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A177189 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

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