A178872 Partial sums of round(4^n/7).

0, 1, 3, 12, 49, 195, 780, 3121, 12483, 49932, 199729, 798915, 3195660, 12782641, 51130563, 204522252, 818089009, 3272356035, 13089424140, 52357696561, 209430786243, 837723144972, 3350892579889, 13403570319555, 53614281278220, 214457125112881

Offset: 0

Author

Mircea Merca, Dec 28 2010

Keywords

• nonn
• less
• easy

Comments

a(n) (prefixed with a 0) and its higher order differences define the following infinite array:

0, 0, 1, 3, 12, 49,..

0, 1, 2, 9, 37, 146,...

1, 1, 7, 28, 109, 439... - Paul Curtz, Jun 08 2011

Examples

			a(3)=0+1+2+9=12.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..500
• Index entries for linear recurrences with constant coefficients, signature (3,3,4).
• Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.

Crossrefs

Cf. A049347, A164346.

Programs

Magma

[Floor((4*4^n+5)/21): n in [0..30]]; // Vincenzo Librandi, May 01 2011

Maple

A178872 := proc(n) add( round(4^i/7),i=0..n) ; end proc:

Mathematica

Join[{a = b = 0}, Table[c = 4^n - a - b; a = b; b = c, {n, 0, 100}]] (* Vladimir Joseph Stephan Orlovsky, Jun 28 2011 *)
Accumulate[Round[4^Range[0,30]/7]] (* or *) LinearRecurrence[{3,3,4},{0,1,3},30] (* Harvey P. Dale, Feb 18 2023 *)

PARI

a(n) = (4^(n+1)+5)\21; \\ Altug Alkan, Oct 05 2017

Formula

a(n) = round((8*4^n+1)/42) = round((4*4^n-4)/21).

a(n) = floor((4*4^n+5)/21).

a(n) = ceiling((4*4^n-4)/21).

a(n) = a(n-3) + 3*4^(n-2) = a(n-3) + A164346(n-2) for n > 2.

a(n) = 3*a(n-1) + 3*a(n-2) + 4*a(n-3) for n > 2.

G.f.: -x/((4*x-1)*(x^2+x+1)).

a(n+1) - 4*a(n) = A049347(n). - Paul Curtz, Jun 08 2011