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Partial sums of round(n^2/19).
1

%I #27 Apr 28 2024 02:08:14

%S 0,0,0,0,1,2,4,7,10,14,19,25,33,42,52,64,77,92,109,128,149,172,197,

%T 225,255,288,324,362,403,447,494,545,599,656,717,781,849,921,997,1077,

%U 1161,1249,1342,1439,1541,1648,1759,1875,1996,2122,2254

%N Partial sums of round(n^2/19).

%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="/A177237/b177237.txt">Table of n, a(n) for n = 0..895</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((n-2)*(n+3)*(2*n+1)/114).

%F a(n) = floor((2*n^3 + 3*n^2 - 11*n + 42)/114).

%F a(n) = ceiling((2*n^3 + 3*n^2 - 11*n - 54)/114).

%F a(n) = round((2*n^3 + 3*n^2 - 11*n)/114).

%F a(n) = a(n-19) + (n+1)*(n-19) + 128, n > 18.

%F From _R. J. Mathar_, Dec 13 2010: (Start)

%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-19) - 3*a(n-20) + 3*a(n-21) - a(n-22).

%F G.f.: x^4*(1+x)*(1 - x + x^2 - x^3 + x^4)*(1 - x + x^2 - x^4 + x^6 - x^7 + x^8)/((1-x)^3 * (1 - x^19)). (End)

%e a(19) = 0 + 0 + 0 + 0 + 1 + 1 + 2 + 3 + 3 + 4 + 5 + 6 + 8 + 9 + 10 + 12 + 13 + 15 + 17 + 19 = 128.

%p seq(round((2*n^3+3*n^2-11*n)/114),n=0..50)

%t Accumulate[Round[Range[0,50]^2/19]] (* _Harvey P. Dale_, Aug 15 2022 *)

%o (Magma) [Floor((2*n^3+3*n^2-11*n+42)/114): n in [0..50]]; // _Vincenzo Librandi_, Apr 29 2011

%o (SageMath)

%o [(2*n^3 +3*n^2 -11*n +42)//114 for n in range(61)] # _G. C. Greubel_, Apr 27 2024

%Y Cf. A177100, A177116.

%K nonn,easy

%O 0,6

%A _Mircea Merca_, Dec 10 2010