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A173645
Partial sums of floor(n^2/11).
1
0, 0, 0, 0, 1, 3, 6, 10, 15, 22, 31, 42, 55, 70, 87, 107, 130, 156, 185, 217, 253, 293, 337, 385, 437, 493, 554, 620, 691, 767, 848, 935, 1028, 1127, 1232, 1343, 1460, 1584, 1715, 1853, 1998, 2150, 2310, 2478, 2654, 2838, 3030, 3230, 3439, 3657, 3884
OFFSET
0,6
LINKS
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,-3,1,0,0,0,0,0,0,0,1,-3,3,-1).
FORMULA
a(n) = round((2*n^3 + 3*n^2 - 23*n - 12)/66).
a(n) = floor((2*n^3 + 3*n^2 - 23*n + 18)/66).
a(n) = ceiling((2*n^3 + 3*n^2 - 23*n - 42)/66).
a(n) = a(n-11) + (n-5)^2 + 6, n > 10.
From R. J. Mathar, Nov 24 2010: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-11) - 3*a(n-12) + 3*a(n-13) - a(n-14).
G.f.: x^4*(x+1)*(x^4 - x^3 + x^2 - x + 1) / ((x^10 + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)*(x-1)^4). (End)
EXAMPLE
a(6) = 6 = 0 + 0 + 0 + 0 + 1 + 2 + 3.
MAPLE
A173645(n):=round((2*n^3+3*n^2-23*n-12)/66)
MATHEMATICA
Accumulate[Floor[Range[0, 50]^2/11]] (* Harvey P. Dale, Sep 23 2015 *)
PROG
(Magma) [ &+[Floor(k^2/11): k in [0..n]]: n in [0..60] ]; // Bruno Berselli, Apr 28 2011
(PARI) vector(60, n, n--; (2*n^3+3*n^2-23*n+18)\66) \\ G. C. Greubel, Jul 02 2019
(Sage) [floor((2*n^3+3*n^2-23*n+18)/66) for n in (0..60)] # G. C. Greubel, Jul 02 2019
CROSSREFS
Sequence in context: A024918 A117245 A011914 * A122047 A177100 A265071
KEYWORD
nonn,easy
AUTHOR
Mircea Merca, Nov 24 2010
STATUS
approved