A123753 Partial sums of A070941.

1, 3, 6, 9, 13, 17, 21, 25, 30, 35, 40, 45, 50, 55, 60, 65, 71, 77, 83, 89, 95, 101, 107, 113, 119, 125, 131, 137, 143, 149, 155, 161, 168, 175, 182, 189, 196, 203, 210, 217, 224, 231, 238, 245, 252, 259, 266, 273, 280, 287, 294, 301, 308, 315, 322, 329, 336, 343

Offset: 0

Author

Reinhard Zumkeller, Oct 12 2006

Keywords

• nonn

Links

• Peter Luschny, Table of n, a(n) for n = 0..10000
• Hsien-Kuei Hwang, S. Janson and T.-H. Tsai, Exact and asymptotic solutions of the recurrence f(n) = f(floor(n/2)) + f(ceiling(n/2)) + g(n): theory and applications, Preprint 2016.
• Hsien-Kuei Hwang, S. Janson and T.-H. Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, 13:4 (2017), #47.

Crossrefs

Cf. A001855, A003314, A033156, A054248, A061168, A083652, A097383, A295508.

Programs

Maple

A123753 := proc(n) local i, J, z; i := n+1: J := i; i := i-1; z := 1;
while 0 <= i do J := J+i; i := i-z; z := z+z od; J end:
seq(A123753(n), n=0..57); # Peter Luschny, Nov 30 2017
# Alternatively:
a := n -> (n+1)*(1 + ilog2(2*n+3)) - 2^ilog2(2*n+3) + 1:
seq(a(n), n=0..57); # Peter Luschny, Dec 02 2017

Mathematica

a[n_] := (n + 1)(1 + IntegerLength[n + 1, 2]) - 2^IntegerLength[n + 1, 2] + 1;
Table[a[n], {n, 0, 57}] (* Peter Luschny, Dec 02 2017 *)

Python

def A123753(n):
    s, i, z = n+1, n, 1
    while 0 <= i: s += i; i -= z; z += z
    return s
print([A123753(n) for n in range(0, 58)]) # Peter Luschny, Nov 30 2017

Python

def A123753(n): return (n+1)*(1+(m:=n.bit_length()))-(1<Chai Wah Wu, Mar 29 2023

Formula

a(n) = A003314(n+1)+1. - Reinhard Zumkeller, Oct 12 2006

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