A178543 Partial sums of round(3^n/5).

0, 1, 3, 8, 24, 73, 219, 656, 1968, 5905, 17715, 53144, 159432, 478297, 1434891, 4304672, 12914016, 38742049, 116226147, 348678440, 1046035320, 3138105961, 9414317883, 28242953648, 84728860944, 254186582833

Offset

0,3

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..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).

Formula

a(n) = round(3^(n+1)/10).

a(n) = floor((3*3^n + 3)/10).

a(n) = ceiling((3*3^n - 3)/10).

a(n) = a(n-4) + 8*3^(n-3), n > 3.

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

G.f.: x/((1-3*x)*(1+x^2)).

a(n) = 3^(n+1)/10 - (-1)^n* A112030(n)/10. - R. J. Mathar, Jan 08 2011

Example

a(4) = round(1/5) + round(3/5) + round(9/5) + round(27/5) + round(81/5) = 0 + 1 + 2 + 5 + 16 = 24.

Maple

seq(round(3^n/10), n=1..25);

Mathematica

Accumulate[Floor[3^Range[0, 30]/5+1/2]] (* Harvey P. Dale, Jul 02 2011 *)

Prog

(Magma) [Round(3^(n+1)/10): n in [0..40]]; // Vincenzo Librandi, Jun 21 2011
(PARI) vector(40, n, n--; (3*(3^n +1)/10)\1) \\ G. C. Greubel, Jan 30 2019
(Sage) [floor(3*(3^n+1)/10) for n in range(40)] # G. C. Greubel, Jan 30 2019

Crossrefs

Cf. A112030.

Sequence in context: A080923 A118264 A006365 * A188175 A046919 A291886

Adjacent sequences: A178540 A178541 A178542 * A178544 A178545 A178546

Keyword

nonn,less,easy

Author

Mircea Merca, Dec 28 2010

Status

approved