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 round(n^2/16).
1

%I #25 Sep 08 2022 08:45:53

%S 0,0,0,1,2,4,6,9,13,18,24,32,41,52,64,78,94,112,132,155,180,208,238,

%T 271,307,346,388,434,483,536,592,652,716,784,856,933,1014,1100,1190,

%U 1285,1385,1490,1600,1716,1837,1964,2096,2234,2378,2528,2684

%N Partial sums of round(n^2/16).

%C The round function is defined here by round(x) = floor(x + 1/2).

%C There are several sequences of integers of the form round(n^2/k) for whose partial sums we can establish identities as following (only for k = 2, ..., 9, 11, 12, 13, 16, 17, 19, 20, 28, 29, 36, 44).

%H Vincenzo Librandi, <a href="/A177189/b177189.txt">Table of n, a(n) for n = 0..885</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.

%F a(n) = round((2*n+1)*(2*n^2 + 2*n + 3)/192).

%F a(n) = floor((n+3)*(2*n^2 - 3*n + 13)/96).

%F a(n) = ceiling((n-2)*(2*n^2 + 7*n + 18)/96).

%F a(n) = round((2*n^3 + 3*n^2 + 4*n)/96).

%F a(n) = a(n-16) + (n+1)*(n-16) + 94, n > 15.

%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) with g.f. x^3*(1 - x + x^2 + x^4 - x^3) / ( (1+x)*(1+x^2)*(1+x^4)*(x-1)^4 ). - _R. J. Mathar_, Dec 13 2010

%e a(16) = 0 + 0 + 0 + 1 + 1 + 2 + 2 + 3 + 4 + 5 + 6 + 8 + 9 + 11 + 12 + 14 + 16 = 94.

%p seq(round((2*n^3+3*n^2+4*n)/96),n=0..50)

%t Accumulate[Round[Range[0,50]^2/16]] (* _Harvey P. Dale_, Mar 16 2011 *)

%o (Magma) [Floor((n+3)*(2*n^2-3*n+13)/96): n in [0..50]]; // _Vincenzo Librandi_, Apr 29 2011

%Y Cf. A177100, A177116.

%K nonn,easy

%O 0,5

%A _Mircea Merca_, Dec 10 2010