OFFSET
1,3
COMMENTS
Complement of A016945. - R. J. Mathar, Feb 25 2008
Nonnegative integers m such that floor(2*m^2/12) = 2*floor(m^2/12). See the Crossrefs field of A265187 for similar sequences. - Bruno Berselli, Dec 08 2015
Also, numbers k such that Fibonacci(k) mod 4 = 0, 1 or 3. - Bruno Berselli, Oct 17 2017
LINKS
David A. Corneth, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).
FORMULA
O.g.f.: x*(x^2+1)*(x^2+x+1)/((x-1)^2*(x^4+x^3+x^2+x+1)). - R. J. Mathar, Feb 25 2008
a(n) = a(n-5) + 6 for n > 5. - R. J. Mathar, Feb 25 2008
a(n) = a(n-1) + a(n-5) - a(n-6) for n > 6. - Harvey P. Dale, Oct 05 2014
From Wesley Ivan Hurt, Aug 16 2016: (Start)
a(n) = n + floor((n-4)/5).
a(n) = (6*n - 4 - ((n+1) mod 5))/5.
a(5k) = 6k-1, a(5k-1) = 6k-2, a(5k-2) = 6k-4, a(5k-3) = 6k-5, a(5k-4) = 6k-6. (End)
Sum_{n>=2} (-1)^n/a(n) = log(2+sqrt(3))/sqrt(3) - log(2)/6. - Amiram Eldar, Dec 17 2021
MAPLE
for n from 0 to 200 do if n mod 6 <> 3 then printf(`%d, `, n) fi: od:
A047263:=n->6*floor(n/5)+[0, 1, 2, 4, 5][(n mod 5)+1]: seq(A047263(n), n=0..100); # Wesley Ivan Hurt, Aug 16 2016
MATHEMATICA
Select[Range[0, 100], Mod[#, 6]!=3&] (* Harvey P. Dale, May 17 2011 *)
LinearRecurrence[{1, 0, 0, 0, 1, -1}, {0, 1, 2, 4, 5, 6}, 90] (* Harvey P. Dale, Oct 05 2014 *)
PROG
(Magma) [n : n in [0..100] | n mod 6 in [0, 1, 2, 4, 5]]; // Wesley Ivan Hurt, Aug 16 2016
(PARI) first(n) = {select(x->(x%6!=3), vector(6*n\5, i, i-1))} \\ David A. Corneth, Oct 17 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Dec 11 1999
EXTENSIONS
More terms from James A. Sellers, Feb 19 2001
STATUS
approved