%I #38 Mar 30 2023 01:49:19
%S 1,3,6,9,13,17,21,25,30,35,40,45,50,55,60,65,71,77,83,89,95,101,107,
%T 113,119,125,131,137,143,149,155,161,168,175,182,189,196,203,210,217,
%U 224,231,238,245,252,259,266,273,280,287,294,301,308,315,322,329,336,343
%N Partial sums of A070941.
%H Peter Luschny, <a href="/A123753/b123753.txt">Table of n, a(n) for n = 0..10000</a>
%H Hsien-Kuei Hwang, S. Janson and T.-H. 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, S. Janson and T.-H. Tsai, <a href="https://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.
%F a(n) = A003314(n+1)+1. - _Reinhard Zumkeller_, Oct 12 2006
%F Let bil(n) = floor(log_2(n)) + 1 for n>0, bil(0) = 0 and b(n) = n + n*bil(n) - 2^bil(n) + 1 then a(n) = b(n+1). (This suggests that '0' be prepended to this sequence.) - _Peter Luschny_, Dec 02 2017
%p A123753 := proc(n) local i, J, z; i := n+1: J := i; i := i-1; z := 1;
%p while 0 <= i do J := J+i; i := i-z; z := z+z od; J end:
%p seq(A123753(n), n=0..57); # _Peter Luschny_, Nov 30 2017
%p # Alternatively:
%p a := n -> (n+1)*(1 + ilog2(2*n+3)) - 2^ilog2(2*n+3) + 1:
%p seq(a(n), n=0..57); # _Peter Luschny_, Dec 02 2017
%t a[n_] := (n + 1)(1 + IntegerLength[n + 1, 2]) - 2^IntegerLength[n + 1, 2] + 1;
%t Table[a[n], {n, 0, 57}] (* _Peter Luschny_, Dec 02 2017 *)
%o (Python)
%o def A123753(n):
%o s, i, z = n+1, n, 1
%o while 0 <= i: s += i; i -= z; z += z
%o return s
%o print([A123753(n) for n in range(0, 58)]) # _Peter Luschny_, Nov 30 2017
%o (Python)
%o def A123753(n): return (n+1)*(1+(m:=n.bit_length()))-(1<<m)+1 # _Chai Wah Wu_, Mar 29 2023
%Y Cf. A001855, A003314, A033156, A054248, A061168, A083652, A097383, A295508.
%K nonn
%O 0,2
%A _Reinhard Zumkeller_, Oct 12 2006