login

Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.

Partial sums of floor(n^3/2).
1

%I #29 Sep 08 2022 08:45:50

%S 0,0,4,17,49,111,219,390,646,1010,1510,2175,3039,4137,5509,7196,9244,

%T 11700,14616,18045,22045,26675,31999,38082,44994,52806,61594,71435,

%U 82411,94605,108105,123000,139384,157352,177004,198441,221769,247095,274531,304190,336190,370650,407694,447447,490039,535601,584269,636180,691476,750300,812800

%N Partial sums of floor(n^3/2).

%C Partial sums of A036487.

%H Vincenzo Librandi, <a href="/A173704/b173704.txt">Table of n, a(n) for n = 0..5000</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_06">Index entries for linear recurrences with constant coefficients</a>, signature (4,-5,0,5,-4,1).

%F a(n) = Sum_{k=0..n} floor(k^3/2).

%F a(n) = round((n^4+2*n^3+n^2-2*n)/8).

%F a(n) = round((n^4+2*n^3+n^2-2*n-1)/8).

%F a(n) = floor((n^4+2*n^3+n^2-2*n)/8).

%F a(n) = ceiling((n-1)*(n+1)*(n^2+2*n+2)/8).

%F a(n) = a(n-2)+(n-1)*(2*n^2-n+2)/2, n>1.

%F From _R. J. Mathar_, Nov 26 2010: (Start)

%F a(n) = 4*a(n-1) - 5*a(n-2) + 5*a(n-4) - 4*a(n-5) + a(n-6).

%F G.f.: -x^2*(4+x+x^2) / ( (1+x)*(x-1)^5 ).

%F a(n) = (n^4 + 2*n^3 + n^2 - 2*n - 1 + (-1)^n)/8. (End)

%e a(4) = floor(1/2) + floor(8/2) + floor(27/2) + floor(64/2) = 49.

%p A173704 := proc(n) (n^4+2*n^3+n^2-2*n-1+(-1)^n)/8 ; end proc:

%t Table[Sum[Floor[k^3/2], {k, 0, n}], {n,0,50}] (* _G. C. Greubel_, Nov 23 2016 *)

%o (Magma) [Round((n^4+2*n^3+n^2-2*n)/8): n in [0..40]]; // _Vincenzo Librandi_, Jun 22 2011

%Y Cf. A036487.

%K nonn

%O 0,3

%A _Mircea Merca_, Nov 25 2010

%E Maple program replaced by _R. J. Mathar_, Nov 26 2010