href="http://BCmathematics.monsite-orange.fr/Asymptotic%20sum%202%5E(n%20mod%20k).pdf">An asymptotic formula for sum_{k=1..n}x^(n mod k)</a> [broken link, <a href="/A198383/a198383.pdf">draft</a>]
%F a(n) = 2^ceiling(n/2) + O(2^(n/3)).
%t Table[Sum[2^Mod[n, k], {k, n}], {n, 44}] (* _Michael De Vlieger_, Oct 19 2015 *)
%o (PARI) a(n) = sum(k=1, n, 2^(n%k))
%Y Cf. A198259.
%K nonn
%O 1,2
%A _Benoit Cloitre_, Oct 24 2011
|
