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
Examples
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.
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).
Programs
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Magma
[&+[-i mod 4: i in [0..n]]: n in [0..70]]; // Bruno Berselli, Jul 09 2014
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Maple
A244953:=n->add(-i mod 4, i=0..n): seq(A244953(n), n=0..50);
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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 *) -
PARI
a(n) = sum(i=0, n, -i % 4); \\ Michel Marcus, Jul 09 2014
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
Comments