login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

Partial sums of ceiling(n^2/15).
1

%I #31 Sep 08 2022 08:45:52

%S 0,1,2,3,5,7,10,14,19,25,32,41,51,63,77,92,110,130,152,177,204,234,

%T 267,303,342,384,430,479,532,589,649,714,783,856,934,1016,1103,1195,

%U 1292,1394,1501,1614,1732,1856,1986,2121,2263,2411,2565,2726,2893

%N Partial sums of ceiling(n^2/15).

%C 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).

%H Vincenzo Librandi, <a href="/A175846/b175846.txt">Table of n, a(n) for n = 0..10000</a>

%H Mircea Merca, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL14/Merca/merca3.html">Inequalities and Identities Involving Sums of Integer Functions</a> J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,1,-2,2,-2,1,-1,2,-1).

%F a(n) = round((2*n+1)*(n^2 + n + 28)/90).

%F a(n) = floor((n+1)*(2*n^2 + n + 56)/90).

%F a(n) = ceiling((2*n^3 + 3*n^2 + 57*n)/90).

%F a(n) = a(n-15) + (n+1)*(n-15) + 92.

%F 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

%F 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

%e a(15) = 0 + 1 + 1 + 1 + 2 + 2 + 3 + 4 + 5 + 6 + 7 + 9 + 10 + 12 + 14 + 15 = 92.

%p seq(round((2*n+1)*(n^2+n+28)/90),n=0..50)

%o (Magma) [Round((2*n+1)*(n^2+n+28)/90): n in [0..60]]; // _Vincenzo Librandi_, Jun 22 2011

%o (PARI) a(n)=(n+1)*(2*n^2+n+56)\90 \\ _Charles R Greathouse IV_, Jul 06 2017

%K nonn,easy

%O 0,3

%A _Mircea Merca_, Dec 05 2010