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%I #29 Sep 08 2022 08:45:50
%S 0,0,1,2,4,8,13,20,29,41,55,72,93,117,145,177,214,255,301,353,410,473,
%T 542,618,700,789,886,990,1102,1222,1351,1488,1634,1790,1955,2130,2315,
%U 2511,2717,2934,3163,3403,3655,3919,4196,4485,4787,5103,5432,5775,6132
%N Partial sums of A056833.
%H Vincenzo Librandi, <a href="/A173721/b173721.txt">Table of n, a(n) for n = 0..5000</a>
%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1,0,0,0,1,-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) = Sum_{k=0..n} round(k^2/7).
%F a(n) = round((2*n^3 + 3*n^2 + n)/42).
%F a(n) = a(n-7) + (n-3)^2 + 4, n > 6.
%F From _R. J. Mathar_, Nov 26 2010: (Start)
%F a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-7) - 3*a(n-8) + 3*a(n-9) - a(n-10).
%F G.f.: x^2*(1+x)*(x^2-x+1)^2 / ( (x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)*(x-1)^4 ).
%F a(n) = n*(n+1)*(2*n+1)/42 - b(n)/7 where b(n) = 0, 1, -2, 0, 2, -1, 0, ... with period 7. (End)
%e a(5) = round(1/7) + round(4/7) + round(9/7) + round(16/7) + round(25/7) = 0+1+1+2+4 = 8.
%p A173721 := proc(n) a := n*(n+1)*(2*n+1)/42 ; b := op( 1+(n mod 7),[0,1,-2,0,2,-1,0]) ; a-b/7 ;end proc:
%p seq(A173721(n),n=0..80) ; # _R. J. Mathar_, Nov 26 2010
%t Table[Round[(2*n^3 + 3*n^2 + n)/42], {n,0,50}] (* _G. C. Greubel_, Nov 29 2016 *)
%o (Magma) a056833:=func< n | Round(n^2/7) >; [ &+[ a056833(j): j in [0..n] ]: n in [0..60] ]; // _Klaus Brockhaus_, Nov 26 2010
%o (PARI) a(n)=(2*n^3+3*n^2+n)\/42 \\ _Charles R Greathouse IV_, Nov 29 2016
%Y Cf. A056833 (nearest integer to n^2/7).
%K nonn,easy
%O 0,4
%A _Mircea Merca_, Nov 26 2010
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