A178420 Partial sums of floor(2^n/3).

0, 1, 3, 8, 18, 39, 81, 166, 336, 677, 1359, 2724, 5454, 10915, 21837, 43682, 87372, 174753, 349515, 699040, 1398090, 2796191, 5592393, 11184798, 22369608, 44739229, 89478471, 178956956, 357913926, 715827867, 1431655749, 2863311514

Offset: 1

Author

Mircea Merca, Dec 21 2010

Keywords

• nonn
• easy

Comments

Essentially the same as A011377: 0 followed by the terms of A011377. - Joerg Arndt, Apr 22 2016

Partial sums of A000975(n-1).

Examples

			a(5) = 0 + 1 + 2 + 5 + 10 = 18.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.
• Index entries for linear recurrences with constant coefficients, signature (3,-1,-3,2).

Crossrefs

Column k=2 of A368296.

Cf. A000975, A178455.

Programs

Magma

[Floor((4*2^n-3*n-4)/6): n in [1..30]]; // Vincenzo Librandi, Jun 23 2011

Maple

seq(round((4*2^n-3*n-4)/6),n=1..50)

Mathematica

f[n_] := Floor[(4 2^n - 3 n - 4)/6]; f[Range[60]] (* Vladimir Joseph Stephan Orlovsky, Jan 29 2011 *)
CoefficientList[Series[x / ((1 + x) (1 - 2 x) (1 - x)^2), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 26 2014 *)
LinearRecurrence[{3,-1,-3,2},{0,1,3,8},40] (* or *) Accumulate[ Table[ Floor[ 2^n/3],{n,40}]] (* Harvey P. Dale, Dec 24 2015 *)

PARI

a(n)=(4<Charles R Greathouse IV, Jul 31 2013

Formula

a(n) = A011377(n-1) for n >= 1. - Joerg Arndt, Apr 22 2016

a(n) = round((8*2^n - 6*n - 9)/12).

a(n) = floor((4*2^n - 3*n - 4)/6).

a(n) = ceiling((4*2^n - 3*n - 5)/6).

a(n) = round((4*2^n - 3*n - 4)/6).

a(n) = a(n-2) + 2^(n-1) - 1, n > 2.

From Bruno Berselli, Jan 15 2011: (Start)

a(n) = (8*2^n - 6*n - 9 + (-1)^n)/12.

G.f.: x^2/((1+x)*(1-2*x)*(1-x)^2). (End)

G.f.: Q(0)/(3*(1-x)^2), where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k - 1/(1 + 1/(2*4^k - 8*x*16^k/(4*x*4^k + 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 21 2013

a(n) = 2*a(n-1) + floor(n/2) for n > 1. - Bruno Berselli, Apr 30 2014

a(n) = floor(2^(n+1)/3) - floor((n+1)/2). - Seiichi Manyama, Dec 22 2023