|
%I #30 Sep 08 2022 08:45:51
%S 0,0,0,0,1,2,4,7,11,16,22,29,38,48,60,74,90,108,128,150,175,202,232,
%T 265,301,340,382,427,476,528,584,644,708,776,848,924,1005,1090,1180,
%U 1275,1375,1480,1590,1705,1826,1952,2084,2222,2366,2516,2672
%N Partial sums of floor(n^2/16).
%C There are several sequences of integers of the form floor(n^2/k) for whose partial sums we can establish identities as following (only for k = 2,...,9,11,12,15,16,24).
%H Vincenzo Librandi, <a href="/A175777/b175777.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1,0,0,0,0,1,-3,3,-1).
%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.
%F a(n) = round((2*n+1)*(2*n^2 + 2*n - 21)/192).
%F a(n) = floor((n-1)*(2*n^2 + 5*n - 15)/96).
%F a(n) = ceiling((n+2)*(2*n^2 - n - 18)/96).
%F a(n) = round(n*(n+4)*(2*n - 5)/96).
%F a(n) = a(n-16) + (n+1)*(n-16) + 90, n > 15.
%F G.f.: x^4*(1 - x + x^2) / ( (1+x)*(x^2+1)*(x^4+1)*(x-1)^4 ). - _R. J. Mathar_, Dec 06 2010
%F a(n)= 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-8) - 3*a(n-9) + 3*a(n-10) - a(n-11). - _R. J. Mathar_, Dec 06 2010
%e a(16) = 0 + 0 + 0 + 0 + 1 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 9 + 10 + 12 + 14 + 16 = 90.
%p seq(round(n*(n+4)*(2*n-5)/96),n=0..50)
%t Accumulate[Table[Floor[n^2/16],{n,0,60}]] (* _Harvey P. Dale_, Dec 13 2010 *)
%o (Magma) [Round((2*n+1)*(2*n^2+2*n-21)/192): n in [0..60]]; // _Vincenzo Librandi_, Jun 22 2011
%o (PARI) concat([0,0,0,0],Vec((1-x+x^2)/((1+x)*(x^2+1)*(x^4+1)*(x-1)^4)+O(x^99))) \\ _Charles R Greathouse IV_, Oct 18 2011
%K nonn,easy
%O 0,6
%A _Mircea Merca_, Dec 04 2010
|
