|
%I #14 Mar 09 2023 13:52:08
%S 0,1,3,5,9,13,18,23,31,39,48,57,68,79,91,103,119,135,152,169,188,207,
%T 227,247,270,293,317,341,367,393,420,447,479,511,544,577,612,647,683,
%U 719,758,797,837,877,919,961,1004,1047,1094,1141,1189
%N a(n) - a(n-1) = A113474(n).
%C First differences are A113474.
%C The following sequences all appear to have the same parity: A003071, A029886, A061297, A092524, A093431, A102393, A104258, A122248, A128975. - _Jeremy Gardiner_, Dec 28 2008
%H Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, <a href="http://140.109.74.92/hk/wp-content/files/2016/12/aat-hhrr-1.pdf">Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications</a>, preprint, 2016.
%H Hsien-Kuei Hwang, Svante Janson, Tsung-Hsi Tsai, <a href="http://doi.org/10.1145/3127585">Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications</a>, ACM Transactions on Algorithms 13:4 (2017), #47.
%H Tanya Khovanova, <a href="http://arxiv.org/abs/1410.2193">There are no coincidences</a>, arXiv preprint 1410.2193 [math.CO], 2014.
%F G.f.: (1/(1-x))*Sum{k=0..infinity, x^(2^k)/((1-x)*(1-x^(2^k)))}-x^2/(1-x)^3.
%F a(n) = Sum_{k=1..n} Sum_{j=0..n} floor(k/2^j) - binomial(n,2).
%F a(n) = A122247(n)-binomial(n,2).
%o (PARI) a(n) = sum(k=1, n, sum(j=0, n, k\2^j)) - binomial(n, 2); \\ _Michel Marcus_, Mar 09 2023
%Y Cf. A113474, A122247.
%K easy,nonn
%O 0,3
%A _Paul Barry_, Aug 27 2006
|
