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A174709
Partial sums of floor(n/6).
15
0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 8, 10, 12, 14, 16, 18, 21, 24, 27, 30, 33, 36, 40, 44, 48, 52, 56, 60, 65, 70, 75, 80, 85, 90, 96, 102, 108, 114, 120, 126, 133, 140, 147, 154, 161, 168, 176, 184, 192
OFFSET
0,8
COMMENTS
Partial sums of A152467.
LINKS
Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions, J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.
FORMULA
a(n) = round(n*(n-4)/12) = round((2*n^2 - 8*n - 1)/24).
a(n) = floor((n-2)^2/12).
a(n) = ceiling((n+1)*(n-5)/12).
a(n) = a(n-6) + n - 5, n > 5.
From R. J. Mathar, Nov 30 2010: (Start)
a(n) = 2*a(n-1) - a(n-2) + a(n-6) - 2*a(n-7) + a(n-8).
G.f.: -x^6 / ( (1+x)*(x^2-x+1)*(1+x+x^2)*(x-1)^3 ).
a(n) = -n/3 + 5/72 + n^2/12 + (-1)^n/24 + A057079(n+5)/6 + A061347(n)/18. (End)
a(6n) = A000567(n), a(6n+1) = 2*A000326(n), a(6n+2) = A033428(n), a(6n+3) = A049451(n), a(6n+4) = A045944(n), a(6n+5) = A028896(n). - Philippe Deléham, Mar 26 2013
a(n) = A008724(n-2). - R. J. Mathar, Jul 10 2015
Sum_{n>=6} 1/a(n) = Pi^2/18 - Pi/(2*sqrt(3)) + 49/12. - Amiram Eldar, Aug 13 2022
EXAMPLE
a(7) = floor(0/6) + floor(1/6) + floor(2/6) + floor(3/6) + floor(4/6) + floor(5/6) + floor(6/6) + floor(7/6) = 0 + 0 + 0 + 0 + 0 + 0 + 1 + 1 = 2.
MAPLE
a(n):=round(n*(n-4)/12)
MATHEMATICA
Table[Sum[Floor[k/6], {k, 0, n}], {n, 0, 25}] (* G. C. Greubel, Dec 13 2016 *)
PROG
(Magma) [Round(n*(n-4)/12): n in [0..60]]; // Vincenzo Librandi, Jun 22 2011
(PARI) a(n)=(n-2)^2\12 \\ Charles R Greathouse IV, Oct 07 2015
CROSSREFS
Sequence in context: A011866 A321152 A292983 * A008724 A237118 A112402
KEYWORD
nonn,easy
AUTHOR
Mircea Merca, Nov 30 2010
STATUS
approved