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A175846
Partial sums of ceiling(n^2/15).
1
0, 1, 2, 3, 5, 7, 10, 14, 19, 25, 32, 41, 51, 63, 77, 92, 110, 130, 152, 177, 204, 234, 267, 303, 342, 384, 430, 479, 532, 589, 649, 714, 783, 856, 934, 1016, 1103, 1195, 1292, 1394, 1501, 1614, 1732, 1856, 1986, 2121, 2263, 2411, 2565, 2726, 2893
OFFSET
0,3
COMMENTS
There are several sequences of integers of the form ceiling(n^2/k) for whose partial sums we can establish identities as following (only for k = 2,...,8,10,11,12, 14,15,16,19,20,23,24).
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((2*n+1)*(n^2 + n + 28)/90).
a(n) = floor((n+1)*(2*n^2 + n + 56)/90).
a(n) = ceiling((2*n^3 + 3*n^2 + 57*n)/90).
a(n) = a(n-15) + (n+1)*(n-15) + 92.
a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + 2*a(n-5) - 2*a(n-6) + a(n-7) - a(n-8) + 2*a(n-9) - a(n-10). - R. J. Mathar, Mar 11 2012
G.f.: x*(x+1)*(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)/((x-1)^4*(x^2 + x + 1)*(x^4 + x^3 + x^2 + x + 1)). - Colin Barker, Oct 26 2012
EXAMPLE
a(15) = 0 + 1 + 1 + 1 + 2 + 2 + 3 + 4 + 5 + 6 + 7 + 9 + 10 + 12 + 14 + 15 = 92.
MAPLE
seq(round((2*n+1)*(n^2+n+28)/90), n=0..50)
PROG
(Magma) [Round((2*n+1)*(n^2+n+28)/90): n in [0..60]]; // Vincenzo Librandi, Jun 22 2011
(PARI) a(n)=(n+1)*(2*n^2+n+56)\90 \\ Charles R Greathouse IV, Jul 06 2017
CROSSREFS
Sequence in context: A177277 A025488 A306473 * A088585 A304712 A175842
KEYWORD
nonn,easy
AUTHOR
Mircea Merca, Dec 05 2010
STATUS
approved