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A244953 a(n) = Sum_{i=0..n} (-i mod 4).

%I A244953 #51 Oct 29 2023 12:50:39
%S A244953 0,3,5,6,6,9,11,12,12,15,17,18,18,21,23,24,24,27,29,30,30,33,35,36,36,
%T A244953 39,41,42,42,45,47,48,48,51,53,54,54,57,59,60,60,63,65,66,66,69,71,72,
%U A244953 72,75,77,78,78,81,83,84,84,87,89,90,90,93,95,96,96,99
%N A244953 a(n) = Sum_{i=0..n} (-i mod 4).
%C A244953 Partial sums of A158459.
%C A244953 Similar to A047271 with every third term repeated.
%H A244953 Vincenzo Librandi, <a href="/A244953/b244953.txt">Table of n, a(n) for n = 0..5000</a>
%H A244953 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,1,-1).
%F A244953 a(n) = Sum_{i=0..n} A158459(i).
%F A244953 From _Bruno Berselli_, Jul 09 2014: (Start)
%F A244953 G.f.: (3 + 2*x + x^2)/((1 + x)*(1 - x)^2*(1 + x^2)).
%F A244953 a(n) = 1 + n + ( 2*(1 + n) - (1 + (-1)^n)*(1 + 2*i^(n*(n+1))) )/4, where i = sqrt(-1).
%F A244953 a(n) = 6 + Sum_{i=1..3}((4-i)*floor((n-i)/4)). (End)
%F A244953 a(n) = a(n-1) + a(n-4) - a(n-5). - _Robert Israel_, Jul 09 2014
%F A244953 a(n) = (3*n + 4 - (n mod 4 - 2)^2)/2. - _Thomas Klemm_, Aug 21 2022
%e A244953 To quickly generate terms of the sequence: start with zero for n=0, then add 3 more for n=1, then add 2 more for n=2, add 1 more..., then add 0..., and repeat.
%p A244953 A244953:=n->add(-i mod 4, i=0..n): seq(A244953(n), n=0..50);
%t A244953 Table[Sum[Mod[-i, 4], {i, 0, n}], {n, 0, 50}]
%t A244953 Table[1 + n + (2 (1 + n) - (1 + (-1)^n) (1 + 2 I^(n (n + 1))))/4, {n, 0, 70}] (* _Bruno Berselli_, Jul 09 2014 *)
%t A244953 LinearRecurrence[{1,0,0,1,-1},{0,3,5,6,6},70] (* _Harvey P. Dale_, Oct 29 2023 *)
%o A244953 (PARI) a(n) = sum(i=0, n, -i % 4); \\ _Michel Marcus_, Jul 09 2014
%o A244953 (Magma) [&+[-i mod 4: i in [0..n]]: n in [0..70]]; // _Bruno Berselli_, Jul 09 2014
%Y A244953 Cf. A158459. Same members as A047271. Similar to A130482.
%K A244953 nonn,easy
%O A244953 0,2
%A A244953 _Wesley Ivan Hurt_, Jul 08 2014