A178459 Partial sums of floor(2^n/31).

0, 0, 0, 0, 1, 3, 7, 15, 31, 64, 130, 262, 526, 1054, 2111, 4225, 8453, 16909, 33821, 67646, 135296, 270596, 541196, 1082396, 2164797, 4329599, 8659203, 17318411, 34636827, 69273660, 138547326, 277094658, 554189322, 1108378650, 2216757307, 4433514621, 8867029249, 17734058505, 35468117017, 70936234042

Offset: 1

Author

Mircea Merca, Dec 22 2010

Keywords

• nonn
• easy

Comments

Partial sums of A119610(n-4).

Examples

			a(29) = 0 + 0 + 0 + 0 + 1 + 2 + 4 + 8 + 16 + 33 + 66 + 132 + 264 + 528 + 1057 + 2114 + 4228 + 8456 + 16912 + 33825 + 67650 + 135300 + 270600 + 541200 + 1082401 + 2164802 + 4329604 + 8659208 + 17318416 = 34636827.
		

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,-2,0,0,1,-3,2).

Crossrefs

Cf. A119610.

Programs

Magma

[Round((10*2^n-31*n-7)/155): n in [1..40]]; // Vincenzo Librandi, Jun 21 2011

Maple

seq(round((10*2^n-31*n-7)/155),n=1..32)

Mathematica

Floor[2^Range[40]/31]//Accumulate (* Harvey P. Dale, May 11 2018 *)

Formula

a(n) = round((10*2^n - 31*n - 7)/155).

a(n) = floor((10*2^n - 31*n + 22)/155).

a(n) = ceiling((10*2^n - 31*n - 36)/155).

a(n) = round((10*2^n - 31*n - 10)/155).

a(n) = a(n-5) + 2^(n-4) - 1, n > 4,

G.f.: -x^5/((x-1)^2*(2*x-1)*(x^4 + x^3 + x^2 + x + 1)). - Colin Barker, Oct 27 2012

From Seiichi Manyama, Dec 22 2023: (Start)

a(n) = Sum_{k=0..n} 2^(n-k) * floor(k/5).

a(n) = floor(2^(n+1)/31) - floor((n+1)/5). (End)