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

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, 39, 41, 42, 42, 45, 47, 48, 48, 51, 53, 54, 54, 57, 59, 60, 60, 63, 65, 66, 66, 69, 71, 72, 72, 75, 77, 78, 78, 81, 83, 84, 84, 87, 89, 90, 90, 93, 95, 96, 96, 99

Offset

0,2

Comments

Partial sums of A158459.

Similar to A047271 with every third term repeated.

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..5000

Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Formula

a(n) = Sum_{i=0..n} A158459(i).

From Bruno Berselli, Jul 09 2014: (Start)

G.f.: (3 + 2*x + x^2)/((1 + x)*(1 - x)^2*(1 + x^2)).

a(n) = 1 + n + ( 2*(1 + n) - (1 + (-1)^n)*(1 + 2*i^(n*(n+1))) )/4, where i = sqrt(-1).

a(n) = 6 + Sum_{i=1..3}((4-i)*floor((n-i)/4)). (End)

a(n) = a(n-1) + a(n-4) - a(n-5). - Robert Israel, Jul 09 2014

a(n) = (3*n + 4 - (n mod 4 - 2)^2)/2. - Thomas Klemm, Aug 21 2022

Example

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.

Maple

A244953:=n->add(-i mod 4, i=0..n): seq(A244953(n), n=0..50);

Mathematica

Table[Sum[Mod[-i, 4], {i, 0, n}], {n, 0, 50}]
Table[1 + n + (2 (1 + n) - (1 + (-1)^n) (1 + 2 I^(n (n + 1))))/4, {n, 0, 70}] (* Bruno Berselli, Jul 09 2014 *)
LinearRecurrence[{1, 0, 0, 1, -1}, {0, 3, 5, 6, 6}, 70] (* Harvey P. Dale, Oct 29 2023 *)

Prog

(PARI) a(n) = sum(i=0, n, -i % 4); \\ Michel Marcus, Jul 09 2014
(Magma) [&+[-i mod 4: i in [0..n]]: n in [0..70]]; // Bruno Berselli, Jul 09 2014

Crossrefs

Cf. A158459. Same members as A047271. Similar to A130482.

Sequence in context: A342269 A364751 A123572 * A076819 A355848 A181757

Adjacent sequences: A244950 A244951 A244952 * A244954 A244955 A244956

Keyword

nonn,easy

Author

Wesley Ivan Hurt, Jul 08 2014

Status

approved