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A178719
Partial sums of (1/5)*floor(6^n/7).
2
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
Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.
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
KEYWORD
nonn,less,easy
AUTHOR
Mircea Merca, Dec 26 2010
STATUS
approved