A178719 Partial sums of (1/5)*floor(6^n/7).

0, 0, 1, 7, 44, 266, 1599, 9597, 57586, 345520, 2073125, 12438755, 74632536, 447795222, 2686771339, 16120628041, 96723768254, 580342609532, 3482055657201, 20892333943215, 125354003659300, 752124021955810, 4512744131734871, 27076464790409237, 162458788742455434, 974752732454732616

Offset

0,4

Comments

Partial sums of A033116.

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..500

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 (7,-5,-7,6).

Formula

a(n) = (1/5)*round((24*6^n - 70*n - 49)/140).

a(n) = (1/5)*floor((12*6^n - 35*n - 12)/70).

a(n) = (1/5)*ceiling((12*6^n - 35*n - 37)/70).

a(n) = (1/5)*round((12*6^n - 35*n - 12)/70).

a(n) = a(n-2) + (6^(n-1) - 1)/5, n > 1.

a(n) = 7*a(n-1) - 5*a(n-2) - 7*a(n-3) + 6*a(n-4), n > 3.

G.f.: x^2 / ( (1+x)*(1-6*x)*(1-x)^2 ).

a(n) = (24*6^n - 70*n + 25*(-1)^n - 49)/700. - Bruno Berselli, Feb 18 2011

a(n) = (floor(6^(n+1)/35) - floor((n+1)/2))/5. - Seiichi Manyama, Dec 22 2023

Example

a(3) = (1/5)*(floor(6^1/7) + floor(6^2/7) + floor(6^3/7) = (1/5)*(0+5+30) = (1/5)*35 = 7.

Maple

A178719 := proc(n) add( floor(6^i/7)/5, i=0..n) ; end proc:

Mathematica

f[n_] := Floor[6^n/7]/5; Accumulate@ Array[f, 22]
CoefficientList[Series[x^2/((1+x)(1-6x)(1-x)^2), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 26 2014 *)

Prog

(Magma) [(1/5)*Floor((12*6^n-35*n-12)/70): n in [0..30]]; // Vincenzo Librandi, Jun 21 2011
(PARI) vector(30, n, n--; (((12*6^n-35*n-12)/70)\1)/5) \\ G. C. Greubel, Jan 24 2019
(Sage) [floor((12*6^n-35*n-12)/70)/5 for n in (0..30)] # G. C. Greubel, Jan 24 2019

Crossrefs

Column k=6 of A368296.

Cf. A033116.

Sequence in context: A216237 A291388 A037531 * A094113 A218992 A190974

Adjacent sequences: A178716 A178717 A178718 * A178720 A178721 A178722

Keyword

nonn,less

Author

Mircea Merca, Dec 26 2010

Status

approved