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
Examples
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.
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).
Programs
-
Magma
[(1/5)*Floor((12*6^n-35*n-12)/70): n in [0..30]]; // Vincenzo Librandi, Jun 21 2011
-
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 *) -
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
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
Comments